Question 4515 Marks
If $e^y= y^x$, prove that $\frac{\text{dy}}{\text{dx}}=\frac{(\log\text{y})^2}{\log\text{y}-1}$
AnswerWe have, $e^y = y^x$
Taking log on both sides,
$\log\text{e}^{\text{y}}=\log\text{y}^\text{x}$
$\Rightarrow\text{y}\log\text{e}=\text{x}\log\text{y}$
$\Rightarrow\text{y}=\text{x}\log\text{y}\ .....(\text{i})$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big(1-\frac{\text{x}}{\text{y}}\Big)=\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big(\frac{\text{y}-\text{x}}{\text{y}}\big)=\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}\log\text{y}}{\text{y}-\text{x}}$
$\Rightarrow\frac{\text{y}\log\text{y}}{\Big(\text{y}-\frac{\text{y}}{\log\text{y}}\Big)}$
[Using equation (i)]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}\log\text{y}(\log\text{y})}{\text{y}\log\text{y}-\text{y}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\log\text{y})^2}{\text{y}(\log\text{y}-1)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\log\text{y})^2}{(\log\text{y}-1)}$
View full question & answer→Question 4525 Marks
If the value of c prescribed in Roll's theorem for the function$\text{f}(\text{x})=2\text{x}(\text{x}-3)^{\text{n}}$ on the interval $\big[0,2\sqrt3\big]$ is $\frac{3}{4},$ write the value of n (a positive integers).
AnswerWe have,
$\text{f}(\text{x})=2\text{x}(\text{x}-3)^{\text{n}}$
Differentiating the given function with respect to x, we get
$\text{f}'(\text{x})=2\big[\text{xn}(\text{x}-3)^{\text{n}-1}+(\text{x}-3)^{\text{n}}\big]$
$\Rightarrow\text{f}'(\text{x})=2(\text{x}-3)^{\text{n}}\Big[\frac{\text{xn}}{(\text{x}-3)}+1\Big]$
$\Rightarrow\text{f}'(\text{c})=2(\text{c}-3)^{\text{n}}\Big[\frac{\text{cn}}{(\text{c}-3)}+1\Big]$
Given:
$\text{f}'\Big(\frac{3}{4}\Big)=0$
$\therefore\ 2-\Big(\frac{9}{4}\Big)^{\text{n}}\Bigg[\frac{\frac{3}{4}\text{n}}{\big(\frac{-9}{4}\big)}+1\Bigg]=0$
$\Rightarrow2-\Big(\frac{9}{4}\Big)^{\text{n}}\Big[\frac{-\text{n}}{3}+1\Big]=0$
$\Rightarrow\Big[\frac{-\text{n}}{3}+1\Big]=0$
$\Rightarrow-\text{n}+3=0$
$\Rightarrow\text{n}=3$
View full question & answer→Question 4535 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$y = x^n + n^x + x^x + n^n$
AnswerWe have, $y = x^n + n^x + x^x + n^x$
$\Rightarrow\text{y}=\text{x}^\text{n}+\text{n}^\text{x}+\text{e}^{\log\text{x}^\text{x}}+\text{n}^\text{n}$
$\Rightarrow\text{y}=\text{x}^\text{n}+\text{n}^\text{x}+\text{e}^{\text{x}\log\text{x}}+\text{n}^\text{n}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})+\frac{\text{d}}{\text{dx}}(\text{n}^\text{x})+\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}\log\text{x}})+\frac{\text{d}}{\text{dx}}(\text{n}^\text{n})$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{e}^{\log\text{x}^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\Big[\text{x}\big(\frac{1}{\text{x}}\big)+\log\text{x}\Big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\big[1+\log\text{x}\big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\big[\log\text{e}+\log\text{x}\big] \\ \big[\because\log_\text{e}\text{e}=1\text{ and }\log\text{A}+\log\text{B}=\log(\text{AB})\big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\log\big(\text{ex}\big)$
View full question & answer→Question 4545 Marks
If $\text{x}=\text{a}\Big(\text{t}+\frac{1}{\text{t}}\Big)\text{ and y}=\text{a}\Big(\text{t}-\frac{1}{\text{t}}\Big),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$
AnswerHere, $ \text{x}=\text{a}\Big(\text{t}+\frac{1}{\text{t}}\Big)$
Differentiating it with respect to t,
$\frac{\text{dx}}{\text{dt}}=\text{a}\frac{\text{d}}{\text{dt}}\Big(\text{t}+\frac{1}{\text{t}}\Big)$
$=\text{a}\Big(\text{t}-\frac{1}{\text{t}^{2}}\Big)$
$\frac{\text{dx}}{\text{dt}}=\text{a}\Big(\frac{\text{t}^{2}-1}{\text{t}^{2}}\Big)\ .....(\text{i})$
And, $\text{y}=\text{a}\Big(\text{t}-\frac{1}{\text{t}}\Big)$
Differetiating it with respect to t,
$\frac{\text{dy}}{\text{dt}}=\text{a}\frac{\text{d}}{\text{dt}}\Big(\text{t}-\frac{1}{\text{t}}\Big)$
$=\text{a}\Big(1+\frac{1}{\text{t}^{2}}\Big)$
$\frac{\text{dy}}{\text{dt}}=\text{a}\Big(\frac{\text{t}^{2}+1}{\text{t}^{2}}\Big)\ .....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\text{a}\frac{(\text{t}^{2}+1)}{\text{t}^{2}}\times\frac{\text{t}^{2}}{\text{a}(\text{t}^{2}-1)}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{t}^{2}+1}{\text{t}^{2}-1}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$
$\Big[\text{Since},\frac{\text{x}}{\text{y}}=\frac{\text{a}(\text{t}^{2}+1)}{\text{t}}\times\frac{\text{t}}{\text{a}(\text{t}^{2}-1)}=\Big(\frac{\text{t}^{2}+1}{\text{t}^{2}-1}\Big)\Big]$
View full question & answer→Question 4555 Marks
If $\text{f(x)}=\begin{cases}\text{ax}^2-\text{b}, & \text{if |x|}<1\\\frac{1}{|\text{x}|}, & \text{if |x|}\geq1\end{cases}$ is differentiable at x = 1, find a, b.
AnswerHere,
$\text{f(x)}=\begin{cases}\text{ax}^2-\text{b}, & \text{if |x|}<1\\\frac{1}{|\text{x}|}, & \text{if |x|}\geq1\end{cases}$
$=\begin{cases}-\frac{1}{\text{x}}, & \text{if |x|}\leq-1\\\text{ax}^2-\text{b}, & \text{if}-1<\text{x}<1\\\frac{1}{\text{x}},&\text{if x}\geq1\end{cases}$
$\text{LHL }=\lim\limits_{\text{x}\rightarrow1^{-}}\text{f(x)}$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}(1-\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\text{a}(1-\text{h})^2-\text{b}$
$= \text{a}- \text{b}$
$\text{RHL }=\lim\limits_{\text{x}\rightarrow1^{+}}\text{f(x)}$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}(1+\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{1+\text{h}}$
Since, f(x) is continuous, so
LHL = RHL
a - b = 1 .......(1)
(LHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-1}{1-\text{h}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}(1-\text{h})^2-\text{b}-1}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}(1-\text{h})^2-(\text{a}-1)-1}{-\text{h}}$
Using equation (1),
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}+\text{ah}^2-2\text{ah}-\text{a}+1-1}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{ah}^2-2\text{ah}}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}(2\text{a}-\text{ah})$
$=2\text{a}$
RHL at x = 1 $=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-{\text{f}(1)}}{1+\text{h}-1}$
$\lim_\limits{\text{h}\rightarrow0}\frac{\frac{1}{1+\text{h}}-1}{\text{h}}$
$\lim_\limits{\text{h}\rightarrow0}\frac{1-1-\text{h}}{(1+\text{h})\text{h}}$
$= -1$
Since f(x) is differentiable at x = 1,
(LHL at x = 1) = (RHL at x = 1)
2a = -1
$\text{a}=\frac{-1}{2}$
Put $\text{a}=\frac{-1}{2}$ in equation (1),
a - b = 1
$\Big(\frac{-1}{2}\Big)-\text{b}=1$
$\text{b}=\frac{-1}{2}-1$
$\text{b}=\frac{-3}{2}$
$\text{a}=\frac{-1}{2}$
View full question & answer→Question 4565 Marks
Differentiate the following functions with respect to x:
$\text{x}^{(\sin\text{x}-\cos\text{x})}+\frac{\text{x}^2-1}{\text{x}^2+1}$
AnswerLet $\text{y}=\text{x}^{(\sin\text{x}-\cos\text{x})}+\Big(\frac{\text{x}^2-1}{\text{x}^2+1}\Big)$
$\text{y}=\text{e}^{\log\text{x}^{\sin\text{c}-\cos\text{x}}}+\Big(\frac{\text{x}^2-1}{\text{x}^2+1}\Big)$
$\text{y}=\text{e}^{(\sin\text{c}-\cos\text{x})\log\text{x}}+\Big(\frac{\text{x}^2-1}{\text{x}^2+1}\Big)$
$\big[\text{Since},\text{e}^{\log\text{a}}=\text{a},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule and quotient rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\text{e}^{(\sin\text{x}-\cos\text{x})\log\text{x}}\Big]+\frac{\text{d}}{\text{dx}}\Big[\frac{\text{x}^2-1}{\text{x}^2+1}\Big]$
$=\text{e}^{(\sin\text{x}-\cos\text{x})\log\text{x}}+\frac{\text{d}}{\text{dx}}\big\{(\sin\text{x}-\cos\text{x})\log\text{x}\big\} \\ +\bigg[\frac{(\text{x}^2+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-1)-(\text{x}^2-1)\frac{\text{d}}{\text{dx}}(\text{x}^2+1)}{(\text{x}^2+1)^2}\bigg]$
$=\text{e}^{\log\text{x}^{(\sin\text{x}-\cos\text{x})}}\Big[(\sin\text{x}-\cos\text{x})\frac{\text{d}}{\text{dx}}(\log\text{x})+(\log\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x})\Big] \\+\Big[\frac{(\text{x}^2+1)(2\text{x})-(\text{x}^2-1)(2\text{x})}{(\text{x}^2+1)^2}\Big]$
$=\text{e}^{(\sin\text{x}-\cos\text{x})}\Big[(\sin\text{x}-\cos\text{x}\big(\frac{1}{\text{x}}\big)+\log\text{x}(\sin\text{x}+\cos\text{x})\Big] \\ +\Big[\frac{2\text{x}^3+2\text{x}-2\text{x}^3+2\text{x}}{(\text{x}^2+1)^2}\Big]$
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin\text{x}-\cos\text{x}}\Big[\frac{(\sin\text{x}-\cos\text{x})}{\text{x}}+\log\text{x}(\sin\text{x}+\cos\text{x})\Big]+\frac{4\text{x}}{(\text{x}^2+1)^2}$
View full question & answer→Question 4575 Marks
Differentiate $\sin^{-1}\Big(2\text{ax}\sqrt{1-\text{a}^2\text{x}^2}\Big)$ with respect to $\sqrt{1-\text{a}^2\text{x}^2},$ if $-\frac{1}{\sqrt{2}}<\text{ax}<\frac{1}{\sqrt{2}}$.
Answer Let $\text{u}=\sin^{-1}\Big(2\text{ax}\sqrt{1-\text{a}^2\text{x}^2}\Big)$
Put $\text{ax} =\sin\theta\Rightarrow\theta=\sin^{-1}(\text{ax})$
$\therefore\text{u}=\sin^{-1}\Big(2\sin\theta\sqrt{1-\sin^{2}\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\sin\theta\cos\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
And
Let, $\text{v}=\sqrt{1-\text{a}^2\text{x}^2}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{a}^2\text{x}^2}}\times\frac{\text{d}}{\text{dx}}\big(1-\text{a}^2\text{x}^2\big)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\Big(\frac{0-2\text{a}^2\text{x}}{2\sqrt{1-\text{a}^2\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-\text{a}^2\text{x}}{\sqrt{1-\text{a}^2\text{x}^2}}\ .....(\text{ii})$
Here,
$-\frac{1}{\sqrt{2}}<\text{ax}<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{1}{\sqrt{2}}<\sin\theta<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{u}=2\theta\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\text{u}=2\sin^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=2\times\frac{1}{\sqrt{1-(\text{ax})^2}}\frac{\text{d}}{\text{dx}}(\text{ax})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2}{1-\text{a}^2\text{x}^2}(\text{a})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2\text{a}}{1-\text{a}^2\text{x}^2}\ .....(\text{iii})$
Dividing equation (iii) by (ii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\Big(\frac{2\text{a}}{\sqrt{1-\text{a}^2\text{x}^2}}\Big)\Big(\frac{\sqrt{1-\text{a}^2\text{x}^2}}{\text{-a}^2\text{x}}\Big)$
$\therefore\frac{\text{du}}{\text{dv}}=-\frac{2}{\text{ax}}$
View full question & answer→Question 4585 Marks
$\text{If y}=\text{e}^{\text{a}\cos^{-1}\text{x}},-1\leq\text{x}\leq1,\ \text{show that}(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-\text{a}^2\text{y}=0.$
Answerit is given that, $\text{y}=\text{e}^{\text{a}\cos^{-1}\text{x}}$
Taking logarithm on both the sides we obtain
$\log\text{y}=\text{a}\cos^{-1}\text{x}\log\text{e}$
$\log\text{y}=\text{a}\cos^{-1}\text{x}$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{a}\times\frac{-1}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{ay}}{\sqrt{1-\text{x}^2}}$$$
By squaring both the sides, we obtain
$\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\frac{\text{a}^2\text{y}^2}{1-\text{x}^2}$
$\Rightarrow\ (1-\text{x}^2)\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{a}^2\text{y}^2$
$(1-\text{x}^2)\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{a}^2\text{y}^2$
Again differentiating both sides with respect to x, we obtain
$\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\frac{\text{d}}{\text{dx}}(1-\text{x}^2)+(1-\text{x}^2)\times\frac{\text{d}}{\text{dx}}\Big[\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]=\text{a}^2\frac{\text{d}}{\text{dx}}(\text{y}^2)$
$\Rightarrow\ \Big(\frac{\text{dy}}{\text{dx}}\Big)^2(-2\text{x})+(1-\text{x}^2)\times2\frac{\text{dy}}{\text{dx}}.\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{a}^2.2\text{y}.\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ -\text{x}\frac{\text{dy}}{\text{dx}}+(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{a}^2.\text{y}\ \Big[\frac{\text{dy}}{\text{dx}}\neq0\Big]$
$\Rightarrow\ (1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-\text{a}^2\text{y}=0$
Hence, proved.
View full question & answer→Question 4595 Marks
Differentiate the following functions from first principles:
$\text{e}^{\sqrt{2\text{x}}}$
AnswerLet $\text{f(x)}=\text{e}^{\sqrt{2\text{x}}}$
$\Rightarrow\text{f}(\text{x}+\text{h})=\text{e}^{\sqrt{2(\text{x}+\text{h})}}$
$\therefore\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0 }\frac{\text{e}^{\sqrt{2(\text{x}+\text{h})}}-\text{e}^{\sqrt{2\text{x}}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{e}^{\sqrt{2\text{x}}}\frac{\text{e}^{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}-1}{\text{h}}$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\big(\text{e}^{2(\text{x}+\text{h})-\sqrt{2\text{x}}}-1\big)}{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}\Bigg)\bigg(\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{\text{h}}\bigg)$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{2}(\text{x}+\text{h})-\sqrt{2\text{x}}}{\text{h}}\ \Big[\text{Since},\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\Big]$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{\text{h}}\times\frac{\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}}{\sqrt{2(\text{x}+\text{x})}+\sqrt{2\text{x}}}$
$[\text{Rationalizing the numerator]}$
$=\text{de}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{2(\text{x}+\text{h})-2\text{x}}{\text{h}\big(\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}\big)}$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{2\text{x}+2\text{h}-2\text{x}}{\text{h}\big(\sqrt{2}(\text{x}+\text{h})+\sqrt{2\text{x}}\big)}$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}}{\text{h}\big(\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}\big)}$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}}{\big(\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}\big)}$
$=\frac{\text{e}^{\sqrt{2\text{x}}}}{\sqrt{2\text{x}}}$
So,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sqrt{2\text{x}}}\big)=\frac{\text{e}^{2\text{x}}}{\sqrt{2\text{x}}}$
View full question & answer→Question 4605 Marks
If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=\frac{\sin^3\text{t}}{\sqrt{\cos2\text{t}}},\text{y}\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}}$
AnswerThe given equations are $\text{x}=\frac{\sin^3\text{t}}{\sqrt{\cos2\text{t}}}\text{ and y}=\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}}$Then, $\frac{\text{dx}}{\text{dt}}= \frac{\text{d}}{\text{dt}}\Big[\frac{\sin^3\text{t}}{\sqrt{\cos2\text{t}}}\Big]$
$=\frac{\sqrt{\cos2\text{t}}.\frac{\text{d}}{\text{dt}}(\sin^3\text{t})-\sin^3\text{t}.\frac{\text{d}}{\text{dt}}\sqrt{\cos2\text{t}}}{\cos2\text{t}}$
$=\frac{\sqrt{\cos2\text{t}}.3\sin^2\text{t}.\frac{\text{d}}{\text{dt}}(\sin\text{t})-\sin^3\text{t}\times\frac{1}{2\sqrt{\cos2\text{t}}}.\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$=\frac{3\sqrt{\cos2\text{t}}.\sin^2\text{t}\cos\text{t}-\frac{\sin^3\text{t}}{2\sqrt{\cos2\text{t}}}.(-2\sin2\text{t})}{\cos2\text{t}}$
$=\frac{3\cos2\text{t}\sin^2\text{t}\cos\text{t}+\sin^3\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}$
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}}\Big]$
$=\frac{\sqrt{\cos2\text{t}}.\frac{\text{d}}{\text{dt}}(\cos^3\text{t})-\cos^3\text{t}.\frac{\text{d}}{\text{dt}}(\sqrt{\cos2\text{t}})}{\cos2\text{t}}$
$=\frac{\sqrt{\cos2\text{t}}\cos^2\text{t}.\frac{\text{d}}{\text{dt}}(\cos\text{t})-\cos^3\text{t}.\frac{1}{2\sqrt{\cos2\text{t}}}.\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$=\frac{3\sqrt{\cos2\text{t}}.\cos^2\text{t}(-\sin\text{t})-\cos^3\text{t}.\frac{1}{2\sqrt{\cos2\text{t}}}.(-2\sin2\text{t})}{\cos2\text{t}}$
$=\frac{-3\cos2\text{t}\cdot\cos^2\text{t}\cdot\sin\text{t}+\cos^3\text{t}.\sin2\text{t}}{\cos2\text{t}\cdot\sqrt{\cos2\text{t}}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dt}}\Big)}=\frac{-3\cos2\text{t}.\cos^2\text{t}.\sin\text{t}+\cos^3\text{t}\sin2\text{t}}{3\cos2\text{t}\sin^2\text{t}\cos\text{t}+\sin^3\text{t}\sin2\text{t}}$
$=\frac{-3\cos2\text{t}.\cos^2\text{t}.\sin\text{t}+\cos^3\text{t}(2\sin\text{t}\cos\text{t})}{3\cos2\text{t}\sin^2\text{t}\cos\text{t}+\sin^3\text{t}(2\sin\text{t}\cos\text{t})}$
$=\frac{\sin\text{t}\cos\text{t}[-3\cos2\text{t}.\cos\text{t}+2\cos^3\text{t}]}{\sin\text{t}\cos\text{t}[3\cos2\text{t}\sin\text{t}+2\sin^3\text{t}]}$
$=\frac{[-3(2\cos^2\text{t}-1)\cos\text{t}+2\cos^3\text{t}]}{[3(1-2\sin^2\text{t})\sin\text{t}+2\sin^3\text{t}]}$ $\begin{bmatrix}\cos2\text{t}=(2\cos^2\text{t}-1). \\\cos2\text{t}=(1-2\sin^2\text{t}) \end{bmatrix}$
$=\frac{-4\cos^3\text{t}+3\cos\text{t}}{3\sin\text{t}-4\sin^3\text{t}}$
$=\frac{-\cos3\text{t}}{\sin3\text{t}}$ $\begin{bmatrix}\cos3\text{t}=4\cos^3\text{t}-3\cos\text{t}. \\\sin3\text{t}=3\sin\text{t}-4\sin^3\text{t} \end{bmatrix}$
$=-\cot3\text{t}$
View full question & answer→Question 4615 Marks
Differentiate the following functions from first principles:
$\text{e}^{\sqrt{\cot\text{x}}}$
AnswerLet $\text{f(x)} =\text{e}^{\sqrt{\cot\text{x}}}$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\text{e}^{\sqrt{\cot\text{x}}}$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})=\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{\cot(\text{x}+\text{h})}}-\text{e}^{\sqrt{\cot\text{x}}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{\cot\text{x}}}\Big(\text{e}^{\sqrt{\cot(\text{x}+\text{h})}-\sqrt{\cot\text{x}}}-1\Big)}{\text{h}}$
$=\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\text{e}^{\sqrt{\cot(\text{x}+\text{h})}-\sqrt{\cot\text{x}}}-1}{{\sqrt{\cot(\text{x}+\text{h})}}-\sqrt{\cot\text{x}}}\bigg)\times\bigg(\frac{\sqrt{\cot(\text{x}+\text{h})}-\sqrt{\cot\text{x}}}{\text{h}}\bigg)$
$=\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{\cot(\text{x}+\text{h})}-\sqrt{\cot\text{x}}}{\text{h}}\times\frac{\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}}{\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}}$
$\Big[\text{Since, } \lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\text{ and rationalizing numerator}\Big]$
$\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\cot(\text{x}+\text{h})-\cot\text{x}}{\text{h}\big(\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}\big)}$
$=\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\cot(\text{x}+\text{h})\cot\text{x}+1}{\cot(\text{x}+\text{h}-\text{x})}}{\text{x}\big(\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}\big)}$
$\Big[\text{Since,} \cot(\text{A}-\text{B})=\frac{\cot\text{A}\cot\text{B}+1}{\cot\text{A}-\cot\text{B}}\Big]$
$=\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\cot(\text{x}+\text{h})\cot\text{x}+1}{\cot(-\text{h})\times\text{h}\big(\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}\big)}$
$=\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\cot(\text{x}+\text{h})\cot\text{x}+1}{\Big(\frac{\text{h}}{\cot\text{h}}\Big)\big(\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}\big)}$
$=\frac{\text{e}^\sqrt{\cot\text{x}}\times(\cot^2\text{x}+1)}{2\sqrt{\cot\text{x}}}\ \Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot\text{x}}}\ \big[\because(1+\cot^2\text{x})=\text{cosec}^2\text{x}\big]$
$\therefore\ \frac{\text{d}}{\text{dx}}\Big(\text{e}^\sqrt{\cot\text{x}}\Big)=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot}\text{x}}$
View full question & answer→Question 4625 Marks
If $\text{y}=\frac{\text{e}^\text{x}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}},$ prove that $\frac{\text{dy}}{\text{dx}}=1-\text{y}^2$
AnswerGivne, $\text{y}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
Differentiate with respect to x,
$\frac{\text{d}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}\Big)$
$=\Bigg[\frac{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)}{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)^2}\Bigg]$
[Using quotient rule and chain rule]
$=\begin{bmatrix} \frac{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)\Big[\text{e}^{\text{x}}-\text{e}^{-\text{x}}\frac{\text{d}}{\text{dx}}(-\text{x})-\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)\Big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\frac{\text{d}}{\text{dx}}(-\text{x})\Big)\Big]}{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)^2} \end{bmatrix}$
$=\begin{bmatrix} \frac{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)-\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)}{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)^2} \end{bmatrix}$
$=\bigg[\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}+2\text{e}^{\text{x}}\times\text{e}^{-\text{x}}-\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}+2\text{e}^{\text{x}}\text{e}^{-\text{x}}}{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)^2}\bigg]$
$\frac{\text{dy}}{\text{dx}}\bigg[\frac{4}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}\bigg]\ .....(\text{i})$
Now,
$1-\text{y}^2=1-\Big(\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}\Big)^2$
$=1-\frac{(\text{e}^{\text{x}}-\text{e}^{-\text{x}})^2}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}$
$=\frac{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2-(\text{e}^{\text{x}}-\text{e}^{-\text{x}})^2}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}$
$=\frac{4}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}$
View full question & answer→Question 4635 Marks
If $\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=\text{a}(\text{x}-\text{y}),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{1-\text{y}^2}}{1-\text{x}^2}$
AnswerWe have, $\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=\text{a}\big(\text{x}-\text{y}\big)$
Let $\text{x}=\sin\text{A},\text{y}=\sin\text{B}$
$\Rightarrow\sqrt{1-\sin^2\text{A}}+\sqrt{1-\sin^2\text{B}}=\text{a}\big(\sin\text{A}-\sin\text{B}\big)$
$\Rightarrow\cos\text{A}+\cos\text{B}=\text{a}\big(\sin\text{A}-\sin\text{B}\big)$
$\Rightarrow\text{a}=\frac{\cos\text{A}+\cos\text{B}}{\sin\text{A}-\sin\text{B}}$
$\Rightarrow\text{a}=\frac{2\cos\frac{\text{A}+\cos\text{B}}{2}\cos\frac{\text{A}-\text{B}}{2}}{2\cos\frac{\text{A}+\text{B}}{2}\sin\frac{\text{A}-\text{B}}{2}}$
$\begin{bmatrix}\because\sin\text{A}-\sin\text{B}=2\cos\frac{\text{A}+\text{B}}{2}\sin\frac{\text{A}-\text{B}}{2} \\ \because\cos \text{A}+\cos\text{B}=2\cos\frac{\text{A}+\text{B}}{2}\cos\frac{\text{A}-\text{B}}{2}\end{bmatrix}$
$\Rightarrow\text{a}=\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)$
$\Rightarrow\cot^{-1}\text{a}=\frac{\text{A}-\text{B}}{2}$
$\Rightarrow2\cot^{-1}\text{a}={\text{A}-\text{B}}$
$\Rightarrow2\cot^{-1}\text{a}=\sin^{-1}\text{x}-\sin^{-1}\text{y}$
$\Big[\because \text{x}=\sin\text{A},\text{y}=\sin\text{B}\Big]$
Differentiating with respect to x, we get
$\frac{\text{d}}{\text{dx}}\big(2\cot^{-1}\text{a}\big)=\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}\big)-\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{y}\big)$
$\Rightarrow0=\frac{1}{\sqrt{1-\text{x}^2}}-\frac{1}{\sqrt{1-\text{y}^2}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{1}{\sqrt{1-\text{y}^2}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{1-\text{y}^2}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sqrt\frac{1-\text{y}^2}{1-\text{x}^2}$
View full question & answer→Question 4645 Marks
Find a point on the curve $y = x^3 + 1$ where the tangent is parallel to the chord joining $(1, 2)$ and $(3, 28).$
AnswerLet, $f(x) = x^2+ 1$ The tangent to the curve is parallel to the chord joining the points $(1, 2)$ and $(3, 28).$
Assume that the chord joins the points $(a, f(a))$ and $(b, f(b)).$
$\therefore a = 1, b = 3$ The polynomial function is everywhere continuous and differentiable.
So, $f(x) = x^2+ 1$ is continuous on $[1, 3]$ and differentiable on $(1, 3).$
Thus, both the conditions of Lagrange's theorem are satisfied.
Concequently, there exists $\text{c}\in(1,3)$
such that $\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}$
Now, $f(x) = x^2+ 1 $
$\Rightarrow f'(x) = 3x^2, $
$f(1) = 2, f(3) = 28$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(3)-\text{f}(1)}{3-1}$
$\Rightarrow3\text{x}^2=\frac{26}{2}$
$\Rightarrow3\text{x}^2=13$
$\Rightarrow\text{x}=\pm\sqrt{\frac{13}{3}}$
Thus, $\text{c}=\sqrt{\frac{13}{3}}$
such that $\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}$
Clearly, $\text{f}(\text{c})=\bigg[\Big(\frac{13}{3}\Big)^{\frac{3}{2}}+1\bigg]$
Thus, $c, f(c)$, i.e. $\bigg(\sqrt{\frac{13}{3}},1+\Big(\frac{13}{3}\Big)^{\frac{3}{2}}\bigg),$ is a point on the given curve where the tangent is parallel to the chord joining the points $(1, 2)$ and $(3, 28).$
View full question & answer→Question 4655 Marks
If $\text{x}=3\cot-2\cos^3\text{t},\text{y}=3\sin\text{t}-2\sin^3\text{t}$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}.$
AnswerGiven,
$\text{x}=3\cot-2\cos^3\text{t},$
$\text{y}=3\sin\text{t}-2\sin^3\text{t}$
Differentiating both w.r.t. t,
$\frac{\text{dx}}{\text{dt}}=-3\sin\text{t}-6\cos^2\text{t}(-\sin\text{t})$
$\frac{\text{dx}}{\text{dt}}=-3\sin\text{t}+6\cos^2\text{t}\sin\text{t}$
And $\text{y}=3\sin\text{t}-2\sin^2\text{t}$
Differentiating both w.r.t. t,
$\frac{\text{dy}}{\text{dt}}=3\cos\text{t}-6\sin^2\text{t}\cos\text{t}$
Now,
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cot-2\sin^2\text{t}\cos\text{t}}{-\sin\text{t}+2\cos^2\text{t}\sin\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cot[1-2\sin^2\text{t]}}{\sin\text{t}[2\cos^2\text{t}-1]}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\cot\text{t}$
Differentiating both w.r.t. x,
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}(\cot\text{x})}{\text{dx}}=-\text{cosec}^2\text{x}$
View full question & answer→Question 4665 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\text{x}^2-3}+(\text{x}-3)^{\text{x}^2}$
AnswerLet $\text{y}=\text{x}^{\text{x}^2-3}+(\text{x}-3)^{\text{x}^2}$
Also, let $\text{u}=\text{x}^{\text{x}^2-3}\text{ and v}=(\text{x}-3)^{\text{x}^2}$
$\therefore \text{y}=\text{u}+\text{v}$
Differentiating both sides with respect to x, we obtain
$\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ .....(\text{i})$
$\text{u}=\text{x}^{\text{x}^2-3}$
$\log\text{u}=(\text{x}^2-3)\log\text{x}$
Differentiating with respect to x, we obtain
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log\text{x}\times\frac{\text{d}}{\text{dx}}\big(\text{x}^2-3\big)+\big(\text{x}^2-3\big)\times\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log\text{x}\times2\text{x}+(\text{x}^2-3)\times\frac{1}{\text{x}}$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}^2-3}\times\Big[\frac{\text{x}^2-3}{\text{x}}+2\text{x}\log\text{x}\Big]$
Also,
$\text{v}=(\text{x}-3)^{\text{x}^2}$
$\therefore\log\text{v}=\log(\text{x}-3)^{\text{x}^2}$
$\Rightarrow\log\text{v}=\text{x}^2\log(\text{x}-3)$
Differentaiting both sides with respect to x, we obtain
$\frac{1}{\text{v}}\times\frac{\text{dv}}{\text{dx}}=\log(\text{x}-3)\times\frac{\text{d}}{\text{dx}}(\text{x}^2)+\text{x}^2\times\frac{\text{d}}{\text{dx}}[\log(\text{x}-3)]$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\log(\text{x}-3)\times2\text{x}+\text{x}^2\times\frac{1}{\text{x}-3}\times\frac{\text{d}}{\text{dx}}(\text{x}-3)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[2\text{x}\log(\text{x}-3)+\frac{\text{x}^2}{\text{x}-3}\times1\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\text{x}-3)^{\text{x}^2}\Big[\frac{\text{x}^2}{\text{x}-3}+2\text{x}\log(\text{x}-3)\Big]$
Substituting the expressions of $\frac{\text{du}}{\text{dx}}$ and $\frac{\text{dv}}{\text{dx}}$ in equation (1), we obtain
$\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}^2-3}\Big[\frac{\text{x}^2-3}{\text{x}}+2\text{x}\log\text{x}\Big] \\ +(\text{x}-3)^{\text{x}^2}\Big[\frac{\text{x}^2}{\text{x}-3}+2\text{x}\log(\text{x}-3)\Big]$
View full question & answer→Question 4675 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{2\text{a}^{\text{x}}}{1-\text{a}^{2\text{x}}}\Big),\text{a}>1, -\infty<\text{x}<0$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{2\text{a}^{\text{x}}}{1-\text{a}^{2\text{x}}}\Big)$
Put $\text{a}^{\text{x}}=\tan\theta$
$\Rightarrow\text{y}=\tan^{-1}\Big\{\frac{2\times\text{a}^\text{x}}{1-(\text{a}^{\text{x}})^2}\Big\}$
$\Rightarrow\text{y}=\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)$
$\Rightarrow\text{y}=\tan^{-1}(\tan2\theta)\ .....(\text{i})$
Here, $-\infty<\text{x}<0$
$\Rightarrow\text{a}^{-\infty}<\text{a}^{\text{x}}<2^{0}$
$\Rightarrow 0<\tan\theta<1$
$\Rightarrow 0<\theta<\frac{\pi}{4}$
$\Rightarrow 0<2\theta<\frac{\pi}{2}$
So, from equation (i),
$\text{y}=2\theta\Big[\text{Since},\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\big(-\frac{\pi}{2},\frac{\pi}{2}\big)\Big]$
$\Rightarrow\text{y}=2\tan^{-1}(\text{a}^{\text{x}})$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{2}{1+(\text{a}^{\text{x}})^2}\frac{\text{d}}{\text{dx}}(\text{a}^{\text{x}})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\times\text{a}^{\text{x}}\log_\text{e}\text{a}}{1+\text{a}^{2\text{x}}}$
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{2\text{a}^{\text{x}}\log_\text{e}\text{a}}{1+\text{a}^{2\text{x}}}$
View full question & answer→Question 4685 Marks
Differentiate $\cos^{-1}(4\text{x}^3-3\text{x})$ with respect to $\tan^{-1}\Big(\frac{\sqrt{1-\text{x}^2}}{\text{x}}\Big),$ if $\frac{1}{2}<\text{x}<1$
AnswerLet, $\text{u}=\cos^{-1}(4\text{x}^3-3\text{x})$
Put, $\text{x}=\cos\theta$
$\Rightarrow\theta=\cos^{-1}\text{x}$
Now, $\text{u}=\cos^{-1}(4\cos^3\theta-3\cos\theta)$
$\Rightarrow \text{u}=\cos^{}-1(\cos3\theta)\ .....(\text{i})$
Let, $\text{v}=\tan^{-1}\Big(\frac{\sqrt{1-\text{x}^2}}{\text{x}}\Big)$
$\Rightarrow\text{v}=\tan^{-1}\bigg(\frac{\sqrt{-1\cos^2\theta}}{\cos\theta}\bigg)$
$\Rightarrow\text{v}=\tan^{-1}\Big(\frac{\sin\theta}{\cos\theta}\Big)$
$\Rightarrow\text{v}=\tan^{-1}(\tan\theta)\ .....(\text{ii})$
Here,
$\frac{1}{2}<\text{x}<1$
$\Rightarrow\frac{1}{2}<\cos<1$
$\Rightarrow0<\theta<\frac{\pi}{3}$
So, from equation (i),
$\text{u}=3\theta\big[\text{Since,} \cos^{-1}(\cos\theta)=\theta,\text{if }\theta\in[0,\pi]\big]$
$\Rightarrow\text{u}=3\cos^{-1}\text{x}$
Differenting it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{-3}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\theta \Big[\text{since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{v}=\cos^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\Big(\frac{-3}{\sqrt{1-\text{x}^2}}\Big)\Big(\frac{-\sqrt{1-\text{x}^2}}{1}\Big)$
$\therefore\frac{\text{du}}{\text{dv}}=3$
View full question & answer→Question 4695 Marks
If $\text{y}=\text{x}^\text{n}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\},$ prove that $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}+(1-2\text{n})\frac{\text{dy}}{\text{dx}}+(1+\text{n}^2)\text{y}=0.$
AnswerWe have,
$\text{y}=\text{x}^\text{n}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\},...(1)$
Differentiating y with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{nx}^{\text{n}-1}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\}+\text{x}^\text{n}\\\Big\{-\text{a}\sin(\log\text{x})\times\frac{1}{\text{x}}+\text{b}\cos(\log\text{x})\times\frac{1}{\text{x}}\Big\}$
$=\frac{\text{n}}{\text{x}}\text{x}^\text{n}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text {x})\}+\text{x}^{\text{n}-1}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}$
$\Rightarrow\frac{\text{n}}{\text{x}}\text{y}+\text{x}^{\text{n}-1}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}\ [\text{from}(1)]$
$\Rightarrow\text{x}^{\text{n}-1}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}=\frac{\text{dy}}{\text{dx}}-\frac{\text{n}}{\text{x}}\text{y}...(2)$
Differentiating $\frac{\text{dy}}{\text{dx}}$ with respect to x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{n}}{\text{x}}\frac{\text{dy}}{\text{dx}}=\frac{\text{ny}}{\text{x}^2}+(\text{n}-1)\text{x}^{\text{n}-2}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}\\+\text{x}^{\text{n}-1}\Big\{-\text{a}\cos(\log\text{x})\times\frac{1}{\text{x}}-\text{b}\sin(\log\text{x})\times\frac{1}{\text{x}}\Big\}$
$=\frac{\text{n}}{\text{x}}\frac{\text{dx}}{\text{dy}}-\frac{\text{ny}}{\text{x}^2}+(\text{n}-1)\frac{\text{x}^{\text{n}-1}}{\text{x}}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}\\-\frac{\text{x}^\text{n}}{\text{x}^2}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\}$
$\frac{\text{n}}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{\text{ny}}{\text{x}^2}+\Big(\frac{\text{n}-1}{\text{x}}\Big)\Big(\frac{\text{dy}}{\text{dx}}-\frac{\text{n}}{\text{x}}\text{y}\Big)-\frac{\text{y}}{\text{x}^2}$
$=\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{n}+\text{n}-1}{\text{x}}\Big)-\frac{(\text{n}+\text{n}^2+\text{n}+1)\text{y}}{\text{x}^2}$
$=\Big(\frac{2\text{n}-1}{\text{x}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{(\text{n}^2+1)\text{y}}{\text{x}^2}$
$\Rightarrow\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}(2\text{n}-1)\frac{\text{dy}}{\text{dx}}+(\text{n}^2+1)\text{y}=0$
Hence, $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}+(1-2\text{n})\text{x}\frac{\text{dy}}{\text{dx}}+(1+\text{n}^2)\text{y}=0$
View full question & answer→Question 4705 Marks
Show that the function $\text{f(x)}\begin{cases}\text{x}^\text{m}\sin(\frac{1}{\text{x}}), &\text{x}\neq0 \\0 ,& \text{x}=0\end{cases}$
Continuous but not diffierentiable at x = 0, if 0 < m < 1
Answer$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^{-}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(-\text{h})^\text{m}\sin\Big(-\frac{1}{\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}(-\text{h})^\text{m}\sin\Big(\frac{1}{\text{h}}\Big)$
$=0\times\text{k}\ [\text{Where}-1\leq\text{k}\leq1]$
$=0$
$\text{RHL }=\lim_\limits{\text{x}\rightarrow0^{+}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0^+}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(+\text{h})^\text{m}\sin\Big(\frac{1}{0+\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0^-}(-\text{h})^\text{m}\sin\Big(\frac{1}{\text{h}}\Big)$
$=0\times\text{k'}\ [\text{When}-1\leq\text{k}'\leq1]$
$=0$
LHL = f(0) = RHL
$\therefore$ f(x) is continuous at x = 0
For differentiable at x = 0
(LHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(0-\text{h})-\text{f}(0)}{(0-\text{h})-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(-\text{h})^\text{m}\sin\Big(-\frac{1}{\text{h}}\Big)}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}(-\text{h})^\text{m-1}\sin\Big(-\frac{1}{\text{h}}\Big)$
= Not definded [Since 0 < m < 1]
(RHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{(0+\text{h})-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^\text{m}\sin\Big(\frac{1}{\text{h}}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}(\text{h})^\text{m-1}\sin\Big(\frac{1}{\text{h}}\Big)$
= Not defined [as 0, m < 1]
$\therefore$ (LHL at x = 0) and (RHL at x = 0) are not defined, so f(x) is continuous but not differentiable at x = 0, when 0 < m < 1.
View full question & answer→Question 4715 Marks
Differentiate $\tan^{-1}\Big(\frac{1+\text{ax}}{1-\text{ax}}\Big)$ with respect to $\sqrt{1+\text{a}^2\text{x}^2}$
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{1+\text{ax}}{1-\text{ax}}\Big)$
Put $\text{ax}= \tan\theta$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{1+\tan\theta}{1-\tan\theta}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\bigg(\frac{\tan\frac{\pi}{4}+\tan\theta}{1-\tan\frac{\pi}{4}\tan\theta}\bigg)$
$\Rightarrow\text{u}=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}+\theta\Big)\Big]$
$\Rightarrow\text{u}=\frac{\pi}{4}+\theta$
$\Rightarrow\text{u}=\frac{\pi}{4}+\tan^{-1}(\text{ax}) [\text{since,}\tan\theta=\text{ax}]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=0+\frac{1}{1+(\text{ax}^2)}\frac{\text{d}}{\text{dx}}(\text{ax})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{\text{a}}{1+\text{a}^2\text{x}^2}\ .....(\text{i})$
Now,
Let, $\text{v}=\sqrt{1+\text{a}^2\text{x}^2}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1+\text{a}^2\text{x}^2}}\frac{\text{d}}{\text{dx}}(1+\text{a}^2\text{x}^2)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1+\text{a}^2\text{x}^2}}(2\text{a}^2\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{\text{a}^2\text{x}}{\sqrt{1+\text{a}^2\text{x}^2}}\ .....(\text{ii})$
Dividing equation (i) by (ii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{\text{a}}{1+\text{a}^2\text{x}^2}\times\frac{\sqrt{1+\text{a}^2\text{x}^2}}{\text{a}^2\text{x}}$
$\frac{\text{du}}{\text{dv}}=\frac{1}{\text{ax}\sqrt{1+\text{a}^2\text{x}^2}}$
View full question & answer→Question 4725 Marks
Finde the value of a and b, if the function f(x) defined by $\text{f(x)}\begin{cases}\text{x}^2+3\text{x}+\text{a}, &\text{x}\leq1\\\text{bx}+2, & \text{x}>1\end{cases}$is differentiable at x = 1.
AnswerGiven that f(x) is differentiable at x = 1, Therefore, f(x) is countinuous at x = 1.
$\lim\limits_{\text{x}\rightarrow1^{-}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^{+}}\text{f(x)}=\text{f(1)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}(\text{x}^2+3\text{x}+\text{a})=\lim\limits_{\text{x}\rightarrow1}(\text{bx}+2)=1+3+\text{a}$
$\Rightarrow1+3+\text{a}=\text{b}+2$
$\Rightarrow\text{a}-\text{b}+2=0\dots(1)$
Again, f(x) is differentiable at x = 1. So,
(LHL at x = 1) = (RHL at x = 1)
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(z)}-\text{f}(1)}{\text{z}-1}=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(z)}-\text{f}(1)}{\text{z}-1}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}^2+3\text{x}+\text{a})-(4+\text{a})}{\text{x}-1}=\lim\limits_{\text{x}\rightarrow0}\frac{(\text{bx}+2)-(4+\text{a})}{\text{x}-1}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2+3\text{x}-4}{\text{x}-1}=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{bx}-2-\text{a})}{\text{x}-1}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}(\text{x}-4)=\lim\limits_{\text{x}\rightarrow1}\frac{\text{b}(\text{x}-1)}{\text{x}-1}$
$\Rightarrow5=\text{b}$
Hence, a = 3 and b = 5.
View full question & answer→Question 4735 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\frac{(\text{x}^2-1)^3(2\text{x}-1)}{\sqrt{(\text{x}-3)(4\text{x}-1)}}$
AnswerWe have, $\text{y}=\frac{(\text{x}^2-1)^3(2\text{x}-1)}{\sqrt{(\text{x}-3)(4\text{x}-1)}}.....(\text{i})$$\text{y}=\frac{(\text{x}^2-1)^3(2\text{x}-1)}{(\text{x}-3)^{\frac{1}{2}}(4\text{x}-1)^{\frac{1}{2}}}$
Taking log on both sides, $\log\text{y}=\log\Bigg[\frac{(\text{x}^2-1)^3(2\text{x}-1)}{(\text{x}-3)^{\frac{1}{2}}(4\text{x}-1)^{\frac{1}{2}}}\Bigg]$$\Rightarrow\log\text{y}=\log(\text{x}^2-1)^3+\log(2\text{x}-1)-\log(\text{x}-3)^{\frac{1}{2}}-\log(4\text{x}-1)^{\frac{1}{2}}$
$\Rightarrow\log\text{y}=\log(\text{x}^2-1)^3+\log(2\text{x}-1)-\frac{1}{2}\log(\text{x}-3)-\frac{1}{2}\log(4\text{x}-1)$
Differentiating with respect to x using chain rule, $\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3\frac{\text{d}}{\text{dx}}\Big\{\log(\text{x}^2-1)\Big\}+\frac{\text{d}}{\text{dx}}\Big\{\log(2\text{x}-1)\Big\}\\-\frac{1}{2}\frac{\text{d}}{\text{dx}}\Big\{\log(\text{x}-3)\Big\}-\frac{1}{2}\Big\{\log(4\text{x}-1)\Big\}$$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3\Big(\frac{1}{\text{x}^2-1}\Big)\frac{\text{d}}{\text{dx}}(\text{x}^2-1)+\frac{1}{(2\text{x}-1)}\frac{\text{d}}{\text{dx}}(2\text{x}-1)\\-\frac{1}{2}\Big(\frac{1}{\text{x}-3}\Big)\frac{\text{d}}{\text{dx}}(\text{x}-3)-\frac{1}{2}\frac{1}{(4\text{x}-1)}\frac{\text{d}}{\text{dx}}(4\text{x}-1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3\Big(\frac{1}{\text{x}^2-1}\Big)(2\text{x})+\frac{1}{2\text{x}-1}(2)-\frac{1}{2}\Big(\frac{1}{\text{x}-3}\Big)(1)-\frac{1}{2}\Big(\frac{1}{4\text{x}-1}\Big)(4)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\Big[\frac{6\text{x}}{\text{x}^2-1}+\frac{2}{2\text{x}-1}-\frac{1}{2(\text{x}-3)}-\frac{2}{4\text{x}-1}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{6\text{x}}{\text{x}^2-1}+\frac{2}{2\text{x}-1}-\frac{1}{2(\text{x}-3)}-\frac{2}{4\text{x}-1}\Big]$
$\Rightarrow\frac{(\text{x}^2-1)^3(2\text{x}-1)}{\sqrt{(\text{x}-3)(4\text{x}-1)}}\Big[\frac{6\text{x}}{\text{x}^2-1}+\frac{2}{2\text{x}-1}-\frac{1}{2(\text{x}-3)}-\frac{2}{4\text{x}-1}\Big]$
[Using equation (i)]
View full question & answer→Question 4745 Marks
If $\text{x}=\cos\text{t}+\log\tan\frac{\text{t}}{2},\text{y}=\sin\text{t},$ Then find the value of $\frac{\text{d}^2\text{y}}{\text{dt}^2}\ \text{and}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}\ \text{at}\ \text{t}=\frac{\pi}{4}.$
Answer$\text{x}=\cos\text{t}+\log\tan\frac{\text{t}}{2},\text{y}=\sin\text{t}$
On differentiating with respect to t, we get
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big(\cos\text{t}+\log\tan\frac{\text{t}}{2}\Big)=-\sin\text{t}+\frac{1}{\tan\frac{\text{t}}{2}}\times\sec^2\frac{\text{t}}{2}\times\frac{1}{2}$
$=-\sin\text{t}+\frac{1}{2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}=-\sin\text{t}+\frac{1}{\sin\text{t}}$
$=\frac{-\sin^2\text{t}+1}{\sin\text{t}}=\frac{-\sin^2\text{t}+1}{\sin\text{t}}$
$=\frac{\cos^2\text{t}}{\sin\text{t}}$
and
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\sin\text{t})=\cos\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dt}^2}=\frac{\text{d}}{\text{dt}}\Big(\frac{\text{dy}}{\text{dt}}\Big)=\frac{\text{d}}{\text{dt}}(\cos\text{t})=-\sin\text{t}$
$\Big(\frac{\text{d}^2\text{y}}{\text{dt}^2}\Big)_{(\text{t}=\frac{\pi}{4})}=-\sin\Big(\frac{\pi}{4}\Big)=-\frac{1}{\sqrt{2}}...(1)$
Also, $\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\cos\text{t}}{\frac{\cos^2\text{t}}{\sin\text{t}}}=\frac{\sin\text{t}}{\cos\text{t}}=\tan\text{t}$
Now, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}(\tan\text{x})$
$\frac{\text{d}}{\text{dt}}(\tan\text{t})\times\frac{\text{dt}}{\text{dx}}=\sec^2\text{t}\times\frac{\sin\text{t}}{\cos^2\text{t}}$
$=\frac{\sin\text{t}}{\cos^4\text{t}}$
$\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)_{\text{t}=\frac{\pi}{4}}=\frac{\sin\Big(\frac{\pi}{4}\Big)}{\cos^4\Big(\frac{\pi}{4}\Big)}=2\sqrt{2}...(2)$
Hence, at $\text{t}=\frac{\pi}{4},\frac{\text{d}^2\text{y}}{\text{dt}^2}=-\frac{1}{\sqrt{2}}\ \text{and}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=2\sqrt{2}$
View full question & answer→Question 4755 Marks
If $\text{x}=\sin^{-1}\Big(\frac{2\text{t}}{1+\text{t}^2}\Big)$ and $\text{y}=\tan^{-1}\Big(\frac{2\text{t}}{1-\text{t}^2}\Big),-1<\text{t}<1,$ prove that $\frac{\text{dy}}{\text{dx}}=1$
AnswerWe have, $\text{x}=\sin^{-1}\Big(\frac{2\text{t}}{1+\text{t}^{2}}\Big)$
Put $\text{t}=\tan\theta$
$ \Rightarrow-1<\tan\theta<1$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
$\Rightarrow-\frac{\pi}{2}<2\theta<\frac{\pi}{2}$
$\therefore\text{x}=\sin^{-1}\Big(\frac{2\tan\theta}{1+\tan^{2}\theta}\Big)$
$\Rightarrow\text{x}=\sin^{-1}(\sin2\theta)$
$\Rightarrow\text{x}=2\theta$
$\Big[\therefore-\frac{\pi}{2}<2\theta<\frac{\pi}{2}\Big]$
$\Rightarrow\text{x}=2(\tan^{-1}\text{t})$
$\big[\therefore \text{t}=\sin\theta\big]$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{2}{1+\text{t}^{2}}\ .....(\text{i})$
Now, $\text{y}=\tan^{-1}\Big(\frac{2\text{t}}{1-\text{t}^{2}}\Big)$
Put $\text{t}=\tan\theta$
$\Rightarrow\text{y}=\tan^{-1}\Big(\frac{2\text{t}}{1-\text{t}^{2}}\Big)$
$\Rightarrow\text{y}=\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan\theta}\Big)$
$\Rightarrow\text{y}=\tan^{-1}(\tan2\theta)$
$\Rightarrow\text{y}=2\theta$
$\Big[\therefore-\frac{\pi}{2}<2\theta<\frac{\pi}{2}\Big]$
$\Rightarrow\text{y}= 2 \tan^{-1}\text{t}$
$\big[\therefore\text{t}=\tan\theta\big]$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{2}{1+\text{t}^{2}}\ .....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2}{1+\text{t}^{2}}\times\frac{1+\text{t}^{2}}{2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1$
View full question & answer→Question 4765 Marks
Differentiate $\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ with respect to $\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ if -1 < x < 1.
AnswerLet $\text{u}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
put $\text{x}=\tan\theta\Rightarrow \theta=\tan^{-1}\text{x},\text{so}$
$\text{u}=\sin^{-1}\Big(\frac{2\tan\theta}{1+\tan^2\theta}\Big)$
$\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let $\text{v}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$
$=\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)$
$\text{v}=\tan^{-1}(\tan2\theta)\ .....(\text{ii})$
Here, $-1<\text{x}<1$
$\Rightarrow-1<\tan\theta<1$
$\Rightarrow -\frac{\pi}{4}<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{u}=2\theta$
$\Big[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\text{u}=2\tan^{-1}\text{x}$
Diffrerentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{2}{(1+\text{x}^2)}\ .....\text{(iii)}$
From equation (ii),
$\text{v}=2\theta$
$\Big[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\text{v}=2\tan^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{2}{1+\text{x}^2}\ .....\text{(iv)}$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2}{1+\text{x}^2}\times\frac{1+\text{x}^2}{2}$
$\frac{\text{du}}{\text{dv}}=1$
View full question & answer→Question 4775 Marks
If $\text{x}=2\cos\theta-\cot2\theta$ and $\text{y}=2\sin\theta-\sin2\theta,$ prove that $\frac{\text{dy}}{\text{dx}}=\tan\big(\frac{3\theta}{2}\big)$
Answer Here, $\text{x}=2\cos\theta-\cos2\theta$
Diffierentiating it with respect to $\theta$ using chain rule,
$\frac{\text{dx}}{\text{d}\theta}=2(-\sin\theta)-(-\sin2\theta)\frac{\text{d}}{\text{d}\theta}(2\theta)$
$=-2\sin\theta+2\sin2\theta$
$\frac{\text{dx}}{\text{d}\theta}=2(\sin2\theta-\sin\theta)...(\text{i})$
And, $\text{y}=2\sin\theta-\sin\theta$
Differentiating it with respect to $\theta$ using chain rule,
$\frac{\text{dt}}{\text{d}\theta}=2\cos\theta-\cos2\theta\frac{\text{d}}{\text{d}\theta}(2\theta)$
$=2\cos\theta-\cos2\theta(2)$
$=2\cos\theta-\cos2\theta(2)$
$\frac{\text{dy}}{\text{d}\theta}=2(\cos\theta-\cos2\theta)...(\text{ii})$
Dividing equation (ii) by equation (i),
$\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{2(\cos\theta-\cos2\theta)}{2(\sin2\theta-\sin\theta)}$
$=\frac{(\cos\theta-\cos2\theta)}{(\sin2\theta-\sin\theta)}$
$\frac{\text{dy}}{\text{dx}}=\frac{-2\sin\big(\frac{\theta+2\theta}{2}\big)\sin\big(\frac{\theta-2\theta}{2}\big)}{2\cos\big(\frac{2\theta+\theta}{2}\big)\sin\big(2\theta-\theta\big)}$
$\Big[\sin\text{A}-\sin\text{B}=2\cos\Big(\frac{\text{A+B}}{2}\Big)\sin\Big(\frac{\text{A+B}}{2}\Big)\Big]$
$\Rightarrow-\frac{\sin\big(\frac{3\theta}{2}\big)\big(\sin\big(\frac{-\theta}{2}\big)\big)}{\cos\big(\frac{3\theta}{2}\big)\sin\big(\frac{\theta}{2}\big)}$
$\Rightarrow-\frac{\sin\big(\frac{3\theta}{2}\big)\big(-\sin\frac{-\theta}{2}\big)}{\cos\big(\frac{3\theta}{2}\big)\sin\big(\frac{\theta}{2}\big)}$
$\Rightarrow\frac{\sin\big(\frac{3\theta}{2}\big)}{\cos\big(\frac{3\theta}{2}\big)}$
$\frac{\text{dy}}{\text{dx}}=\tan\big(\frac{3\theta}{2}\big)$
View full question & answer→Question 4785 Marks
Find the values of a and b so that the function $\text{f(x)}\begin{cases}\text{x}^2+3\text{x}+\text{a}, & \text{if x}\leq1\\\text{bx}+2, & \text{if x} > 1\end{cases}$ is differentiable at each $\text{x}\in\text{R}.$
AnswerIt is given that the function is differentiable at each $\text{x}\in\text{R}$ and every differentiable function is continuous.
Therefore,
Given: $\text{f(x)}=\begin{Bmatrix}\text{x}^2+3\text{x}+\text{a}, & \text{if x}\leq1\\\text{bx}+2, & \text{if x} > 1 \end{Bmatrix}$
Therefore,
f(x) is continuous at x = 1.
Therefore,
$\lim_\limits{\text{x}\rightarrow1^{-}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^{+}}\text{f(x)}=\text{f}(1)$
Implies that $\lim_\limits{\text{x}\rightarrow1}\text{x}^2+3\text{x}+\text{a}=\lim_\limits{\text{x}\rightarrow1}\text{bx}+2=\text{a}+4$ [Using def. of f(x)]
Implies that a + 4 = b + 2 = a + 4 ......(1)
Since, f(x) is differentiable at x = 1. Therefore,
(LHL at x = 1) = (RHL at x = 1)
$\lim_\limits{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\text{f(c)}-\text{f}(1)}{\text{x}-1}$
Implies that $\lim_\limits{\text{x}\rightarrow1}\frac{\text{x}^2+3\text{x}+\text{a}-\text{a}-4}{\text{x}-1}=\lim_\limits{\text{x}\rightarrow1}\frac{\text{bx}+2-4-\text{a}}{\text{x}-1}$ [Using def. of f(x)]
Implies that $\lim_\limits{\text{x}\rightarrow1}\frac{(\text{x}+4)(\text{x-1})}{\text{x}-1}=\lim_\limits{\text{x}\rightarrow1}\frac{\text{bx}-2-\text{a}}{\text{x}-1}$
Implies that $\lim_\limits{\text{x}\rightarrow1}\frac{(\text{x}+4)(\text{x}-1)}{\text{x}-1}=\lim_\limits{\text{x}\rightarrow1}\frac{\text{bx}-\text{b}}{\text{x}-1}$ [Using (1)]
Implies that $\lim_\limits{\text{x}\rightarrow1}\frac{(\text{x}+4)(\text{x}-1)}{\text{x}-1}=\lim_\limits{\text{x}\rightarrow1}\frac{\text{b}(\text{x}-1)}{\text{x}-1}$
Implies that 5 = b
From (1), we have
a + 4 = b + 2
Implies that a + 4 = 5 + 2
Implies that a = 7 - 4
Implies that a = 3
Hence, a = 3, b = 5.
View full question & answer→Question 4795 Marks
If $\text{x}=\text{a}\sin\text{t}-\text{b}\cos\text{t},\text{y}=\text{a}\cos\text{t}+\text{b}\sin\text{t},$ Prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{x}^2+\text{y}^2}{\text{y}^2}$
AnswerWe have$\text{x}=\text{a}\sin\text{t}-\text{b}\cos\text{t},\text{y}=\text{a}\cos\text{t}+\text{b}\sin\text{t},$
On differentiating with respect to t, we get
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{a}\sin\text{t}-\text{b}\cos\text{t})=\text{a}\cos\text{t}+\text{b}\sin\text{t}$
and
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{a}\cos\text{t}+\text{b}\sin\text{t})=-\text{a}\sin\text{t}+\text{b}\cos\text{t}$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-\text{a}\sin\text{t}+\text{b}\cos\text{t}}{\text{a}\cos+\text{b}\sin\text{t}}$
Therefore
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}\Big(\frac{-\text{a}\sin\text{t}+\text{b}\cos\text{t}}{\text{a}\cos\text{t}+\text{b}\sin\text{t}}\Big)$
$=\frac{\text{d}}{\text{dt}}\Big(\frac{-\text{a}\sin\text{t}+\text{b}\cos\text{t}}{\text{a}\sin\text{t}+\text{b}\sin\text{t}}\Big)\times\frac{\text{dt}}{\text{dx}}$
$=\frac{(\text{a}\cos\text{t}+\text{b}\sin\text{t})\frac{\text{d}}{\text{dt}}(-\text{a}\sin\text{t}+\text{b}\cos\text{t})-(-\text{a}\sin\text{t}+\text{b}\cos\text{t})\frac{\text{d}}{\text{dt}}(\text{a}\cos\text{t}+\text{b}\sin\text{t})}{(\text{a}\cos\text{t}+\text{b}\sin\text{t})^2}$
$$$=\frac{(\text{a}\cos\text{t}+\text{b}\sin\text{t})(\text{a}\cos\text{t}+\text{b}\sin\text{t})(-\text{a}\sin\text{t}+\text{b}\cos\text{t})-(-\text{a}\sin\text{t}+\text{b}\cos\text{t})}{(\text{a}\cos\text{t}+\text{b}\sin\text{t})^2}$
$=\frac{-(\text{a}\cos\text{t}+\text{b}\sin\text{t})^2-(-\text{a}\sin\text{t}+\text{b}\cos\text{t})^2}{(\text{a}\cos\text{t}+\text{b}\sin\text{t})}$
$=\frac{-(\text{a}\cos\text{t}+\text{b}\sin\text{t})^2-(\text{a}\sin\text{t}+\text{b}\cos\text{t})^2}{(\text{a}\cos\text{t}+\text{b}\sin\text{t})}$
$=\frac{-\text{y}^2-\text{x}^2}{\text{y}^3}$
View full question & answer→Question 4805 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\sin\text{ xy}+\cos(\text{x}+\text{y})=1$
AnswerWe have, $\sin\text{ xy}+\cos(\text{x}+\text{y})=1$Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}\sin\text{xy}+\frac{\text{d}}{\text{dx}}\cos(\text{x}+\text{y})=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\cos \text{xy}\frac{\text{d}}{\text{dx}}(\text{xy})-\sin(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})=0$
[Using chain rule]
$\Rightarrow\cos(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]-\sin(\text{x}+\text{y})\Big[1+\frac{\text{dy}}{\text{dx}}\Big]=0$
$\Rightarrow\cos(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)\Big]-\sin(\text{x}+\text{y})+\sin(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}\cos(\text{xy})\frac{\text{dy}}{\text{dx}}+\text{y}\cos(\text{xy})-\sin(\text{x}+\text{y})+\sin(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\big[\text{x}\cos(\text{xy})+\sin(\text{x}+\text{y})\big]\frac{\text{dy}}{\text{dx}} \\ =\big[\sin(\text{x}+\text{y})-\text{y}\cos(\text{xy})\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big[\frac{\sin(\text{x}+\text{y})-\text{y}\cos\text{xy}}{\text{x}\cos\text{xy}+\sin(\text{x}+\text{y})}\Big]$
View full question & answer→Question 4815 Marks
Differentiate the following functions with respect to x:
$\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}$
AnswerLet $\text{y}=\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)^{\frac{1}{2}}$
$=\frac{\text{1}}{\text{2}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)$
$=\frac{\text{1}}{\text{2}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)^{\frac{1}{2}}\Big[\frac{(1-\sin\text{x})(\cos\text{x})-(1+\sin\text{x})(-\cos\text{x})}{(1-\sin\text{x})^2}\Big]$
$=\frac{\text{1}}{\text{2}}\frac{(1+\sin\text{x})^\frac{1}{2}}{(1-\sin\text{x})^\frac{1}{2}}\Big[\frac{\cos\text{x}-\cos\text{x}+\cos\text{x}\sin\text{x}+\sin\text{x}\cos\text{x}}{(1-\sin\text{x})^2}\Big]$
$=\frac{1}{2}\times\frac{2\cos\text{x}}{\sqrt{1+\sin\text{x}}(1-\sin\text{x})^\frac{3}{2}}$
$=\frac{\cos\text{x}}{\sqrt{1+\sin\text{x}}(1-\sin\text{x})^\frac{3}{2}}$
$=\frac{\cos\text{x}}{\sqrt{1+\sin\text{x}}\sqrt{1-\sin\text{x}}(1-\sin\text{x})}$
$=\frac{\cos\text{x}}{\sqrt{1-\sin^2\text{x}}(1-\sin\text{x})}$
$=\frac{\cos\text{x}}{\cos\text{x}(1-\sin\text{x})}$ $\big[\text{Using }1-\sin^2\text{x}=\cos^2\text{x}\big] $
$=\frac{1}{(1-\sin\text{x})}\times\frac{(1+\sin\text{x})}{(1+\sin\text{x})}$
Thus, $ \frac{\text{dy}}{\text{dx}}=\frac{1}{\cos^2\text{x}}+\frac{\sin\text{x}}{\cos^2\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\sec^2\text{x}+\tan\text{x}\sec\text{x}$
$\frac{\text{dy}}{\text{dx}}=\sec\text{x}[\tan\text{x}+\sec\text{x}] $
View full question & answer→Question 4825 Marks
Is $|\sin\text{x}|$ differentible? What about $\cos|\text{x}|?$
Answer Let, d(x) = |sin x|
$\sin\text{x}=0,$ for $\text{x}=\text{n}\pi,$
$|\sin\text{x}|=\begin{cases}-\sin\text{x}\ (2\text{m}-1)\pi<\text{x}<2\text{mx},&\text{where m}\in\text{Z}\\\sin\text{x}\ 2\text{mx}<\text{x}<(2\text{m}+1)\pi,&\text{where m}\in\text{Z}\\-\sin\text{x}\ (2\text{m}+1)\pi<\text{x}<2(\text{m}+1)\pi,&\text{where m}\in\text{Z}\end{cases}$
$(\text{LHL at x}=2\text{mx})=\lim_\limits{\text{x}\rightarrow2\text{mx}^{-}}\frac{\text{f(x)}-\text{f}(2\text{mx})}{\text{x}-2\text{mx}}$
$=\lim_\limits{\text{x}\rightarrow2\text{mx}^{-}}\frac{-\sin(\text{x}-0)}{\text{x}-2\text{mx}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-\sin(2\text{mx}-\text{h})}{2\text{mx}-\text{h}-2\text{mx}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{h})}{\text{h}}=-1$
$(\text{RHL at x}=2\text{mx})=\lim_\limits{\text{x}\rightarrow2\text{mx}^{+}}\frac{\text{f(x)}-\text{f}(2\text{mx})}{\text{x}-2\text{mx}}$
$=\lim_\limits{\text{x}\rightarrow2\text{mx}^{+}}\frac{\sin(\text{x)}-0}{\text{x}-2\text{mx}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(2\text{mx}+\text{h})}{2\text{mx}+\text{h}-2\text{mx}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{h})}{\text{h}}=1$
Here, $\text{LHL}\neq\text{RHL}$ So, function is not differentiable at $\text{x}=2\text{m}\pi,$ where, $\text{m}\in\text{Z}\ \dots(1)$
$[\text{LHL at x}=(2\text{m}+1)\pi]=\lim_\limits{\text{x}\rightarrow(2\text{m}+1)\pi^{-}}\frac{\text{f(x)}-\text{f}[(2\text{m}+1)\pi]}{\text{x}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow(2\text{m}+1)\pi^{-}}\frac{\sin(\text{x})-0}{\text{x}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\sin[(2\text{m}+1)\pi-\text{h}]}{(2\text{m}+1)\pi-\text{h}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\sin(\text{h})}{\text{h}}=-1$
$[\text{RHL at x}=(2\text{m}+1)\pi]=\lim_\limits{\text{x}\rightarrow(2\text{m}+1)\pi^{+}}\frac{\text{f(x)}-\text{f}[(2\text{m}+1)\pi]}{\text{x}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow(2\text{m}+1)\pi^{+}}\frac{-\sin(\text{x})-0}{\text{x}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{-\sin[(2\text{m}+1)\pi+\text{h}]}{(2\text{m}+1)\pi+\text{h}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\sin(\text{h})}{\text{h}}=1$
Here, $\text{LHL}\neq\text{RHL}.$
So, function is not differentiable at $\text{x}=(2\text{m}+1)\pi,$ where, $\text{m}\in\text{Z}\ \dots(2)$
From, (1) and (2), we get
$\text{f(x)}=|\sin\text{x}|$ is not differentiable at $\text{x}=\text{n}\pi$
We know that,
$\cos|\text{x}|=\cos\text{x}$ For all $\text{x}\in\text{R}$
Also we know that cos x is differentiable at all real points.
Therefore, cos |x| is differentiable everywhere.
View full question & answer→Question 4835 Marks
Find a point on the curve $y = x^2 + x$, where the tangent is parallel to the chord joining $(0, 0)$ and $(1, 2)$.
AnswerLet, $f(x) = x^2 + x$ The tangent to the curve is parallel to the chord joining the point (0, 0) and (1, 2). Assume that the chord joins the points (a, f(a)) and (b, f(b)). $\therefore$ a = 0, b = 1 The polynomial function is everywhere continuous and differentiable. So, $f(x) = x^2 + x$ is continuous on [0, 1] and differentiable on (0, 1). Thus, both the conditions of Lagrange's theorem are satisfied. Concequently, there exists $\text{c}\in(0,1)$ such that $\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
Now, $f(x) = x2 + x \Rightarrow f'(x) = 2x + 1, f(1) = 2, f(0) = 0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow2\text{x}+1=\frac{2-0}{1-0}$
$\Rightarrow2\text{x}=1$
$\Rightarrow\text{x}=\frac{1}{2}$Thus, $\text{c}=\frac{1}{2}\in(0,1)$ such that $\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}.$
Clearly,
$\text{f}(\text{c})=\Big(\frac{1}{2}\Big)^2+\frac{1}{2}=\frac{3}{4}.$
Thus, (c, f(c)), i.e. $\Big(\frac{1}{2},\frac{3}{4}\Big),$ is a point on the given curve where the tangent is parallel to the chord joining the points (4, 0) and (5, 1).
View full question & answer→Question 4845 Marks
If $\sin(\text{xy})+\frac{\text{y}}{\text{x}}=\text{x}^2-\text{y}^2,$ find $\frac{\text{dy}}{\text{dx}}$
Answer We have, $\sin(\text{xy})+\frac{\text{y}}{\text{x}}=\text{x}^2-\text{y}^2$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin\text{ xy})+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{\text{d}}{\text{dx}}(\text{x}^2)-\frac{\text{d}}{\text{dx}}(\text{y}^2)$
$\Rightarrow \cos(\text{xy})\frac{\text{d}}{\text{dx}}(\text{xy})+\Bigg\{\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\frac{\text{d}}{\text{dx}}(\text{x})}{\text{x}^2}\Bigg\}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \cos(\text{xy})\Big\{\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big\}+\Bigg\{\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}(1)}{\text{x}^2}\Bigg\}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \cos(\text{xy})\Big\{\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)\Big\}+\frac{1}{\text{x}^2}\Big(\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\Big)=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \text{x}\cos(\text{xy})\frac{\text{dy}}{\text{dx}}+\text{y}\cos(\text{xy})+\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}^2}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\Big\{\text{x}\cos(\text{xy})+\frac{1}{\text{x}}+2\text{y}\Big\}=\frac{\text{y}}{\text{x}^2}-\text{y}\cos(\text{xy})+2\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}^2\cos(\text{xy})+1+2\text{xy}}{\text{x}}\Big\}=\frac{1}{\text{x}^2}\big(\text{y}-\text{x}^2\text{y}\cos(\text{xy})+2\text{x}^2\big)$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{2\text{x}^3+\text{y}-\text{x}^2\text{y}\cos(\text{xy})}{\text{x}\big(\text{x}^2\cos(\text{xy})+1+2\text{xy}\big)}$
View full question & answer→Question 4855 Marks
Differentiate the functions given in Exercise:
$(\log\text{x})^\text{x}+\text{x}^{\log\text{x}}$
AnswerLet $\text{y}=(\log\text{x})^\text{x}+\text{x}^{\log\text{x}}=\text{u}+\text{v }\text{ where u}=(\log\text{x})^\text{x}\text{ and v}=\text{x}^{\log\text{x}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ \dots\text{(i)}$
Now $\text{u}=(\log\text{x})^\text{x}\ \Rightarrow\ \log\text{u}=\log(\log\text{x})^\text{x}=\text{x}\log(\log\text{x})$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{u}=\frac{\text{d}}{\text{dx}}[\text{x}\log(\log\text{x})]\ \Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}[\log(\log\text{x})]+\log(\log\text{x})\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}\log\text{x}+\log(\log\text{x}).1$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{1}{\log\text{x}}\frac{1}{\text{x}}+\log(\log\text{x})\ \Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}\Big[\frac{1}{\log\text{x}}+\log(\log\text{x})\Big]$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=(\log\text{x)}^\text{x}\Big[\frac{1}{\log\text{x}}+\log(\log\text{x})\Big]\ \dots\text{(ii)}$
Again $\text{v}=\text{x}^{\log\text{x}}\ \Rightarrow\ \log\text{v}=\log\text{x}^{\log\text{x}}=\log\text{x}.\log\text{x}=(\log\text{x})^2$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{v}=\frac{\text{d}}{\text{dx}}(\log\text{x})^2\ \Rightarrow\ \frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=2\log\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\ \frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=2\log\text{x}.\frac{1}{\text{x}}\ \Rightarrow\ \frac{\text{dv}}{\text{dx}}=\text{v}\Big(\frac{2}{\text{x}}\log\text{x}\Big)=\text{x}^{\log\text{x}}\frac{2}{\text{x}}\log\text{x}$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=2\text{x}^{\log\text{x}-1}\log\text{x}\ \dots\text{(iii)}$
Putting the values from eq. (ii) and (iii) in eq. (i),
$\frac{\text{dy}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\frac{1}{\log\text{x}}+\log(\log\text{x})\Big]+2\text{x}^{\log\text{x}-1}\log\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\frac{1+\log\text{x}\log(\log\text{x})}{\log\text{x}}\Big]+2\text{x}^{\log\text{x}-1}\log\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=(\log\text{x})^{\text{x}-1}(1+\log\text{x}\log(\log\text{x}))+2\text{x}^{\log\text{x}-1}\log\text{x}$
View full question & answer→Question 4865 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\cos^{-1}\frac{1}{\sqrt{1+\text{t}^2}}\text{ and y}=\sin^{-1}\frac{\text{t}}{\sqrt{1+\text{t}^2}},\text{t}\in\text{R}$
Answer We have, $\text{x}=\cos^{-1}\Big(\frac{1}{\sqrt{1+\text{t}^{2}}}\Big)$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{-1}{\sqrt{1-\Big(\frac{1}{\sqrt{1+\text{t}^{2}}}\Big)^{2}}}\frac{\text{d}}{\text{dt}}\Big(\frac{1}{\sqrt{1+\text{t}}^{2}}\Big)$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{-1}{\sqrt{1-\frac{1}{(1+\text{t}^{2})}}}\left\{\frac{-1}{2(1+\text{t}^{2})^\frac{3}{2}}\right\}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{(1+\text{t}^{2})^{\frac{1}{2}}}{\sqrt{1+\text{t}^{2}-1}}\times\frac{1}{2(1+\text{t}^{2})^\frac{3}{2}}(2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{t}}{\sqrt{\text{t}^{2}}\times(1+\text{t}^{2})}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{1}{1+\text{t}^{2}}...(\text{i})$
Now, $\text{y}=\sin^{-1}\Big(\frac{1}{\sqrt{1+\text{t}^{2}}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{1}{\sqrt{1-\Big(\frac{1}{\sqrt{1+\text{t}^{2}}}\Big)^{2}}}\frac{\text{d}}{\text{dt}}\Big(\frac{1}{\sqrt{1+\text{t}^{2}}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{1}{\sqrt{1-\frac{1}{(1+\text{t}^{2})}}}\left\{\frac{-1}{2(1+\text{t}^{2})\frac{3}{2}}\right\}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{(1+\text{t}^{2})^{\frac{1}{2}}}{\sqrt{1+\text{t}^{2}-1}}\times\frac{-1}{2(1+\text{t}^{2})^{\frac{3}{2}}}(2\text{t})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-1}{2\sqrt{\text{t}^{2}}\times(1+\text{t}^{2})}(2\text{t})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-1}{1+\text{t}^{2}}....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{1}{(1+\text{t}^{2})}\times\frac{(1+\text{t}^{2})}{-1}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-1$
View full question & answer→Question 4875 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\Big(\frac{1-\text{x}^{2\text{n}}}{1+\text{x}^{2\text{n}}}\Big), <\text{x}<\infty$
Answer Let $\text{y}=\cos^{-1}\Big(\frac{1-\text{x}^{2\text{n}}}{1+\text{x}^{2\text{n}}}\Big)$
Put $\text{x}=\tan\theta,\text{ so}$
$\text{y}=\cos^{-1}\bigg(\frac{1-(\text{x}^{\text{n}})^2}{1+(\text{x}^{\text{n}})^2}\bigg)$
$=\cos^{-1}\Big(\frac{1-\tan^2\theta}{1+\tan^2\theta}\Big)$
$\text{y}=\cos^{-1}(\cos2\theta)\ .....(\text{i})$
Here, $0<\text{x}<\infty$
$\Rightarrow 0<\text{x}^{\text{n}}<\infty$
$\Rightarrow 0<\theta<\frac{\pi}{2}$
$\Rightarrow 0<(2\theta)<\pi$
So, from equation (i),
$\text{y}=2\theta\big[\text{Since}, \cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi]\big]$
$\text{y}=2\tan^{-1}\big(\text{x}^{\text{n}}\big)$
Differantiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=2\Big(\frac{1}{1+(\text{x}^{\text{n}})^2}\Big)\frac{\text{d}}{\text{dx}}(\text{x}^{\text{n}})$
$=\frac{2}{1+\text{x}^{2\text{n}}}\times(\text{nx}^{\text{n}-1})$
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{nx}^{\text{n}-1}}{1+\text{x}^{2\text{n}}}$
View full question & answer→Question 4885 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{2^{\text{x}+1}}{1-4^{\text{x}}}\Big),-\infty<\text{x}<0$
Answer Let $\text{y}=\tan^{-1}\Big\{\frac{2^{\text{x}+1}}{1-4^{\text{x}}}\Big\}$
Put $2\text{x}=\tan\theta,\text{ so}$
$\text{y}=\tan^{-1}\Big\{\frac{2^{\text{x}}\times2}{1-(2^\text{x})^2}\Big\}$
$=\tan^{-1}\Big\{\frac{2\tan\theta}{1-\tan^2\theta}\Big\}$
$\text{y}=\tan^{-1}\big\{\tan(2\theta)\big\}\ .....(\text{i})$
Here, $-\infty<\text{x}<0$
$\Rightarrow 2^{-\infty}<2^{\text{x}}<2^{0}$
$\Rightarrow 0<2^{\text{x}}<1$
$\Rightarrow 0< \theta< \frac{\pi}{4}$
$\Rightarrow 0 < (2\theta) <\frac{\pi}{2}$
From equation (i),
$\text{y}=2\theta \Big[\text{Since}, \tan^{-1}(\tan\theta)=\theta, \text{ if }\theta\in \Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\text{y}=2\tan^{-1}(2^\text{x})$
Differentiate it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{2}{1+(2^\text{x})^2}\frac{\text{d}}{\text{dx}}(2^\text{x})$
$=\frac{2\times2^\text{x}\log2}{1+4^{\text{x}}}$
$\frac{\text{dy}}{\text{dx}}=\frac{2^{\text{x}+1}\log2}{1+4^{\text{x}}}$
View full question & answer→Question 4895 Marks
Find the derivative of the function given by $f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$ and hence find $f′(1).$
AnswerGiven: $\text{f(x)}=(1+\text{x})(1+\text{x}^2)(1+\text{x}^4)(1+\text{x}^8)\ \dots\text{(i)}$
$\Rightarrow\ \log\text{f(x)}=\log\big\{(1+\text{x})(1+\text{x}^2)(1+\text{x}^4)(1+\text{x}^8)\big\}$
$\Rightarrow\ \log\text{f(x)}=\log(1+\text{x})+\log(1+\text{x}^2)+\log(1+\text{x}^4)+\log(1+\text{x}^8)$
$\Rightarrow\ \frac{\text{1}}{\text{f(x)}}\frac{\text{d}}{\text{dx}}\text{f(x)}=\frac{\text{1}}{1+\text{x}}\frac{\text{d}}{\text{dx}}(1+\text{x})+\frac{\text{1}}{\text{1+x}^2}\frac{\text{d}}{\text{dx}}(1+\text{x}^2)+\frac{\text{1}}{\text{1+x}^4}\frac{\text{d}}{\text{dx}}(1+\text{x}^4)+\frac{\text{1}}{\text{1+x}^8}\frac{\text{d}}{\text{dx}}(1+\text{x}^8)$
$\Rightarrow\ \frac{1}{\text{f(x)}}\text{f}'\text{(x)}=\frac{1}{1+\text{x}}.1+\frac{1}{1+\text{x}^2}.2\text{x}+\frac{1}{1+\text{x}^4}.4\text{x}^3+\frac{1}{1+\text{x}^8}8\text{x}^7$
$\Rightarrow\ \text{f}'\text{(x)}=\text{f}\text{(x)}\Big[\frac{1}{1+\text{x}}+\frac{2\text{x}}{1+\text{x}^2}+\frac{4\text{x}^3}{1+\text{x}^4}+\frac{8\text{x}^7}{1+\text{x}^8}\Big]$
Putting the value of f(x) from eq. (i),
$\text{f}'\text{(x)}=(1+\text{x})(1+\text{x}^2)(1+\text{x}^4)(1+\text{x}^8)\Big[\frac{1}{1+\text{x}}+\frac{2\text{x}}{1+\text{x}^2}+\frac{4\text{x}^3}{1+\text{x}^4}+\frac{8\text{x}^7}{1+\text{x}^8}\Big]$
$\Rightarrow\ \text{f}'\text{(x)}=(1+1)(1+1^2)(1+1^4)(1+1^8)\Big[\frac{1}{1+1}+\frac{2\times1}{1+1^2}+\frac{4\times1^3}{1+1^4}+\frac{8\times1^7}{1+1^8}\Big]$
$\Rightarrow\ \text{f}'(1)=(2)(2)(2)(2)\Big[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\Big]$
$\Rightarrow\ \text{f}'(1)=16\Big[\frac{15}{2}\Big]=8\times15=120$
View full question & answer→Question 4905 Marks
If $\text{y}=\frac{1}{2}\log\Big(\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\Big),$ Prvoe that $\frac{\text{dy}}{\text{dx}}=2\text{ cosec }2\text{x}$
Answer Given, $\text{y}=\frac{1}{2}\log\Big(\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\Big)$
$\Rightarrow\text{y}=\frac{1}{2}\log\Big(\frac{2\sin^2\text{x}}{2\cos^2\text{x}}\Big)$
$\begin{bmatrix} \text{Since}, 1-\cos2\text{x}=2\sin^2\text{x}, \\ 1+\cos2\text{x}=2\cos^2\text{x} \end{bmatrix}$
$\Rightarrow\text{y}=\frac{1}{2}\log\big(\tan^2\text{x}\big)$
$\Rightarrow\text{y}=\frac{2}{2}\log\tan\text{x}$
$\big[\text{Since}, \log\text{a}^{\text{b}}=\text{b}\log\text{a}\big]$
$\Rightarrow\text{y}=\log\tan\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=(\log\tan\text{x})$
$=\frac{1}{\tan\text{x}}\times\frac{\text{d}}{\text{dx}}(\tan\text{x})$
[Using chain rule]
$=\frac{\sec^2\text{x}}{\tan\text{x}}$
$=\frac{1}{\cos^2\text{x}\times\frac{\sin\text{x}}{\cos\text{x}}}$
$=\frac{1}{\sin\text{x}\cos\text{x}}$
$=\frac{2}{2\sin\text{x}\cos\text{x}}$
$=\frac{2}{\sin2\text{x}}\Big[\text{Since},\frac{1}{\sin\text{x}=\text{cosec x}}\Big]$
So,
$\frac{\text{dy}}{\text{dx}}=2\text{ cosec }2\text{x}$
View full question & answer→Question 4915 Marks
If $\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)+\sec^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big), 0<\text{x}<1$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2}$
Answer Let $\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)+\sec^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
$\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
Put $\text{x}=\tan\theta$
$\text{y}=\sin^{-1}\Big(\frac{2\tan\theta}{1+\tan^2\theta}\Big)+\cos^{-1}\Big(\frac{1-\tan^2\theta}{1+\tan^2\theta}\Big)$
$\text{y}=\sin^{-1}(\sin2\theta)+\cos^{-1}(\cos2\theta)\ .....(1)$
Here, $0<\text{x}<1$
$\Rightarrow 0<\tan\theta<1$
$\Rightarrow 0<\theta<\frac{\pi}{4}$
$\Rightarrow0<(2\theta)<\frac{\pi}{2}$
So, from equation (i),
$\text{y}=2\theta+2\theta$
$\begin{bmatrix} \text{Since}, \sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in \Big[-\frac{\pi}{2},\frac{\pi}{2}\Big] \\ \cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi] \end{bmatrix}$
$\text{y}=4\tan^{-1}\text{x}\ [\text{Since, x}=\tan\theta]$
Diffrentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2}$
View full question & answer→Question 4925 Marks
Discuss the continuity and differentiability of,$\text{f(x)}=\begin{cases}(\text{x}-\text{c})\cos\Big(\frac{1}{\text{x}-\text{c}}\Big), & \text{x}\neq 0\\0, & \text{x}= 0\end{cases}$
Answer $\text{f(x)}=\begin{cases}(\text{x}-\text{c})\cos\Big(\frac{1}{\text{x}-\text{c}}\Big), & \text{x}\neq 0\\0, & \text{x}= 0\end{cases}$
(LHL at x = c) $=\lim_\limits{\text{x}\rightarrow\text{c}^{-}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(\text{c}-\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}-\text{h}-\text{c}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}-\text{h}\cos\Big(-\frac{1}{\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}-\text{h}\cos\Big(\frac{1}{\text{h}}\Big)$
$=0$
(RHL at x = c) $=\lim_\limits{\text{x}\rightarrow\text{c}^{+}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{c}+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(\text{c}+\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}+\text{h}-\text{c}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}-\text{h}\cos\Big(\frac{1}{\text{h}}\Big)$
$=0$
f(c) = 0
Since, LHL = f(x) = RHL at x = c
⇒ f(x) is continuous at x = c
(LHL at x = c) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{c}-\text{h})-\text{f(c)}}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{c}-\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}-\text{h}-\text{c}}\Big)-0}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\cos\Big(-\frac{1}{\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}\cos\Big(\frac{1}{\text{h}}\Big)$
(RHL at x = c) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{c}+\text{h})-\text{f(c)}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{c}+\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}+\text{h}-\text{c}}\Big)-0}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}\cos\Big(\frac{1}{\text{h}}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\cos\Big(\frac{1}{\text{h}}\Big)$
(LHL at x = c) = (RHL at x = c)
So,
f(x) is differentiable and continuous at x = c.
View full question & answer→Question 4935 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\big(1-2\text{x}^2\big),0<\text{x}<1$
Answer Let $\text{y}=\sin^{-1}\big\{1-2\text{x}^2\big\}$
Let $\text{x}=\sin\theta,\text{ So},$
$\text{y}=\sin^{-1}\big(1-2\sin^2\theta\big)$
$=\sin^{-1}(\cos2\theta)$
$\text{y}=\sin^{-1}\Big\{\sin\Big(\frac{\pi}{2}-2\theta\Big)\Big\}\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow 0<\sin\theta<1$
$\Rightarrow 0<\theta<\frac{\pi}{2}$
$\Rightarrow 0<2\theta<\pi$
$\Rightarrow 0> -2\theta>-\pi$
$\Rightarrow\ \frac{\pi}{2}>\Big(\frac{\pi}{2}-2\theta\Big)>\frac{\pi}{2}-\pi$
$\Rightarrow\ \frac{\pi}{2}>\Big(\frac{\pi}{2}-2\theta\Big)>\Big(-\frac{\pi}{2}\Big)$
So, from equatuion (i),
$\text{y}=\frac{\pi}{2}-2\theta$
$\Big[\text{Since}, \sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}\frac{\pi}{2}-2\sin^{-1}\text{x}\big[\text{Since x}=\sin\theta\big]$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=0-2\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\frac{\text{dy}}{\text{dx}}=-\frac{2}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 4945 Marks
If $\text{y}=\sin^{-1}\Big(\frac{\text{x}}{1+\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big), 0<\text{x}<\infty$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2}$
Answer Let $\text{y}=\sin^{-1}\Big(\frac{\text{x}}{\sqrt{1+\text{x}^2}}\Big)+\cos^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big)$
Put $\text{x}=\tan\theta$
$\therefore\text{y}=\sin^{-1}\Big(\frac{\tan\theta}{\sqrt{1+\tan^2\theta}}\Big)+\cos^{-1}\Big(\frac{1}{\sqrt{1+\tan^2\theta}}\Big)$
$\Rightarrow \text{y}=\sin^{-1}\bigg(\frac{\frac{\sin\theta}{\cos\theta}}{\sec\theta}\bigg)+\cos^{-1}\Big(\frac{1}{\sec\theta}\Big)$
$\Rightarrow\text{y}=\sin^{-1}\bigg(\frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos\theta}}\bigg)+\cos^{-1}(\cos\theta)$
$\Rightarrow\text{y}=\sin^{-1}(\sin\theta)+\cos^{-1}(\cos\theta)\ .....(\text{i})$
Here, $0<\text{x}<\infty$
$\Rightarrow 0<\tan\theta<\infty$
$\Rightarrow 0 <\theta<\frac{\pi}{2}$
So, from equation (i),
$\text{y}=\theta+\theta$
$\begin{bmatrix} \text{Since}, \sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in \Big[-\frac{\pi}{2},\frac{\pi}{2}\Big] \\ \cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi] \end{bmatrix}$
$\Rightarrow\text{y}=2\theta$
$\Rightarrow\text{y}=2\tan^{-1}\text{x}\ \big[\text{Since, x}=\tan\theta\big]$
Differentiate it with respect to x,
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2}{1+\text{x}^2}$
View full question & answer→Question 4955 Marks
If $x^y + y^x = (x + y)^{x+y}$, find $\frac{\text{dy}}{\text{dx}}$
AnswerHere,
$x^y + y^x = (x + y)^{x+y}$
$\text{e}^{\log\text{x}^\text{y}}+\text{e}^{\log\text{y}^\text{x}}=\text{e}^{\log(\text{x}+\text{y})^{(\text{x}+\text{y})}}$
$\text{e}^{\text{y}\log\text{x}}+\text{e}^{\text{x}\log\text{y}}=\text{e}^{{(\text{x}+\text{y})}\log(\text{x}+\text{y})}$
$\big[\text{Since},\text{e}^{\log\text{a}}=\text{a},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule, product rule,
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{y}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{y}}\big)=\frac{\text{d}}{\text{dx}}^{(\text{x}+\text{y})\log(\text{x}+\text{y})}$
$\Rightarrow\text{e}^{\text{y}\log\text{x}}\Big[\text{y}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big] \\ +\text{e}^{\text{x}\log\text{y}}\Big[\text{x}\frac{\text{d}}{\text{dx}}\log\text{y}+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ =\text{e}^{(\text{x}+\text{y})\log(\text{x}+\text{y})}\frac{\text{d}}{\text{dx}}\big[(\text{x}+\text{y})\log(\text{x}+\text{y})\big]$
$\Rightarrow\text{e}^{\text{y}\log\text{x}}\Big[\text{y}\big(\frac{1}{\text{x}}\big)+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\text{e}^{\log\text{x}}\Big[\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}(1)\Big] \\ =\text{e}^{\log(\text{x}+\text{y})^{(\text{x}+\text{y})}}\Big[(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})+\log(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})\Big]$
$\Rightarrow\text{x}^\text{y}\Big[\frac{\text{y}}{\text{x}}+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\text{y}^\text{x}\Big[\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big] \\ =(\text{x}+\text{y})^{(\text{x}+\text{y})}\Big[(\text{x}+\text{y})\frac{1}{(\text{x}+\text{y})}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})+\log(\text{x}+\text{y})\Big(1+\frac{\text{dy}}{\text{dx}}\Big)\Big]$
$\Rightarrow\text{x}^\text{y}\times\frac{\text{y}}{\text{x}}+\text{x}^{\text{y}}\log\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}^\text{x}\times\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\text{y}^\text{x}\log\text{y} \\ =(\text{x}+\text{y})^{(\text{x}+\text{y})}\Big[1\times\Big(1+\frac{\text{dy}}{\text{dx}}\Big)+\log(\text{x}+\text{y})\Big(1+\frac{\text{dy}}{\text{dx}}\Big)\Big]$
$\Rightarrow\text{x}^{\text{y}-1}\times\text{y}+\text{x}^\text{y}\log\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}^{\text{x}-1}\times\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}^{\text{x}}\log\text{y} \\ =(\text{x}+\text{y})^{(\text{x}+\text{y})}+(\text{x}+\text{y})^{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}} \\ +(\text{x}+\text{y})^{(\text{x}+\text{y})}\log(\text{x}+\text{y})+(\text{x}+\text{y})^{(\text{x}+\text{y})}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big[\text{x}^{\text{y}}\log\text{x}+\text{xy}^{\text{x}-1}-(\text{x}+\text{y})^{\text{x}+\text{y}}(1+\log(\text{x}+\text{y}))\Big] \\ =(\text{x}+\text{y})^{\text{x}+\text{y}}(1+\log(\text{x}+\text{y}))-\text{x}^{\text{y}-1}\times\text{y}-\text{y}^{\text{x}}\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\bigg[\frac{(\text{x}+\text{y})^{\text{x}+\text{y}}(1+\log(\text{x}+\text{y}))-\text{x}^{\text{y}-1}\times\text{y}-\text{y}^\text{x}\log\text{y}}{\text{x}^\text{y}\log\text{x}+\text{xy}^{\text{x}-1}-(\text{x}+\text{y})^{\text{x}+\text{y}}(1+\log(\text{x}+\text{y}))}\bigg]$
View full question & answer→Question 4965 Marks
Differentiate the following functions from first principles:
$\text{e}^{\cos\text{x}}$
Answer Let $\text{f(x)}=\text{e}^{\cos\text{x}}$
$\Rightarrow\text{f}(\text{x}+\text{h})=\text{e}^{\cos(\text{x}+\text{h})}$
$\therefore\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\cos(\text{x}+\text{h})}-\text{e}^{\cos\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{e}^{\cos\text{x}}\Big[\frac{\text{e}^{\cos(\text{x}+\text{h})-\cos\text{x}}-1}{\text{h}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{ e}^{\cos\text{x}}\Big[\frac{\text{e}^{\cos(\text{x}+\text{h})-\cos\text{x}}-1}{\cos(\text{x}+\text{h})-\cos\text{x}}\Big]\times\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{ e}^{\cos\text{x}}\times\Big(\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\text{h}}\Big)$
$\Big[\text{Since},\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\ \text{e}^{\cos\text{x}}\times\bigg(\frac{-2\sin\frac{\text{x}+\text{h}+\text{x}}{2}\times\sin\frac{\text{x}+\text{h}-\text{x}}{2}}{\text{h}}\bigg)$
$\begin{bmatrix} \cos\text{A}-\cos\text{B}=-2\sin\frac{\text{A}+\text{B}}{2}\sin\frac{\text{A}-\text{B}}{2} \end{bmatrix}$
$=\text{e}^{\cos\text{x}}\lim\limits_{\text{h}\rightarrow0}\frac{-\sin\Big(\frac{2\text{x}+\text{h}}{2}\Big)}{2}\times\frac{\sin\Big(\frac{\text{h}}{2}\Big)}{\frac{\text{h}}{2}}$
$=\text{e}^{\cos\text{x}}\lim\limits-2\sin\Big(\frac{2\text{x}+\text{h}}{2}\Big)\times\frac{1}{2}$
$\Big[\text{Since},\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=\text{e}^{\cos\text{x}}(-\sin\text{x})$
$=-\sin\text{xe}^{\cos\text{x}}$
Hence,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cos\text{x}}\big)=-\sin\text{xe}^{\cos\text{x}}$
View full question & answer→Question 4975 Marks
Differentiate $\sin^{-1}\sqrt{1-\text{x}^2}$ with respect to $\cot^2\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big),$ if 0 < x < 1.
Answer Let $\text{u}=\sin^{-1}\Big(\sqrt{1-\text{x}^2}\Big)$
Put $\text{x}=\cos\theta\Rightarrow\theta=\cos^{-1}\text{x},\text{ so}$
$\text{u}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
And,
Let $\text{v}=\cot^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)$
$=\cot^{-1}\Big(\frac{\cot\theta}{\sqrt{1-\cos^2\theta}}\Big)$
$=\cot^{-1}\Big(\frac{\cos\theta}{\sin\theta}\Big)$
$\text{v}=\cot^{-1}(\cot\theta)\ .....(\text{ii})$
Here, $0<\text{x}<1$
$\Rightarrow0<\cos\theta<1$
$\Rightarrow0<\theta<\frac{\pi}{2}$
So, from equation (i),
$\text{u}=\theta\Big[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{u}=\cos^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\theta\big[\text{Since,}\cot^{-1}(\cot\theta)=\theta,\text{if }\theta\in(0,\pi) \big]$
$\text{v}=\cos^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-1}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{-1}$
$\frac{\text{du}}{\text{dv}}=1$
View full question & answer→Question 4985 Marks
If $x^x + y^x = 1$, prove that $\frac{\text{dy}}{\text{dx}}=-\Big\{\frac{\text{x}^\text{x}(1+\log\text{x})+\text{y}^\text{x}\times\log\text{y}}{\text{x}\times\text{y}^{\text{x}-1}}\Big\}$
AnswerHere,
$x^x + y^x = 1$
$\text{e}^{\log\text{x}^\text{x}}+\text{e}^{\log\text{y}^\text{x}}=1$
$\text{e}^{\text{x}\log\text{x}}+\text{e}^{\text{x}\log\text{y}}=1$
$\big[\text{Since},\text{e}^{\log\text{a}}=\text{a}.\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule and chain rule,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{y}}\big)=\frac{\text{d}}{\text{dx}}(1)$
$\text{e}^{\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})+\text{e}^{\text{x}\log\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=0$
$\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\text{e}^{\log\text{y}^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]=0$
$\text{x}^\text{x}\Big[\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)\Big]+\text{y}^\text{x}\Big[\text{x}\Big(\frac{1}{\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\log\text{y}(1)\Big]=0$
$\text{x}^\text{x}[1+\log\text{x}]+\text{y}^\text{x}\Big(\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big)=0$
$\text{y}^\text{x}\times\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}=-\big[\text{x}^\text{x}(1+\log\text{x})+\text{y}^\text{x}\log\text{y}\big]$
$\big(\text{xy}^{\text{x}-1}\big)\frac{\text{dy}}{\text{dx}}=-\big[\text{x}^\text{x}(1+\log\text{x})+\text{y}^\text{x}\log\text{y}\big]$
$\frac{\text{dy}}{\text{dx}}=-\Big[\frac{\text{x}^\text{x}(1+\log\text{x})+\text{y}^\text{x}\log\text{y}}{\text{xy}^{\text{x}-1}}\Big]$
View full question & answer→Question 4995 Marks
If $\text{y}=\text{a}\{\text{x}+\sqrt{\text{x}^2+1}\}^\text{n}+\text{b}\{\text{x}-\sqrt{\text{x}^2+1}\}^{-\text{n},}$ prove that $(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}-\text{n}^2\text{y}=0.$
Answer We have
$\text{y}=\text{a}\{\text{x}+\sqrt{\text{x}^2+1}\}^\text{n}+\text{b}\{\text{x}-\sqrt{\text{x}^2+1}\}^{-\text{n},}$
Differentiating y with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{an}\{\text{x}+\sqrt{\text{x}^2+1}\}^{\text{n}-1}\Big(1+\frac{1}{2\sqrt{\text{x}^2+1}\times2\text{x}}\Big)-\text{bn}\{\text{x}-\sqrt{\text{x}^2+1}\}^{-\text{n}-1}\Big(1-\frac{1}{2\sqrt{\text{x}^2+1}\times2\text{x}}\Big)$
$=\text{an}\{\text{x}+\sqrt{\text{x}^2+1}\}^{\text{n}-1}\Big(1+\frac{\text{x}}{\sqrt{\text{x}^2+1}}\Big)-\text{bn}\{\text{x}-\sqrt{\text{x}^2+1}\}^{-\text{n}-1}\Big(1-\frac{\text{x}}{\sqrt{\text{x}^2+1}}\Big)$
$=\text{an}\{\text{x}+\sqrt{\text{x}^2+1}\}^{\text{n}-1}\Big(\frac{\sqrt{\text{x}^2+1}+\text{x}}{\sqrt{\text{x}^2+1}}\Big)-\text{bn}\{\text{x}-\sqrt{\text{x}^2+1}\}^{\text{n}-1}\Big(\frac{\sqrt{\text{x}^2+1}-\text{x}}{\sqrt{\text{x}^2+1}}\Big)$
$=\text{an}\{\text{x}+\sqrt{\text{x}^2+1}\}^{\text{n}-1}\Big(\frac{\text{x}+\sqrt{\text{x}^2+1}}{\sqrt{\text{x}^2+1}}\Big)+\text{bn}\{\text{x}-\sqrt{\text{x}^2+1}\}^{-\text{n}-1}\Big(\frac{\text{x}\sqrt{\text{x}^2+1}}{\sqrt{\text{x}^2+1}}\Big)$
$=\Big\{\text{a}\{\text{x}+\sqrt{\text{x}^2+1}\}^\text{n}\Big(\frac{\text{n}}{\sqrt{\text{x}^2+1}}\Big)+\text{b}\{\text{x}-\sqrt{\text{x}^2+1}\Big\}^{-\text{n}}\Big(\frac{\text{n}}{\sqrt{\text{x}^2+1}}\Big)$
$=\Big(\frac{\text{n}}{\sqrt{\text{x}^2+1}}\Big)\text{y}$
$\Rightarrow\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}}=\text{ny}$
Squaring both sides, we get
$(\text{x}^2+1)\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{n}^2\text{y}^2...(2)$
Differentiating (2) with respect to x , we get
$(\text{x}^2+1)2\frac{\text{dy}}{\text{dx}}\times\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{n}^2\Big(2\text{y}\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow(\text{x}^2+1)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{n}^2(\text{y})$
$\Rightarrow(\text{x}^2+1)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{n}^2\text{y}=0$
Hence, $(\text{x}^2+1)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{n}^2\text{y}=0$
View full question & answer→Question 5005 Marks
If $\text{x}=\text{a}(\cos\text{t}+\log\tan\frac{\text{t}}{2})\ \text{and}\ \text{y}=\text{a}(\sin\text{t}),$evaluate $\frac{\text{d}^2\text{y}}{\text{dx}^2}\ \text{at}\ \text{t}=\frac{\pi}{3}.$
Answer We have,
$\text{x}=\text{a}(\cos\text{t}+\log\tan\frac{\text{t}}{2})\ \text{and}\ \text{y}=\text{a}(\sin\text{t}),$
On differentiating with respect to t, we get
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\text{a}\Big(\cos\text{t}+\log\tan\frac{\text{t}}{2}\Big)\Big]=\text{a}\Bigg(-\sin\text{t}+\frac{1}{\tan\frac{\text{t}}{2}}\times\sec^2\frac{\text{t}}{2}\times\frac{1}{2}\Bigg)$
$=\text{a}\Bigg(-\sin\text{t}+\frac{1}{2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}\Bigg)=\text{a}\Big(-\sin\text{t}+\frac{1}{\sin\text{t}}\Big)$
$=\text{a}\Big(\frac{-\sin^2\text{t}+1}{\sin\text{t}}\Big)=\text{a}\Big(\frac{\cos^2\text{t}}{\sin\text{t}}\Big)$
And
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{a}\sin\text{t})=\text{a}\cos\text{t}$
Now, $\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dX}}{\text{dt}}}=\frac{\text{a}\cos\text{t}}{\text{a}\frac{\cos^2\text{t}}{\sin\text{t}}}=\tan\text{t}$
Therefore
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}(\tan(\text{t}))$
$=\frac{\text{d}}{\text{dt}}(\tan(\text{t}))\times\frac{\text{dt}}{\text{dx}}=\sec^2\text{t}\times\frac{\sin\text{t}}{\text{a}\cos^2\text{t}}$
$=\Big(\frac{\sin\text{t}}{\text{a}\cos^4\text{t}}\Big)$
$\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)_{\text{t}=\frac{\pi}{3}}=\Bigg(\frac{\sin\Big(\frac{\pi}{3}\Big)}{\text{a}\cos^4\Big(\frac{\pi}{3}\Big)}\Bigg)=\frac{\frac{\sqrt{3}}{2}}{\text{a}\Big(\frac{1}{16}\Big)}=\frac{8\sqrt{3}}{\text{a}}$
Hence, $\text{at}\ \text{t}=\frac{\pi}{3},\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{8\sqrt{3}}{\text{a}}$
View full question & answer→