Question 5015 Marks
Differentiate the following functions from first principles:
$\log\cos\text{x}$
Answer Let $\text{f(x)} = \log \cos \text{x}$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\log\cos(\text{x}+\text{h})$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})=\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\cos(\text{x}+\text{h})-\log\cos\text{x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log^{\log(\text{x}+\text{h})}_{\cos\text{x}}}{\text{h}}\ \Big[\because\ \log\text{A}-\log\text{B}=\log\Big(\frac{\text{A}}{\text{B}}\Big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\Big[1+\left\{\frac{\cos(\text{z}+\text{h})}{\cos\text{z}}-1\right\}\Big]}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\left\{1+\frac{\cos(\text{x}+\text{h})-\cos\text{z}}{\cos\text{z}}\right\}}{\left\{\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\cos\text{x}}\right\}}\times\lim\limits_{\text{h}\rightarrow0}\left\{\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\cos\text{x}}\right\}$
$=1\times\lim\limits_{\text{h}\rightarrow0}\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\cos\text{x}\times\text{h}}\ \Big[\because\lim\limits_{\text{h}\rightarrow0}\frac{\log(1+\text{x})}{\text{x}}=1\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-2\sin\Big(\frac{\text{x}+\text{h}+\text{x}}{2}\Big)\sin\Big(\frac{\text{x}+\text{h}+\text{x}}{2}\Big)}{\cos\text{x}\times\text{h}}$
$=-2\lim\limits_{\text{h}\rightarrow0}\frac{\sin\Big(\frac{2\text{x}+\text{h}}{2}\Big)\times\sin\big(\frac{\text{h}}{2}\big)}{2\cos\text{x}\times\big(\frac{\text{x}}{2}\big)}$
$=\frac{-2\sin\text{x}}{2\cos\text{x}}\Big[\because \lim\limits\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=-\tan\text{x}$
So,
$\frac{\text{d}}{\text{dx}}(\log\cos\text{x})=-\tan\text{x}$
View full question & answer→Question 5025 Marks
If $\text{y}^\text{x}=\text{e}^{\text{x}-\text{e}},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{y})^2}{\log\text{y}}$
Answer We have, $\text{y}^\text{x}=\text{e}^{\text{y}-\text{x}},$
$\Rightarrow\ \log\text{y}^\text{x}=\log^{\text{y}-\text{x}}$
$\Rightarrow\ \text{x}\log\text{y}=\text{y}-\text{x}\log_\text{e}=(\text{y}-\text{x})$ $[\because\log_\text{e}=1]$
$\Rightarrow\ \log\text{y}=\frac{(\text{y}-\text{x})}{\text{x}}\ \ \dots(\text{i})$
Now, differentiating w.r.t. x, we get
$\frac{\text{d}}{\text{dy}}\log\text{y}\cdot\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\frac{(\text{y}-\text{x})}{\text{x}}$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\frac{\text{x}\cdot\frac{\text{d}}{\text{dx}}(\text{y}-\text{x})-(\text{y}-\text{x})\cdot\frac{\text{d}}{\text{dx}}\text{x}}{\text{x}^2}$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}\Big(\frac{\text{dy}}{\text{dx}}-1\Big)-(\text{y}-\text{x})}{\text{x}^2}$
$\Rightarrow\ \frac{\text{x}^2}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}-\text{x}-\text{y}+\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\Big(\frac{\text{x}^2}{\text{y}}-\text{x}\Big)=-\text{y}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{y}^2}{\text{x}^2-\text{xy}}=\frac{-\text{y}^2}{\text{x}(\text{x}-\text{y})}$
$=\frac{\text{y}^2}{\text{x}(\text{y}-\text{x})}\cdot\frac{\text{x}}{\text{x}}=\frac{\text{y}^2}{\text{x}^2}\cdot\frac{1}{\frac{(\text{y}-\text{x})}{\text{x}}}$
$=\frac{(1+\log\text{y})^2}{\log\text{y}}$ $\Big[\because\log\text{y}=\frac{\text{y}-\text{x}}{\text{x}}\log\text{y}=\frac{\text{y}}{\text{x}}-1\Rightarrow1+\log\text{y}=\frac{\text{y}}{\text{x}}\Big]$
Hence proved.
View full question & answer→Question 5035 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\Big(\frac{\text{x}+\sqrt{1-\text{x}^2}}{\sqrt{2}}\Big),-1<\text{x}<1$
Answer Let $\text{y}=\cos^{-1}\Big(\frac{\text{x}+\sqrt{1-\text{x}^2}}{\sqrt{2}}\Big)$
Put $\text{x}=\cos\theta$
$\text{y}=\sin^{-1}\Big\{\frac{\cos\theta+\sqrt{1-\cos^2\theta}}{\sqrt{2}}\Big\}$
$\text{y}=\cos^{-1}\Big\{\frac{\cos\theta+\sin\theta}{\sqrt{2}}\Big\}$
$\text{y}=\cos^{-1}\Big\{\cos\theta\Big(\frac{1}{\sqrt{2}}\Big)+\sin\theta\Big(\frac{1}{\sqrt{2}}\Big)\Big\}$
$\text{y}=\cos^{-1}\Big\{\cos\theta\cos\frac{\pi}{2}+\sin\theta\sin\frac{\pi}{2}\Big\}$
$\text{y}=\cos^{-1}\Big\{\cos\Big(\theta-\frac{\pi}{4}\Big)\Big\}\ .....(\text{i})$
Here, $-1<\text{x}<1$
$\Rightarrow -1<\cos\theta<1$
$\Rightarrow\frac{3\pi}{4}<\theta<\frac{5\pi}{4}$
$\Rightarrow\Big(\frac{3\pi}{4}-\frac{\pi}{4}\Big)<\Big(\theta-\frac{\pi}{4}\Big)<\frac{5\pi}{4}-\frac{\pi}{4}$
$\Rightarrow\Big(\frac{\pi}{4}\Big)<\Big(\theta-\frac{\pi}{4}\Big)<\pi$
So, from equation (i),
$\text{y}=\Big(\theta-\frac{\pi}{4}\Big)\ \big[\text{Since}, \cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi]\big]$
$\text{y}=\cos^{-1}\text{x}-\frac{\pi}{4}\big[\text{Since, x}=\sin\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\sqrt{1-\text{x}^2}}+0$
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 5045 Marks
Differentiate $(\cos\text{x})^{\sin\text{x}}$ with respect to $(\sin\text{x})^{\cos\text{x}}$
Answer Let, $\text{u} = (\cos)^{\sin\text{x}}$
Taking log on both sides,
$\log\text{u} = \log(\cos\text{x})^{\sin \text{x}}$
$\Rightarrow \log\text{u} = \sin \text{x}\log(\cos\text{x})$
Differentiating it with respect to x using rule,
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\sin \text{x}\frac{\text{d}}{\text{dx}}(\log\cos\text{x})+\log \cos \text{x}\frac{\text{d}}{\text{dx}}(\sin\text{x})$
[using product rule]
$\Rightarrow\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\sin \text{x}\big(\frac{1}{\cos\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}(\cos\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}[(\tan\text{x})\times(-\sin\text{x})+\log\log\text{x}(\cos\text{x})]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\cos\text{x})^{\sin\text{x}}[\cos\text{x}\log\cos\text{x}-\sin\text{x}\tan\text{x}]\ .....\text{(i)} $
Let, $\text{v = }(\sin\text{x})^{\cos\text{x}}$
Taking log on both sides,
$\log\text{v}=\log(\sin\text{x})^{\cos\text{x}}$
$\Rightarrow\log\text{v}=\cos\text{x}\log(\sin\text{x})$
Differentiating it with respect to x using chain rule,
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\cos\text{x}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x}) $
[using product rule]
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\cos\text{x}\big(\frac{1}{\sin\text{x}}\big)\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}(-\sin\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}[\cot\text{x}(\cos \text{x})-\sin\text{x}\log\sin\text{x}]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\sin\text{x})^{\cos\text{x}}[\cot \text{x} (\cos \text{x})-\sin \text{x}\log\sin\text{x}]\ ...(\text{ii})$
Dividing equation (i) by (ii)
$\therefore\frac{\text{du}}{\text{dv}}=\frac{(\cos\text{x})^{\sin\text{x}}[\cos \text{x}\log\cos\text{x}-\sin\text{x}\tan\text{x ]}}{(\sin\text{x})^{\cot\text{x}}[\cot \text{x}(\cos\text{x})-\sin\text{x}\log\sin\text{x}]}$
View full question & answer→Question 5055 Marks
Differentiate the following functions with respect to x:
$\log\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}$
Answer Let, $\text{y}=\log\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}$
$\Rightarrow\ \text{y}=\log\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)^\frac{1}{2}$
$\Rightarrow\ \text{y}=\frac{1}{2}\log\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big) \big[\text{Using }\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big\{\frac{1}{2}\log\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)\Big\}$
$=\frac{1}{2}\times\frac{1}{\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)}\times\frac{\text{d}}{\text{dx}}\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)$
[Using chain rule]
$=\frac{1}{2}\Big(\frac{1+\cos\text{x}}{1-\cos\text{x}}\Big)\bigg[\frac{(1+\cos\text{x})\frac{\text{d}}{\text{dx}}(1-\cos\text{x})-(1-\cos\text{x})\frac{\text{d}}{\text{dx}}(1+\cos\text{x})}{(1+\cos\text{x})^2}\bigg]$
$=\frac{1}{2}\Big(\frac{1+\cos\text{x}}{1-\cos\text{x}}\Big)\bigg[\frac{(1+\cos\text{x})(\sin\text{x})-(1-\cos\text{x})(-\sin\text{x})}{(1+\cos\text{x})^2}\bigg]$
$=\frac{1}{2}\Big(\frac{1+\cos\text{x}}{1-\cos\text{x}}\Big)\Big[\frac{\sin\text{x}+\sin\text{x}\cos\text{x}+\sin\text{x}-\sin\text{x}\cos\text{x}}{(1+\cos\text{x})^2}\Big]$
$=\frac{1}{2}\Big(\frac{1+\cos\text{x}}{1-\cos\text{x}}\Big)\Big[\frac{2\sin\text{x}}{(1+\cos\text{x})^2}\Big]$
$=\frac{\sin\text{x}}{(1-\cos\text{x})(1+\cos\text{x})}$
$=\frac{\sin\text{x}}{1-\cos^2\text{x}}$
$=\frac{\sin\text{x}}{\sin^2\text{x}}$
$=\frac{1}{\sin\text{x}}$
$=\text{cosec x}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\log\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}\Big)=\text{cosec x}$
View full question & answer→Question 5065 Marks
Show that the function $\text{f(x)}=\begin{cases}|2\text{x}-3||\text{x}|, & \text{x}\geq1\\\sin\Big(\frac{\pi\text{x}}{2}\Big),& \text{x}>1\end{cases}$ is continuous but not differentiable at x = 1.
Answer Given: $\text{f(x)}=\begin{cases}|2\text{x}-3||\text{x}|, & \text{x}\geq1\\\sin\Big(\frac{\pi\text{x}}{2}\Big),& \text{x}>1\end{cases}$
Continuity at x = 1:
(LHL at x = 1) = $\lim_\limits{\text{x}\rightarrow1^{-}}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1-\text{h})=\lim_\limits{\text{h}\rightarrow0}\sin\Big(\frac{\pi(1-\text{h})}{2}\Big)=\sin\frac{\pi}{2}=1$
(RHL at x = 1) $=\lim_\limits{\text{x}\rightarrow1^{+}}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1+\text{h})$
$\Rightarrow\ \lim_\limits{\text{h}\rightarrow0}|2(1+\text{h})-3|[1+\text{h}]=\lim_\limits{\text{h}\rightarrow0}|2(1+\text{h})-3|=1$
Hence, (LHL at x = 1) = (RHL at x = 1)
Differentiable at x = 1:
(LHL at x = 1) $=\lim_\limits{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
(LHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{1-\text{h}-1}$
(LHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{-\text{h}}$
(LHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big(\frac{\pi(1-\text{h})}{2}\Big)-1}{-\text{h}}$
(LHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\cos\frac{\pi\text{h}}{2}-1}{-\text{h}}$
(LHL at x = 1) $=-\frac{\pi}{2}\lim_\limits{\text{h}\rightarrow0}\frac{\cos\frac{\pi\text{h}}{2}-1}{\frac{\pi}{2}\text{h}}=0$
(RHL at x = 1) $=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
(RHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{1+\text{h}-1}$
(RHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{\text{h}}$
(RHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{-(2(1+\text{h})-3)-1}{\text{h}}$
(RHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{-2\text{h}}{\text{h}}=-2$
$\text{LHL}\neq\text{RHL}$
Hence, the function is continuous but not differentiable at x = 1.
View full question & answer→Question 5075 Marks
Discuss the continuity and differentiability of the f(x) = |x| + |x - 1| in the interval (-1, 2).
Answer f(x) = |x| + |x - 1| in the interval (-1, 2).
$\text{f(x)}=\begin{cases}\text{x}+\text{x}+1 & -1<\text{x}<0\\1 & 0\leq\text{x}\leq1\\-\text{x}-\text{x}+1&1<\text{x}<2\end{cases}$
$\text{f(x)}=\begin{cases}2\text{x}+1 & -1<\text{x}<0\\1 & 0\leq\text{x}\leq1\\-2\text{x}+1&1<\text{x}<2\end{cases}$
We know that a polynomial and a constant function is continuous and differentiable everywhere.
So, f(x) is continuous and differentiable for $\text{x}\in(-1,0),\text{x}\in(0,1)$ and (1, 2).
We need to check continuitly and differentiability at x = 0 and x = 1
Continuity at x = 0
$\lim_\limits{\text{x}\rightarrow0^{-}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^{-}}2\text{x}+1=1$
$\lim_\limits{\text{x}\rightarrow0^{+}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^{+}}1=1$
$\text{f(0)}=1$
$\lim_\limits{\text{x}\rightarrow0^{+}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^{-}}\text{f(x)}=\text{f(x)}$
$\therefore$ f(x) is continuous at x = 0.
Continuity at x = 1
$\lim_\limits{\text{x}\rightarrow1^{-}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^{-}}1=1$
$\lim_\limits{\text{x}\rightarrow1^{+}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^{+}}1=1$
$\text{f(x)}=1$
$\lim_\limits{\text{x}\rightarrow1^{-}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^{+}}\text{f(x)}=\text{f}(1)$
$\therefore$ f(x) is continuous at x = 1.
View full question & answer→Question 5085 Marks
Differentiate the following functions with respect to x:
$\frac{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}}$
Answer We have, $\frac{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}}$
By rationalising we get,
$\frac{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}}\times\frac{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}$
$=\frac{\big(\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}\big)^2}{\big(\sqrt{\text{x}^2+1}\big)^2-\big(\sqrt{\text{x}^2-1}\big)^2}$
$=\frac{\big(\sqrt{\text{x}^2+1}\big)^2+\big(\sqrt{\text{x}^2-1}\big)^2+2\big(\sqrt{\text{x}^2+1}\big)\big(\sqrt{\text{x}^2-1}\big)}{\text{x}^2+1-\text{x}^2+1}$
$=\frac{\text{x}^2+1+\text{x}^2-1+2\sqrt{\text{x}^4-1}}{2}$
$=\frac{2\text{x}^2+2\sqrt{\text{x}^4-1}}{2}$
$=\text{x}^2+\sqrt{\text{x}^4-1}$
Now, Let $\text{y}=\text{x}^2+\sqrt{\text{x}^4-1}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\sqrt{\text{x}^4-1}\big)$
$=2\text{x}+\frac{1}{2\sqrt{\text{x}^4-1}}\times\frac{\text{d}}{\text{dx}}(\text{x}^4-1)$
$=2\text{x}+\frac{1}{2\sqrt{\text{x}^4-1}}\times(4\text{x}^3)$
$=2\text{x}+\frac{2\text{x}^3}{\sqrt{\text{x}^4-1}}$
View full question & answer→Question 5095 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\frac{3\text{at}}{1+\text{t}^2}\text{ and y}=\frac{3\text{at}^2}{1+\text{t}^2}$
Answer Here, $\text{x}=\frac{3\text{at}}{1+\text{t}^{2}}$
Differentiating it with respect to t using quotiont rule,
$\frac{\text{dx}}{\text{dt}}= \bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(3\text{at})-3\text{at}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})}\bigg]$
$\frac{\text{dx}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})(3\text{a})-3\text{at}(2\text{t})}{(1+\text{t}^{2})}\bigg]$
$\frac{\text{dx}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})(3\text{a})-3\text{at}(2\text{t})}{(1+\text{t}^{2})}\bigg]$
$\frac{\text{dx}}{\text{dt}}=\Big[\frac{3\text{a}-3\text{at}^{2}}{(1-\text{t}^{2})^{2}}\Big]$ And,
$\frac{\text{dx}}{\text{dt}}=\frac{3\text{a}(1-\text{t}^{2})}{(1+\text{t}^{2})^{2}}.....(\text{i})$
$\text{y}=\frac{3\text{at}^{2}}{(1+\text{t}^{2})}$
Differentiating it with respect to t using quotient rule,
$\frac{\text{dy}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(3\text{at}^{2})-3\text{at}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})^{2}}\bigg]$
$\frac{\text{dy}}{\text{dt}}=\Big[\frac{(1+\text{t}^{2})(6\text{at})-(3\text{at}^{2})(2\text{t})}{(1+\text{t}^{2})^{2}}\Big]$
$\frac{\text{dy}}{\text{dt}}=\Big[\frac{6\text{at}+6\text{at}^{3}-6\text{at}^{3}}{(1+\text{t}^{2})^{2}}\Big]$
$\frac{\text{dy}}{\text{dt}}=\frac{6\text{at}}{(1+\text{t}^{2})^{2}}....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{6\text{at}}{(1+\text{t}^{2})^{2}}\times\frac{(1+\text{t}^{2})^{2}}{3\text{a}(1-\text{t}^{2})}$
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{t}}{1-\text{t}^{2}}$
View full question & answer→Question 5105 Marks
If $y^x + x^y + x^x = a^b$, find $\frac{\text{dy}}{\text{dx}}$
AnswerGiven that $y^x + x^y + x^x = a^b$
Putting $u = y^x, v = x^y$ and $w = x^x$, we get $u + v + w = a^b$
Therefore $\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}+\frac{\text{dw}}{\text{dx}}=0\ .....(\text{i})$
Now, $u = y^x$. Taking logrithm on both sides, we have
$\log\text{u}=\text{x}\log\text{y}$
Differentiating both sides w.r.t. x, we have
$\frac{1}{\text{u}}\times\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$=\text{x}\frac{1}{\text{y}}\times\frac{\text{dy}}{\text{dx}}+\log\text{y}\times1$
So, $\frac{\text{du}}{\text{dx}}=\text{u}\Big(\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big)=\text{y}^\text{x}\Big[\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big]\ .....(\text{ii})$
Also $v = x^y$
Taking logarithm on both sides, we have
$\log\text{v}=\text{y}\log\text{x}$
Differentiating both sides w.r.t. x, we have
$\frac{1}{\text{v}}\times\frac{\text{dv}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{dy}}{\text{dx}}$
$=\text{y}\times\frac{1}{\text{x}}+\log\text{x}\times\frac{\text{dy}}{\text{dx}}$
So, $\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\frac{\text{y}}{\text{x}}+\log\text{x}\times\frac{\text{dy}}{\text{dx}}\Big]$
$=\text{x}^\text{y}\Big[\frac{\text{y}}{\text{x}}+\log\text{x}\times\frac{\text{dy}}{\text{dx}}\Big]$
Again $w = x^x$
Taking logarithm on both sides, we have
$\log\text{w}=\text{x}\log\text{x}$
Differentiating both sides w.r.t x, we have
$\frac{1}{\text{w}}\times\frac{\text{dw}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{x}}{\text{dx}}(\text{x})$
$=\text{x}.\frac{1}{\text{x}}+\log\text{x}\times1$
i.e. $\frac{\text{dw}}{\text{dx}}=\text{w}(1+\log\text{x})$
$=\text{x}^\text{x}(1+\log\text{x})\ .....(\text{iv})$
From (i), (ii), (iii), (iv), we have
$\text{y}^\text{x}\Big(\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big)+\text{x}^\text{y}\Big(\frac{\text{y}}{\text{x}}+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big) \\ +\text{x}^\text{x}(1+\log\text{x})=0$
$\big(\text{x}\times\text{y}^{\text{x}-1}+\text{x}^\text{y}\times\log\text{x}\big) \\ \frac{\text{dy}}{\text{dx}}=-\text{x}^\text{x}(1+\log\text{x})-\text{y}\times\text{x}^{\text{y}-1}-\text{y}^\text{x}\log\text{y}=0$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{-\big[\text{y}^\text{x}\log\text{y}+\text{y}\times\text{x}^{\text{y}-1}+\text{x}^\text{x}(1+\log\text{x})\big]}{\text{x}\times\text{y}^{\text{x}-1}+\text{x}^\text{y}\log\text{y}}$
View full question & answer→Question 5115 Marks
If the functions f(x), defined below is continuous at x = 0, find the value of k.
$\text{f(x)}=\begin{cases}\frac{1-\cos2\text{x}}{2\text{x}^2},&\text{x}<0\\\text{k},&\text{x}=0\\\frac{\text{x}}{|\text{x}|},&\text{x}>0\end{cases}$
Answer Given, $\text{f(x)}=\begin{cases}\frac{1-\cos2\text{x}}{2\text{x}^2},&\text{x}<0\\\text{k},&\text{x}=0\\\frac{\text{x}}{|\text{x}|},&\text{x}>0\end{cases}$
$\text{f(x)}=\begin{cases}\frac{1-\cos2\text{x}}{2\text{x}^2},&\text{x}<0\\\text{k},&\text{x}=0\\1,&\text{x}>0\end{cases}$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow 0}\Big(\frac{1-\cos2(-\text{h})}{2(-\text{h})^2}\Big)$
$=\lim_\limits{\text{h}\rightarrow 0}\Big(\frac{1-\cos2\text{h}}{2\text{h}^2}\Big)$
$=\frac{1}{2}\lim_\limits{\text{h}\rightarrow 0}\Big(\frac{2\sin^2\text{h}}{\text{h}^2}\Big)$
$=\frac{2}{2}\lim_\limits{\text{h}\rightarrow 0}\Big(\frac{\sin^2\text{h}}{\text{h}^2}\Big)$
$=\frac{2}{2}\lim_\limits{\text{h}\rightarrow 0}\Big(\frac{\sin^2\text{h}}{\text{h}}\Big)^2$
$=1\times1$
$=1$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow 0}\text{f(h)}=\lim_\limits{\text{h}\rightarrow 0}(1)=1$
Also, $\text{f}(0)=\text{k}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow 0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow 0^+}=\text{f}(0)$
$\Rightarrow1=1=\text{k}$
Hence, the required value of k is 1
View full question & answer→Question 5125 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{e}^{\theta}\Big(\theta+\frac{1}{\theta}\Big)\text{ and y}=\text{e}^{-\theta}\Big(\theta-\frac{1}{\theta}\Big)$
Answer We have, $\text{x}=\text{e}^\theta\Big(\theta+\frac{1}{\theta}\Big)$
$\frac{\text{dx}}{\text{d}\theta}=\text{e}^\theta\frac{\text{d}}{\text{d}\theta}\Big(\theta+\frac{1}{\theta}\Big)+\Big(\theta+\frac{1}{\theta}\Big)\frac{\text{d}}{\text{d}\theta}(\text{e}^\theta)$
[Using product rule]
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{e}^\theta\Big(1-\frac{1}{\theta^{2}}\Big)+\Big(\frac{\theta^{2}+1}{\theta}\Big)\text{e}^{\theta}$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{e}^{\theta}\Big(1-\frac{1}{\theta^{2}}+\frac{\theta^{2}+1}{\theta}\Big)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{e}^\theta\Big(\frac{\theta^{2}-1+\theta^{3}+\theta}{\theta{2}}\Big)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\frac{\text{e}^\theta(\theta^{3}+\theta^{2}+\theta-1)}{\theta^{2}}\ .....(\text{i})$
And, $\text{y}=\text{e}^\theta\Big(\theta-\frac{1}{\theta}\Big)$
View full question & answer→Question 5135 Marks
Differentiate the following functions with respect to x:
$\frac{3\text{x}^2\sin\text{x}}{\sqrt{7-\text{x}^2}}$
AnswerLet $\text{y}=\frac{3\text{x}^2\sin\text{x}}{\sqrt{7-\text{x}^2}}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\bigg\{\frac{3\text{x}^2\sin\text{x}}{(\sqrt{7-\text{x}^2})^\frac{1}{2}}\bigg\}$
$=\frac{(7-\text{x}^2)^\frac{1}{2}\times\frac{\text{d}}{\text{dx}}(3\text{x}^3\sin\text{x})-(3\text{x}^2\sin\text{x})\frac{\text{d}}{\text{dx}}(7-\text{x}^2)^\frac{1}{2}}{\Big[(7-\text{x}^2)^\frac{1}{2}\Big]^2}$
[Using quotient rule, chain rule and product rule]
$\Bigg[\frac{(7-\text{x}^2)^\frac{1}{2}\times3\Big[\text{x}^2\frac{\text{d}}{\text{dx}}(\sin\text{x})+\sin\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^2)\Big]-3\text{x}^2\sin\text{x}\times\frac{1}{2}(7-\text{x}^2)\times\frac{\text{d}}{\text{dx}}(7-\text{x}^2)}{(7-\text{x}^2)}\Bigg]$
$\Bigg[\frac{(7-\text{x}^2)^\frac{1}{2}3(\text{x}^2\cos\text{x}+2\text{x}\sin\text{x})-3\text{x}^2\sin\text{x}\times\frac{1}{2}(7-\text{x}^2)^\frac{-1}{2}(-2\text{x})}{(7-\text{x}^2)}\Bigg]$
$=\Bigg[\frac{(7-\text{x}^2)^\frac{1}{2}\times3(\text{x}^2\cos+2\text{x}\sin\text{x})}{(7-\text{x}^2)}+\frac{3\text{x}^2\sin\text{x}(7-\text{x}^2)^\frac{-1}{2}}{(7-\text{x}^2)}\Bigg]$
$\bigg[\frac{6\text{x}\sin\text{x}+3\text{x}^2\cos\text{x}}{\sqrt{(7-\text{x}^2)}}+\frac{3\text{x}^3\sin\text{x}}{(7-\text{x}^2)^\frac{3}{2}}\bigg]$
So,
$\frac{\text{d}}{\text{dx}}\Big(\frac{3\text{x}^2\sin\text{x}}{\sqrt{7-\text{x}^2}}\Big)\bigg[\frac{6\text{x}\sin\text{x}+3\text{x}^2\cos\text{x}}{\sqrt{(7-\text{x}^2)}}+\frac{3\text{x}^3\sin\text{x}}{(7-\text{x}^2)^\frac{3}{2}}\bigg]$
View full question & answer→Question 5145 Marks
If $\text{x}\sqrt{1+\text{y}}+\text{y}\sqrt{1+\text{x}}=0,$ prove that $(1+\text{x})^2\frac{\text{dx}}{\text{dx}}+1=0$
AnswerWe have $\text{x}\sqrt{1+\text{y}}+\text{y}\sqrt{1+\text{x}}=0$
$\Rightarrow\text{x}\sqrt{1+\text{y}}=-\text{y}\sqrt{1+\text{x}}$
Squaring both sides, we get,
$\Rightarrow\big(\text{x}\sqrt{1+\text{y}}\big)^2=(-\text{y}\sqrt{1+\text{x}}\big)^2$
$\Rightarrow\text{x}^2\big(1+\text{y}\big)=\text{y}^2(1+\text{x}\big)$
$\Rightarrow\text{x}^2+\text{x}^2\text{y}=\text{y}^2+\text{y}^2\text{x}$
$\Rightarrow\text{x}^2-\text{y}^2=\text{y}^2\text{x}-\text{x}^2\text{y}$
$\Rightarrow(\text{x}-\text{y})(\text{x}+\text{y})=\text{x}\text{y}(\text{y}-\text{x})$
$\Rightarrow(\text{x}+\text{y})=-\text{x}\text{y}$
$\Rightarrow\text{y}+\text{x}\text{y}=-\text{x}$
$\Rightarrow\text{y}(1+\text{x})=-\text{x}$
$\Rightarrow\text{y}=\frac{-\text{x}}{(1+\text{x})}$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\bigg[\frac{-(1+\text{x})\frac{\text{d}}{\text{dx}}(\text{x})-(-\text{x})\frac{\text{d}}{\text{dx}}(\text{x}+1)}{(1+\text{x})^2}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big[\frac{-(1+\text{x})(1)+\text{x}(1)}{(1+\text{x})^2}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big[\frac{-1-\text{x}+\text{x}}{(1+\text{x})^2}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1}{(1+\text{x})^2}$
$\Rightarrow(1+\text{x})^2\frac{\text{dy}}{\text{dx}}=-1$
$\Rightarrow(1+\text{x})^2\frac{\text{dy}}{\text{dx}}+1=0$
View full question & answer→Question 5155 Marks
Differentiate the following w.r.t. x:
$(\text{x}+1)^2(\text{x}+2)^3(\text{x}+3)^4$
AnswerLet $\text{y}=(\text{x}+1)^2(\text{x}+2)^3(\text{x}+3)^4$
$\therefore\ \log\text{y}=\log\big\{(\text{x}+1)^2\cdot(\text{x}+2)^3(\text{x}+3)^4\big\}$
$=\log(\text{x}+1)^2+\log(\text{x}+2)^3+\log(\text{x}+3)^4$
and $\frac{\text{d}}{\text{dy}}\log\text{y}\cdot\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[2\log(\text{x}+1)\big]\\+\frac{\text{d}}{\text{dx}}\big[3\log(\text{x}+2)\big]+\frac{\text{d}}{\text{dx}}\big[4\log(\text{x}+3)\big]$
$\frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\frac{2}{(\text{x}+1)}\cdot\frac{\text{d}}{\text{dx}}(\text{x}+1)+3\cdot\frac{1}{(\text{x}+2)}\cdot\\\frac{\text{d}}{\text{dx}}(\text{x}+2)+4\cdot\frac{1}{(\text{x}+3)}\cdot\frac{\text{d}}{\text{dx}}(\text{x}+3)$$\Big[\because\frac{\text{d}}{\text{dx}}(\log\text{x})=\frac{1}{\text{x}}\Big]$
$=\Big[\frac{2}{\text{x}+1}+\frac{3}{\text{x}+2}+\frac{4}{\text{x}+3}\Big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{2}{\text{x}+1}+\frac{3}{\text{x}+2}+\frac{4}{\text{x}+3}\Big]$
$=(\text{x}+1)^2\cdot(\text{x}+2)^3\cdot(\text{x}+3)^4\Big[\frac{2}{\text{x}+1}+\frac{3}{\text{x}+2}+\frac{4}{\text{x}+3}\Big]$
$=(\text{x}+1)^2\cdot(\text{x}+2)^3\cdot(\text{x}+3)^4$
$\bigg[\frac{2(\text{x}+2)(\text{x}+3)+3(\text{x}+1)(\text{x}+3)+4(\text{x}+1)(\text{x}+2)}{(\text{x}+1)(\text{x}+2)(\text{x}+3)}\bigg]$
$=\frac{(\text{x}+1)^2(\text{x}+2)^3(\text{x}+3)^4}{(\text{x}+1)(\text{x}+2)(\text{x}+3)}$
$\big[2(\text{x}^2+5\text{x}+6)+3(\text{x}^2+4\text{x}+3)+4(\text{x}^2+3\text{x}+2)\big]$
$=(\text{x}+1)(\text{x}+2)^2(\text{x}+3)^3$
$\big[2\text{x}^2+10\text{x}+12+3\text{x}^2+12\text{x}+9+4\text{x}^2+12\text{x}+8\big]$
$=(\text{x}+1)(\text{x}+2)^2(\text{x}+3)^3\big[9\text{x}^2+34\text{x}+29\big]$
View full question & answer→Question 5165 Marks
Differentiate w.r.t. x the function in Exercise:
$\text{x}^{\text{x}^2-3}+(\text{x}-3)^{\text{x}^2},$ for x > 3
AnswerLet 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}}\ \ \dots(1)$
$\text{u}=\text{x}^{\text{x}^2-3}$
$\therefore\ \log\text{u}=\log(\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}}\cdot\frac{\text{du}}{\text{dx}}=\log\text{x}.\frac{\text{d}}{\text{dx}}(\text{x}^2-3)+(\text{x}^2+3)\cdot\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log\text{x}\cdot2\text{x}+(\text{x}^2-3)\cdot\frac{1}{\text{x}}$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}^2-3}\cdot\Bigg[\frac{\text{x}^2-3}{\text{x}}+2\text{x}\log\text{x}\Bigg]$
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)$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{v}}\cdot\frac{\text{dv}}{\text{dx}}=\log(\text{x}-3)\cdot\frac{\text{d}}{\text{dx}}(\text{x}^2)+\text{x}^2\cdot\frac{\text{d}}{\text{dx}}\big[\log(\text{x}-3)\big]$
$\Rightarrow\ \frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\log(\text{x}-3)\cdot2\text{x}+\text{x}^2\cdot\frac{1}{\text{x}-3}\cdot\frac{\text{d}}{\text{dx}}(\text{x}-3)$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\text{v}\Bigg[2\text{x}\log(\text{x}-3)+\frac{\text{x}^2}{\text{x}-3}\cdot1\Bigg]$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=(\text{x}-3)^{\text{x}^2}\Bigg[\frac{\text{x}^2}{\text{x}-3}+2\text{x}\log(\text{x}-3)\Bigg]$
Substituting the expressions of $\frac{\text{du}}{\text{dx}}\text{ and }\frac{\text{dv}}{\text{dx}}$ in equation (1), we obtain
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x}^2-3}\Bigg[\frac{\text{x}^2-3}{\text{x}}+2\text{x}\text{logx}\Bigg]+(\text{x}-3)^\text{x} \Bigg[\frac{\text{x}^{\text{x}}}{\text{x}-3}+2\text{x}\log(\text{x}-3)\Bigg]$
View full question & answer→Question 5175 Marks
If $\text{y}=\big\{\log_{\cos\text{x}}\sin\text{x}\big\}\big\{\log_{\sin\text{x}}\cos\text{x}\big\}^{-1}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big),$ find $\frac{\text{dy}}{\text{dx}}$ at $\text{x}=\frac{\pi}{4}$
AnswerWe have, $\text{y}=\big\{\log_{\cos\text{x}}\sin\text{x}\big\}\big\{\log_{\sin\text{x}}\cos\text{x}\big\}^{-1}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
$\Rightarrow \text{y}\big\{\log_{\cos\text{x}}\sin\text{x}\big\}\big\{\log_{\cos\text{x}}\sin\text{x}\big\}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big) \\ \big[\because\log_{\text{a}}\text{b}=(\log_\text{b}\text{a})^{-1}\big]$
$\Rightarrow \text{y}=\Big[\frac{\log\sin\text{x}}{\log\cos\text{x}}\Big]^2+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\ \Big[\because \log_\text{a}\text{b}=\frac{\log\text{b}}{\log\text{a}}\Big]$
Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\frac{\log\sin\text{x}}{\log\sin\text{x}}\Big]^2+\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big\}$
$\Rightarrow\frac{\text{d}}{\text{dx}}=2\Big[\frac{\log\sin\text{x}}{\log\sin\text{x}}\Big]\frac{\text{d}}{\text{dx}}\Big(\frac{\log\sin\text{x}}{\log\sin\text{x}}\Big)+\frac{1}{\sqrt{-1\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)^2}}\times\frac{\text{d}}{\text{dx}}\Big[\frac{2\text{x}}{1+\text{x}^2}\Big]$
$\Rightarrow\frac{\text{d}}{\text{dx}}=2\Big[\frac{\log\sin\text{x}}{\log\sin\text{x}}\Big]\bigg[\frac{(\log\cos\text{x})\frac{\text{d}}{\text{dx}}(\log\sin\text{x})-\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\log\cos\text{x})}{(\log\cos\text{x})^3}\bigg] \\ +\Big[\frac{(1+\text{x}^2)}{\sqrt{1+\text{z}^2-2\text{x}^2}}\Big]\Big[\frac{(1+\text{x}^2)(2)-(2\text{x})(2\text{x})}{(1+\text{x}^2)^3}\Big]$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\Big[\frac{\log\sin\text{x}}{\log\cos\text{x}}\Big]\bigg[\frac{\log\cos\times\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})-\log\sin\text{x}\times\frac{1}{\cos\text{x}}\frac{\text{d}}{\text{dx}}(\cos\text{x})}{(\log\cos\text{x})^3}\bigg] \\ +\Big[\frac{(1+\text{x}^2)}{\sqrt{1+\text{x}^4-2\text{x}^2}}\Big]\Big[\frac{(1+\text{x}^2)(2)-(2\text{x})(2\text{x})}{(1+\text{x}^2)^2}\Big]$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\Big[\frac{\log\sin\text{x}}{\log\cos\text{x}}\Big]\bigg[\frac{\log\cos\text{x}\Big(\frac{\cos\text{x}}{\sin\text{x}}\Big)+\log\sin\text{x}\times\Big(\frac{\sin\text{x}}{\cos\text{x}}\Big)}{(\log\cos\text{x})^2}\bigg] \\ +\Big[\frac{1+\text{x}^2}{\sqrt{(1-\text{x}^2)^3}}\Big]\Big[\frac{2+2\text{x}^2-4\text{x}^2}{(1+\text{x}^2)^2}\Big]$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\frac{\log\sin\text{x}}{(\log\cos\text{x})^3}(\cos\text{x}\log\cos\text{x}+\tan\text{x}\log\sin\text{x})+\frac{2}{1+\text{x}^2}$
Put $\text{x}=\frac{\pi}{4}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\Bigg\{\frac{\log\sin\frac{\pi}{4}}{\big(\log\cos\frac{\pi}{4}\big)^3}\Bigg\} \\ \Big(\cot\frac{\pi}{4}\log\cos\frac{\pi}{4}+\tan\frac{\pi}{4}\log\sin\frac{\pi}{4}\Big)+2\bigg\{\frac{1}{1+\big(\frac{\pi}{4}\big)^2}\bigg\}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\Bigg\{\frac{1}{\big(\log\frac{1}{\sqrt{2}}\big)^2}\Bigg\} \\ \Big(1\times\log\frac{1}{\sqrt{2}}+1\times\log\frac{1}{\sqrt{2}}\Big)+2\Big(\frac{16}{16+\pi^2}\Big)$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\frac{2\log\Big(\frac{1}{\sqrt{2}}\Big)}{\Big\{\log\Big(\frac{1}{\sqrt{2}}\Big)^2\Big\}}+\frac{32}{16+\pi^2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=4\frac{1}{\log\Big(\frac{1}{\sqrt{2}}\Big)}+\frac{32}{16+\pi^2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=4\frac{1}{-\frac{1}{2}\log2}+\frac{32}{16+\pi^2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=-\frac{8}{\log2}+\frac{32}{16+\pi^2}$
So, $\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{a}=\frac{\pi}{4}}=8\Big[\frac{4}{16+\pi^2}-\frac{1}{\log2}\Big]$
View full question & answer→Question 5185 Marks
Differentiate $\cos^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ with respect to $\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ if $0<\text{x}<1$
AnswerLet $\text{u}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
Put $\text{x}=\tan\theta,$
$\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}= \cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$=\cos^{-1}\Big(\frac{1-\tan^2\theta}{1+\tan^2\theta}\Big)$
$\text{v}=\cos^{-1}(\cos2\theta)\ .....(\text{ii})$
Here, 0 < x < 1
$0<\tan\theta<1$
$\Rightarrow 0<\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}\ \big[\text{Since, x}=\tan\frac{\pi}{2}\big]$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{2}{1+\text{x}^2}\ .....(\text{iii})$
From equation (ii),
$\text{v}=2 \theta \ \big[ \text{since,}\cos^{-1}(\cos\theta)=\theta,\text{if }\theta\in[0,\pi]\big]$
$\text{v}=2\tan^{-1}\text{x}[{\text{since,x}=\tan\theta}]$
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 5195 Marks
Show that $\text{f(x)}=\begin{cases}12\text{x}-13, & \text{if x}\leq3\\2\text{x}^2+5, & \text{if x} > 3\end{cases}$ is differentiable at x = 3. Also, find f(3).
AnswerGiven: $\text{f(x)}=\begin{cases}12\text{x}-13, & \text{if x}\leq3\\2\text{x}^2+5, & \text{if x} > 3\end{cases}$
We have to show that the given function is differentiable at x = 3.
We have,
(LHL at x = 3) $=\lim_\limits{\text{x}\rightarrow3^{-}}\frac{\text{f(x)}-\text{f}(3)}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}\frac{12\text{(x)}-13-23}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}\frac{12\text{x}-36}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}\frac{12(\text{x}-3)}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}12$
$= 12$
(RHL at x = 3) $=\lim_\limits{\text{x}\rightarrow3^{+}}\frac{\text{f(x)}-\text{f}(3)}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}\frac{2\text{x}^2+5-23}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}\frac{2\text{x}^2-18}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}\frac{2(\text{x}^2-9)}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}2(\text{x}+3)$
$=2\times6$
$=12$
Thus, (LHL at x = 3) = (RHL at x = 3) = 12.
So, f(x) is differentiable at x = 3 and f(3) = 12.
View full question & answer→Question 5205 Marks
Differentiate the following w.r.t. x:
$\tan^{-1}\bigg(\frac{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}\bigg),-1<\text{x}<1,\text{ x}\neq0$
AnswerLet $\text{y}=\tan^{-1}\bigg(\frac{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}\bigg)$
Substituting $\text{x}^2=\cos2\theta,$ we get
$\text{y}=\tan^{-1}\bigg(\frac{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}{\sqrt{1+\cos2\theta}-\sqrt{1-\cos2\theta}}\bigg)$
$=\tan^{-1}\bigg(\frac{\sqrt{1+2\cos^2\theta-1}+\sqrt{1-1+2\sin^2\theta}}{\sqrt{1+2\cos^2\theta-1}-\sqrt{1-1+2\sin^2\theta}}\bigg)$
$=\tan^{-1}\bigg(\frac{\sqrt{2}\cos\theta+\sqrt{2}\sin\theta}{\sqrt{2}\cos\theta-\sqrt{2}\sin\theta}\bigg)$
$=\tan^{-1}\bigg[\frac{\sqrt{2}(\cos\theta+\sin\theta)}{\sqrt{2}(\cos\theta-\sin\theta)}\bigg]$
$=\tan^{-1}\Big(\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\Big)$
$=\tan^{-1}\Bigg(\frac{\frac{\cos\theta+\sin\theta}{\cos\theta}}{\frac{\cos\theta-\sin\theta}{\cos\theta}}\Bigg)$
$=\tan^{-1}\Big(\frac{1+\tan\theta}{1-\tan\theta}\Big)$
$=\tan^{-1}\tan\Big(\frac{\pi}{4}+\theta\Big)$ $\Big[\because\ \tan(\text{a+b})=\frac{\tan\text{a}+\tan\text{b}}{1-\tan\text{a}\cdot\tan\text{b}}\Big]$
$\therefore\ \text{y}=\frac{\pi}{4}+\theta=\frac{\pi}{4}+\frac{1}{2}\cos^{-1}\text{x}^2$ $\Big[\because\ 2\theta=\cos^{-1}\text{x}^2\Rightarrow\theta=\frac{1}{2}\cos^{-1}\text{x}^2\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=0+\frac{1}{2}\cdot\frac{-1}{\sqrt{1-\text{x}^4}}\cdot\frac{\text{d}}{\text{dx}}\text{x}^2$
$=\frac{1}{2}\cdot\frac{-2\text{x}}{\sqrt{1-\text{x}^4}}=\frac{-\text{x}}{\sqrt{1-\text{x}^4}}$
Find $\frac{\text{dy}}{\text{dx}}$ of each of the functions expressed in parametric form.
View full question & answer→Question 5215 Marks
Differentiate the following functions with respect to x:
$\log\big\{\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big\}$
AnswerLet $\text{y}=\log\big\{\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big\}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\log\big\{\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big\}\Big]$
$=\frac{1}{\big[\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big]}\frac{\text{d}}{\text{dx}}\Big[\text{x}+2+\big(\text{x}^2+4\text{x}+1\big)^\frac{1}{2}\Big]$
[Using chain rule]
$=\frac{1}{\big[\text{x}+2+\sqrt{\text{x}^4+4\text{x}+1}\big]}\times\Big[1+0+\frac{1}{2}\big(\text{x}^2+4\text{x}+1\big)^\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x}^2+4\text{x}+1)\Big]$
$=\frac{1+\frac{(2\text{x}+4)}{2\big(\sqrt{\text{x}^2+4\text{x}+1}\big)}}{\big[\text{x}+2+\sqrt{\text{x}^4+4\text{x}+1}\big]}$
$=\frac{\sqrt{\text{x}^2+4\text{x}+1}+\text{x}+2}{\big[\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big]\times\sqrt{\text{x}^2+4\text{x}+1}}$
$=\frac{1}{\sqrt{\text{x}^2+4\text{x}+1}}$
So,
$\frac{\text{d}}{\text{dx}}\Big[\log\big\{\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big\}\Big]=\frac{1}{\sqrt{\text{x}^2+4\text{x}+1}}$
View full question & answer→Question 5225 Marks
If $\text{y}\sqrt{1-\text{x}^2}+\text{x}\sqrt{1-\text{y}^2}=1,$ prove that $\frac{\text{dy}}{\text{dx}}=-\sqrt{\frac{1-\text{y}^2}{1-\text{x}^2}}$
AnswerWe have, $\text{y}\sqrt{1-\text{x}^2}+\text{x}\sqrt{1-\text{y}^2}=1$
Let $\text{x}=\sin\text{A},\text{y}=\sin\text{B}$
$\Rightarrow\sin\text{B}\sqrt{1-\sin^2\text{A}}+\sin\text{A}\sqrt{1-\sin^2\text{B}}=1$
$\Rightarrow\sin\text{B}\cos\text{A}+\sin\text{A}\cos\text{B}=1$
$\big[\because\sin(\text{x}+\text{y})=\sin\text{x}\cos\text{y}+\cos\text{x}\sin\text{y}\big]$
$\Rightarrow\sin\big(\text{A}+\text{B}\big)=1$
$\Rightarrow\text{A}+\text{B}=\sin^{-1}(1)$
$\Rightarrow\sin^{-1}\text{x}+\sin^{-1}\text{y}=\frac{\pi}{2} \big[\because\text{x}=\sin\text{A},\text{y}=\sin\text{B}\big]$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}\big)+\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{y}\big)=\frac{\text{d}}{\text{dx}}\Big(\frac{\pi}{2}\Big)$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}+\frac{1}{\sqrt{1-\text{y}^2}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\sqrt\frac{1-\text{y}^2}{1-\text{x}^2}$
View full question & answer→Question 5235 Marks
Differentiate the function given in Exercise:
$(\text{x}\cos\text{x})^\text{x}+(\text{x}\sin\text{x})^{\frac{1}{\text{x}}}$
AnswerLet $\text{y}=(\text{x}\cos\text{x})^\text{x}+(\text{x}\sin\text{x})^{\frac{1}{\text{x}}}$
Putting $\text{u}=(\text{x}\cos\text{x})^\text{x}\text{and v}(\text{x}\sin\text{x})^{\frac{1}{\text{x}}},\text{we have }\ \ \text{y}=\text{u}+\text{v}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ \dots\text{(i)}$
Now $\text{u}=(\text{x}\cos\text{x})^\text{x}\ \Rightarrow\ \log\text{u}=\log(\text{x}\cos\text{x})^2=\text{x}\log(\text{x}\cos\text{x})$
$\Rightarrow\ \log\text{u}=\text{x}(\log\text{x}+\log\cos\text{x})\ \Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{u}=\frac{\text{d}}{\text{dx}}\Big\{\text{x}(\log\text{x}+\log\cos\text{x})\Big\}$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\Big[\frac{1}{\text{x}}+\frac{1}{\cos\text{x}}(-\sin\text{x})\Big]+(\log\text{x}+\log\cos\text{x}).1$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}[1-\text{x}\tan\text{x}+\log(\text{x}\cos\text{x})]$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=(\text{x}\cos\text{x})^\text{x}[1-\text{x}\tan\text{x}+\log(\text{x}\cos\text{x})]\ \dots\ \text{(ii)}$
Again $\text{v}=(\text{x}\sin\text{x})^{\frac{1}{\text{x}}}\ \Rightarrow\ \log\text{v}=\log(\text{x}\sin\text{x})^{\frac{1}{\text{x}}}=\frac{1}{\text{x}}\log(\text{x}\sin\text{x})$
$\Rightarrow\ \log\text{v}=\frac{1}{\text{x}}(\log\text{x}+\log\sin\text{x})\ \Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{v}=\frac{\text{d}}{\text{dx}}\Big\{\frac{1}{\text{x}}(\log\text{x}+\log\sin\text{x})\Big\}$
$\Rightarrow\ \frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\frac{1}{\text{x}}\Big[\frac{1}{\text{x}}+\frac{1}{\sin\text{x}}.\cos\text{x}\Big]+(\log\text{x}+\log\sin\text{x})\Big(\frac{-1}{\text{x}^2}\Big)$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\text{v}\Big[\frac{1}{\text{x}^2}+\frac{\cot\text{x}}{\text{x}}-\frac{\log(\text{x}\sin\text{x})}{\text{x}^2}\Big]$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=(\text{x}\sin\text{x})^{\frac{1}{\text{x}}}\Big[\frac{1}{\text{x}^2}+\frac{\cot\text{x}}{\text{x}}-\frac{\log(\text{x}\sin\text{x})}{\text{x}^2}\Big]\ \dots\text{(iii)}$
Putting the values from eq. (ii) and (iii) in eq. (i),
$\frac{\text{dy}}{\text{dx}}=(\text{x}\cos\text{x})^\text{x}[1-\text{x}\tan\text{x}+\log(\text{x}\cos\text{x})]+(\text{x}\sin\text{x})^{\frac{1}{\text{x}}}\Big[\frac{1}{\text{x}^2}+\frac{\cot\text{x}}{\text{x}}-\frac{\log(\text{x}\sin\text{x}}{\text{x}^2}\Big]$
View full question & answer→Question 5245 Marks
If $\text{y}\sin(\text{x}^\text{x}),$ prove that $\frac{\text{dy}}{\text{dx}}=\cos(\text{x}^\text{x})\times\text{x}^\text{x}(1+\log\text{x})$
AnswerLet $\text{y}=\sin(\text{x}^\text{x})\ .....(\text{i})$
Also, Let $\text{u}=\text{x}^\text{x}\ .....(\text{ii})$
Taking log on both sides,
$\Rightarrow\log\text{u}=\log\text{x}^\text{x}$
$\Rightarrow\log\text{u}=\text{x}\log\text{x}$
Differentiating both sides with respect to x,
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\big(\frac{1}{\text{x}}\big)+\log\text{x}(1)$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=1+\log\text{x}$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}(1+\log\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})\ .....(\text{iii})$
[Using equation (ii)]
Now, using equation (ii) in equation (i),
$\text{y}=\sin\text{u}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin\text{u})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\cos\text{u}\frac{\text{du}}{\text{dx}}$
Using equation (ii) and (iii),
$\frac{\text{dy}}{\text{dx}}=\cos(\text{x}^\text{x})\times\text{x}^\text{x}(1+\log\text{x})$
View full question & answer→Question 5255 Marks
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Answer
- Let a be an arbitrary real number then $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{a}^{+}}\text{f(x)}\Rightarrow^{\ \ \text{lim}}_{\text{x}\rightarrow\text{a}^{+}}\cos\text{x} \Rightarrow^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\cos\text{(a + h)}$
$\Rightarrow\ ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}(\cos\text{a}\cos\text{ h} - \sin\text{a} \sin \text{h})= \cos \text{a}\ ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\ \cos\text{h}- \sin\text{a}\ ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\sin\text{h}$
$= \cos\text{a} \times 1 - \sin\text{a}\times0 = \cos \text{a} = \text{f(a)}$
$\therefore \ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}}\text{f(x)} = \text{f(a)}$ for all $\text{a}\in\text{R}$
Therefore, f(x0 is continuous at x = a.
Since, a is an arbitrary real number, therefore $\cos\text{x}$ is continuous.
- $\text{f(x)} = \text{cosec x}= \frac{1}{\sin \text x}$ and domain $\text{x} = \text{R} - (\text{x}\pi), \text{x}\in \text{I}$
$\Rightarrow ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{a}}\frac{1}{{\sin}\text{x}} = \frac{1}{^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\sin(\text{a + h})}=\frac{1}{{^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}(\sin}\text{a}\cos\text{h} + \cos\text{a}\sin\text{h})}$$=\frac{1}{{\sin}\text{a}\cos{0} + \cos\text{a}\sin0}$
$= \frac{1}{{\sin}\text{a}(1)+ \cos\text{a}(0)} = \frac{1}{\sin\text{a}} = \text{f(a)}$
Therefore, f(x) is continuous at x = a.
Since, a is an arbitrary real number, therefore, f(x) = cosec x is continuous.
- $\text{f(x)} = \sec \text{x}= \frac{1}{\cos \text x}$ and domain $\text{x} = \text{R} -(2\text{x} + 1) \frac\pi{2}, \text{x}\in \text{I}$
$\Rightarrow ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{a}}\frac{1}{{\cos}\text{x}} = \frac{1}{^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\cos(\text{a + h})}=\frac{1}{{^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}(\cos}\text{a}\cos\text{h} - \sin\text{a}\sin\text{h})}$$=\frac{1}{{\cos}\text{a}\cos{0} - \sin\text{a}\sin0}$
$= \frac{1}{{\cos}\text{a}(1)- \sin\text{a}(0)} = \frac{1}{\cos\text{a}} = \text{f(a)}$
Therefore, f(x) is continuous at x = a.
Since, a is an arbitrary real number, therefore, f(x) = sec x is continuous.
- $\text{f(x)} = \cot \text{x}= \frac{1}{\tan \text x}$ and domain $\text{x} = \text{R} - (\text{x}\pi), \text{x}\in \text{I}$
$\Rightarrow^{\ \ \text{lim}}_{\text{x}\rightarrow\text{a}}\frac{1}{{\tan}\text{x}}= \frac{1}{\lim\limits_{\text{h}\rightarrow0}\tan(\text{a + h})}=\frac{1}{\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\tan\text{a}+\tan\text{h}}{1-\tan\text{a}\tan\text{h}}\Big)}$$=\frac{1}{\frac{\tan\text{a}+0}{1-\tan\text{a}\tan0}}$
$\frac{1 - 0}{\tan\text{a}} = \frac{1}{\tan \text{a}} = \text{f(a)}$
Therefore, f9x) is continuous at x = a.
Since, a is an arbitrary real number, therefore, $\text{f(x)} = \cot \text{x}$ is continuous. View full question & answer→Question 5265 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\tan\text{x})^{\cot\text{x}}+(\cot\text{x})^{\tan\text{x}}$
AnswerHere,
$\text{y}=(\tan\text{x})^{\cot\text{x}}+(\cot\text{x})^{\tan\text{x}}$
$\text{y}=\text{e}^{\log(\tan\text{x})^{\cot\text{x}}}+\text{e}^{\log(\cot\text{x})^{\tan\text{x}}}$
$\big[\text{Since},\log_\text{e}\text{e}=1,\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$\text{y}=\text{e}^{\cot\text{x}\log\tan\text{x}}+\text{e}^{\tan\text{x}\log(\cot\text{x})}$
Differentiating it with respect to x using rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cot\text{x}\log\tan\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan\text{x}\log\cot\text{x}}\big)$
$=\text{e}^{\cot\text{x}\log\tan\text{x}}\frac{\text{d}}{\text{dx}}(\cot\text{x}\log\tan\text{x})+\text{e}^{\tan\text{x}\log\cot\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x}\log\cot\text{x})$
$=\text{e}^{\log(\tan\text{x})^{\cot\text{x}}}\Big[\cot\text{x}\frac{\text{d}}{\text{dx}}\log\tan\text{x}+\log\tan\text{x}\frac{\text{d}}{\text{dx}}\cot\text{x}\Big] \\ +\text{e}^{\log(\cot\text{x})^{\tan\text{x}}}\Big[ \tan\text{x}\frac{\text{d}}{\text{dx}} \log\cot\text{x}+\log\cot\text{x}\frac{\text{d}}{\text{dx}}(\tan\text{x})\Big]$
$=(\tan\text{x})^{\cot\text{x}}\Big[\cot\text{x}\times\Big(\frac{1}{\tan\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log\tan\text{x}(-\text{cosec}^2\text{x})\Big] \\ +(\cot\text{x})^{\tan\text{x}}\Big[\tan\text{x}\big(\frac{1}{\cot\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cot\text{x})+\log\cot\text{x}\big(\sec^2\text{x}\big)\Big]$
$=(\tan\text{x})^{\cot\text{x}}\Big[(1)\big(\sec^2\text{x}\big)-\text{cosec}^2\text{x}\log\tan\text{x}\Big] \\ +(\cot)^{\tan\text{x}}\Big[(1)\big(-\text{cosec}^2\text{x}\big)+\sec^2\text{x}\log\cot\text{x}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\tan\text{x})^{\cot\text{x}}\Big[\sec^{2\text{x}}-\text{cosec}^2\text{x}\log\tan\text{x}\Big] \\ +(\cot)^{\tan\text{x}}\Big[\sec^2\text{x}\log\cot\text{x}-\text{cosec}^2\text{x}\Big]$
View full question & answer→Question 5275 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big\}$
AnswerLet $\text{y}=\cos^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big\}$
Let $\text{x}=\text{a}\cot\theta$
$\Rightarrow\ \text{y}=\cos^{-1}\Big\{\frac{\text{a}\cot\theta}{\sqrt{\text{a}^2\cot^2\theta+\text{a}^2}}\Big\}$
$\Rightarrow\text{y}=\cos^{-1}\Big\{\frac{\text{a}\cot\theta}{\sqrt{\text{a}^2(\cot^2\theta+1)}}\Big\}$
$\Rightarrow\ \text{y}=\sin^{-1}\Big(\frac{\text{a}\cot\theta}{\text{a cosec}\theta}\Big)$
$\Rightarrow\ \text{y}=\cos^{-1}\Bigg(\frac{\frac{\cos\theta}{\sin\theta}}{\frac{1}{\sin\theta}}\Bigg)$
$\Rightarrow\ \text{y}=\cos^{-1}(\cos\theta)$
$\Rightarrow\ \text{y}=\theta$
$\Rightarrow\ \text{y}=\cot^{-1}\big(\frac{\text{x}}{\text{a}}\big)\ \big[\text{Since, x}=\text{a}\cot\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\big(\frac{\text{x}}{\text{a}}\big)^2}\frac{\text{d}}{\text{dx}}\big(\frac{\text{x}}{\text{a}}\big)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{a}^2}{\text{a}^2+\text{x}^2}\times\big(\frac{1}{\text{a}}\big)$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{a}}{\text{a}^2+\text{x}^2}$
View full question & answer→Question 5285 Marks
Find the value of 'a' for which the function f defined by
$\text{f}\text{(x)}=\begin{cases}\text{a}\sin\frac{\pi}{2}(\text{x}+1),& \text{x}\leq0 \\\frac{\tan\text{x-sin}\text{x}}{\text{x}^3} &\text{x} > 0\end{cases}$ is discontinuous at x = 0.
AnswerSince f(x) is continuous at x = 0, L.H.L = R.H.L.
Thus, we have
$\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0^-}\text{a}\sin\frac{\pi}{2}(\text{x}+1)=\lim\limits_{\text{x} \rightarrow 0^+}\frac{\tan\text{x}-\sin\text{x}}{\text{x}^3}$
$\Rightarrow\text{a}\times1=\lim\limits_{\text{x} \rightarrow 0}\frac{\tan\text{x}-\sin\text{x}}{\text{x}^3}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\sin\text{x}}{\cos\text{x}}-\sin\text{x}}{\text{x}^3}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\sin\text{x}}{\text{x}}\Big(\frac{1}{\cos\text{x}}-1\Big)}{\text{x}^2}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\sin\text{x}}{\text{x}}\Big(\frac{1-\cos\text{x}}{\cos\text{x}}\Big)}{\text{x}^2}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}\times\lim\limits_{\text{x} \rightarrow 0}\frac{1}{\cos\text{x}}\times\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{x}}{\text{x}^2}$
$\Rightarrow\text{a}=1\times1\times\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{x}}{\text{x}^2}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{x}}{\text{x}^2}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos^2\text{x}}{\text{x}^2(1+\cos\text{x)}}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin^2\text{x}}{\text{x}^2(1+\cos\text{x)}}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin^2\text{x}}{\text{x}^2}\times\lim\limits_{\text{x} \rightarrow 0}\frac{1}{1+\cos\text{x}}$
$\Rightarrow\text{a}=1\times\lim\limits_{\text{x} \rightarrow 0}\frac{1}{1+\cos\text{x}}$
$\Rightarrow\text{a}=1\times\frac{1}{1+1}$
$\Rightarrow\text{a}=\frac{1}{2}$
View full question & answer→Question 5295 Marks
If $\sqrt{\text{y}+\text{x}}+\sqrt{\text{y}-\text{x}}=\text{c},$ show that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\sqrt{\frac{\text{y}^2}{\text{x}^2}-1}$
AnswerHere,
$\sqrt{\text{y}+\text{x}}+\sqrt{\text{y}-\text{x}}=\text{c}$
Differentiating with respect to x,
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sqrt{\text{y}+\text{x}})+\frac{\text{d}}{\text{dx}}\sqrt{\text{y}-\text{x}}=\frac{\text{d}}{\text{dx}}(\text{c})$
$\Rightarrow\frac{1}{2\sqrt{\text{y}+\text{x}}}\frac{\text{d}}{\text{dx}}(\text{y}+\text{x})+\frac{1}{2\sqrt{\text{y}-\text{x}}}\frac{\text{d}}{\text{dx}}(\text{y}-\text{z})=0$
$\Rightarrow \frac{1}{2\sqrt{\text{y}+\text{x}}}\Big(\frac{\text{dy}}{\text{dx}}+1\Big)+\frac{1}{2\sqrt{\text{y}-\text{x}}}\Big(\frac{\text{dy}}{\text{dx}}-1\Big)=0$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\Big(\frac{1}{2\sqrt{\text{y}+\text{x}}}\Big)+\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{2\sqrt{\text{y}-\text{x}}}\Big)=\frac{1}{2\sqrt{\text{y}-\text{x}}}-\frac{1}{2\sqrt{\text{y}+\text{x}}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\times\Big[\frac{1}{\sqrt{\text{y}+\text{x}}}+\frac{1}{\sqrt{\text{y}-\text{x}}}\Big]=\frac{1}{2}\Big[\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}-\text{x}}\sqrt{\text{y}+\text{x}}}\Big]$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\Big[\frac{\sqrt{\text{y}-\text{x}}-\sqrt{\text{y}+\text{x}}}{\sqrt{\text{y}+\text{x}}\sqrt{\text{y}-\text{x}}}\Big]=\Big[\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}-\text{x}}\sqrt{\text{y}+\text{x}}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}-\text{x}}+\sqrt{\text{y}-\text{x}}}\times\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}+\text{x}}}$
[Rationalizing the denominator]
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{(\text{y}+\text{x})+(\text{y}-\text{x})-2\sqrt{\text{y}+\text{x}}\sqrt{\text{y}-\text{x}}}{\text{y}+\text{x}-\text{y}+\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}-2\sqrt{\text{y}^2-\text{x}^2}}{2\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{2\text{x}}-\frac{2\sqrt{\text{y}^2-\text{x}^2}}{2\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\sqrt{\frac{\text{y}^2-\text{x}^2}{\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\sqrt{\frac{\text{y}^2}{\text{x}^2}-1}$
View full question & answer→Question 5305 Marks
Determine the values of a, b, c for which the function
$\text{f}\text{(x)}=\begin{cases}\frac{\sin\text{(a}+1)\text{x}+\sin\text{x}}{\text{x}}, &\text{for}\text{ x}<0,&\\\text{ c},&\text{for x}=0\\\frac{\sqrt{\text{x}+\text{bx}^2}-\sqrt{\text{x}}}{\text{bx}^\frac{3}{2}},&\text{for x}>0\end{cases}$ is continuous at x = 0.
AnswerThe given function can be rewritten as,
$\text{f}\text{(x)}=\begin{cases}\frac{\sin\text{(a}+1)\text{x}+\sin\text{x}}{\text{x}}, &\text{for}\text{ x}<0,&\\\text{ c},&\text{for x}=0\\\frac{\sqrt{\text{x}+\text{bx}^2}-\sqrt{\text{x}}}{\text{bx}^\frac{3}{2}},&\text{for x}>0\end{cases}$
$\Rightarrow\text{f}\text{(x)}=\begin{cases}\frac{\sin\text{(a}+1)\text{x}+\sin\text{x}}{\text{x}}, &\text{for}\text{ x}<0,&\\\text{ c},&\text{for x}=0\\\frac{\sqrt{1+\text{bx}}-1}{\text{bx}},&\text{for x}>0\end{cases}$
We observe
$\text{(LHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\Big[\frac{-\sin(\text{a}+1)\text{h}-\sin(-\text{h})}{\text{h}}\Big]=\lim\limits_{\text{h} \rightarrow 0}\Big[\frac{-\sin(\text{a}+1)\text{h}}{\text{h}}-\frac{\sin\text{h}}{\text{h}}\Big]$
$=-(\text{a}-1)\lim\limits_{\text{h} \rightarrow 0}\Big[\frac{-\sin(\text{a}+1)\text{h}}{(\text{a}+1)\text{h}}\Big]-\lim\limits_{\text{h} \rightarrow 0}\frac{\sin\text{h}}{\text{h}}=-\text{a}-1$
$\text{(RHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\sqrt{1+\text{bh}}-1}{\text{bh}}\Big)=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\text{bh}}{\text{bh}(\sqrt{1+\text{bh}}+1)}\Big)=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{1}{\sqrt{1+\text{bh}}+1}\Big)=\frac{1}{2}$
And, f(0) = c
If f(x) is continuous at x = 0, then
$=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x )}=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x )}=\text{f}(0)$
$\Rightarrow-\text{a}-1=\frac{1}{2}=\text{c}$
$\Rightarrow-\text{a}-1=\frac{1}{2}$ and $\text{c}=\frac{1}{2}$
$\Rightarrow\text{a}=\frac{-3}{2},\text{c}=\frac{1}{2}$
Now, $\frac{\sqrt{1+\text{bx}}-1}{\text{bx}}$ exists only if $\text{bx}\neq0\Rightarrow\text{b}\neq0.$
$\therefore\text{b}\in\text{R}-\{0\}$
View full question & answer→Question 5315 Marks
Verify the hypothesis and conclusion of Lagrange's mean value theorem for the function
$\text{f}(\text{x})=\frac{1}{4\text{x}-1},1\leq\text{x}\leq4.$
AnswerThe given function $\text{f}(\text{x})=\frac{1}{4\text{x}-1}.$ Clearly, f(x) is does not exist for x = 0 Since for each $\text{x}\in[1,4],$ the function attains a unique definite value, f(x) is continuous on [1, 4].Also, $\text{f}'(\text{x})=\frac{-4}{(4\text{x}-1)^2}$ exists for all $\text{x}\in[1,4],$
Thus, both the conditions of Lagrange's mean value theorem are verified. Concequently, there exists some $\text{c}\in[1,4]$ such that$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(1)}{4-1}$
$=\frac{\text{f}(4)-\text{f}(1)}{3}$ Now, $\text{f}(\text{x})=\frac{1}{4\text{x}-1}\Rightarrow\text{f}'(\text{x})=\frac{-4}{(4\text{x}-1)^2}$ $\text{f}(4)=\frac{1}{15},\text{f}(1)=\frac{1}{3}$ $\therefore\ \text{f}'(\text{x})=\frac{\text{f}(4)-\text{f}(1)}{4-1}$ $\Rightarrow\text{f}'(\text{x})=\frac{\frac{1}{15}-\frac{1}{3}}{4-1}==\frac{-4}{45}$ $\Rightarrow\frac{-4}{(4\text{x}-1)^2}=\frac{-4}{45}$ $\Rightarrow(4\text{x}-1)^2=45$ $\Rightarrow16\text{x}^2-8\text{x}-44=0$ $\Rightarrow4\text{x}^2-2\text{x}-11=0$ $\Rightarrow\text{x}=\frac{1}{4}\big(1+3\sqrt{5}\big)$ Thus, $\text{c}=\frac{1}{4}\big(1+3\sqrt{5}\big)\in(1,4)$ such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(1)}{4-1}.$ Hence, Lagrange's theorem is verified.
View full question & answer→Question 5325 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\sin\text{x}}+\big(\sin\text{x}\big)^\text{x}$
AnswerLet $\text{y}=\text{x}^{\sin\text{x}}+(\sin\text{x})^\text{x}$
Also, let $\text{u}=\text{x}^{\sin\text{x}}\text{ and v}=(\sin\text{x})^\text{x}$
$\therefore\text{y}=\text{u}+\text{v}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ .....(\text{i})$
$\text{u}=\text{x}^{\sin\text{x}}$
$\Rightarrow\log\text{u}=\log\big(\text{x}^{\sin\text{x}}\big)$
$\Rightarrow\log\text{u}=\sin\text{x}\log\text{x}$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin\text{x})\times\log\text{x}+\sin\text{x}\times\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[\cot\text{x}\log\text{x}+\sin\text{x}\times\frac{1}{\text{x}}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^{\sin\text{x}}\Big[\cos\text{x}\log\text{x}+\frac{\sin\text{x}}{\text{x}}\Big]\ .....(\text{ii})$
$\text{v}=(\sin\text{x})^\text{x}$
$\Rightarrow\log\text{v}=\log(\sin\text{x})^\text{x}$
$\Rightarrow\log\text{v}=\text{x}\log(\sin\text{x})$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x})\times\log(\sin\text{x})+\text{x}\times\frac{\text{d}}{\text{dx}}\big[\log(\sin\text{x})\big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\log(\sin\text{x})+\text{x}\times\frac{1}{\sin\text{x}}\times\frac{\text{d}}{\text{dx}}(\sin\text{x})\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\sin\text{x})^\text{x}\Big[\log\sin\text{x}+\frac{\text{x}}{\sin\text{x}}\cos\text{x}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}(\sin\text{x})^\text{x}\big[\log\sin\text{x}+\text{x}\cot\text{x}\big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\sin\text{x})^\text{x}\big[\log\sin\text{x}+\text{x}\cot\text{x}\big]\ .....(\text{iii})$
From (i), (ii) and (iii), we obtain
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin\text{x}}\Big(\cos\text{x}\log\text{x}+\frac{\sin\text{x}}{\text{x}}\big)+(\sin\text{x})^\text{x}\big[\log\sin\text{x}+\text{x}\cot\text{x}\big]$
View full question & answer→Question 5335 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big(\frac{\text{x}+\sqrt{1-\text{x}^3}}{\sqrt{2}}\Big),-1<\text{x}<1$
AnswerLet $\text{y}=\sin^{-1}\Big(\frac{\text{x}+\sqrt{1-\text{x}^3}}{\sqrt{2}}\Big)$
Put $\text{x}=\text{a}\sin\theta,\text{ So}$
$=\sin^{-1}\Big\{\frac{\sin\theta+\sqrt{1-\sin^2\theta}}{\sqrt{2}}\Big\}$
$=\sin^{-1}\Big\{\frac{\sin\theta+\cos\theta}{\sqrt{2}}\Big\}$
$=\sin^{-1}\Big\{\sin\theta\Big(\frac{1}{\sqrt{2}}\Big)+\cos\theta\Big(\frac{1}{\sqrt{2}}\Big)\Big\}$
$=\sin^{-1}\Big\{\sin\theta\cos\frac{\pi}{4}+\cos\theta\sin\frac{\pi}{4}\Big\}$
$\text{y}=\sin^{-1}\Big\{\sin\Big(\theta+\frac{\pi}{4}\Big)\Big\}\ .....(\text{i})$
Here, $-1<\text{x}<1$
$\Rightarrow\ -1<\sin\theta<1$
$\Rightarrow -\frac{\pi}{2}<\theta<\frac{\pi}{2}$
$\Rightarrow\Big(-\frac{\pi}{2}+\frac{\pi}{4}\Big)<\Big(\frac{\pi}{4}+\theta\Big)<\frac{3\pi}{4}$
So, from equation (i),
$\text{y}=\theta+\frac{\pi}{4}\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{ as }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\sin^{-1}\text{x}+\frac{\pi}{4} \big[\text{Since},\sin\theta=\text{x}\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}+0$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 5345 Marks
If $\text{xy}\log(\text{x}+\text{y})=1,$ prove that $\frac{\text{dx}}{\text{dx}}=-\frac{\text{y}(\text{x}^2\text{y}+\text{x}+\text{y})}{\text{x}(\text{xy}^2+\text{x}+\text{y})}$
AnswerHere,
$\text{xy }\log(\text{x}+\text{y})=1$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}\big[\text{xy}\log(\text{x}+\text{y})\big]=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\text{xy}\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})=0$
[Using chain rule and product rule]
$\Rightarrow\text{xy}\times\Big(\frac{1}{\text{x}+\text{y}}\Big)\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})(1)=0$
$\Rightarrow\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\Big(1+\frac{\text{dy}}{\text{dx}}\Big)+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})(1)=0$
$\Rightarrow\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)+\text{x}\Big(\frac{1}{\text{xy}}\Big)\frac{\text{dy}}{\text{dx}}+\text{y}\Big(\frac{1}{\text{xy}}\Big)=0$
$\Big[\text{Since from equation (i)}\log(\text{x}+\text{y})=\frac{1}{\text{xy}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}}{\text{x}+\text{y}}+\frac{1}{\text{y}}\Big]=-\Big[\frac{1}{\text{x}}+\frac{\text{xy}}{\text{x}+\text{y}}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}^2+\text{x}+\text{y}}{(\text{x}+\text{y})\text{y}}\Big]=-\Big[\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}(\text{x}+\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}=-\Big(\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}(\text{x}+\text{y})}\Big)\Big(\frac{(\text{x}+\text{y})\text{y}}{\text{xy}^2+\text{x}+\text{y}}\Big)$
$=-\frac{\text{y}}{\text{x}}\Big(\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}+\text{y}+\text{xy}^2}\Big)$
So,
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}\Big(\frac{\text{x}^2\text{y}+\text{x}+\text{y}}{\text{xy}^2+\text{x}+\text{y}}\Big)$
View full question & answer→Question 5355 Marks
Find the value of a and b so that the function f(x) defind by $\text{f(x)}=\begin{cases}\text{x}+\text{a}\sqrt{2}\sin\text{x},&\text{if }0\leq\text{x}<\frac{\pi}{4}\\2\text{x}\cot\text{ x}+\text{b},&\text{if }\frac{\pi}{4}\leq\text{x}<\frac{\pi}{2}\\\text{a}\cos2\text{x}-\text{b}\sin\text{x},&\text{if }\frac{\pi}{2}\leq\text{x}\leq\pi\end{cases}$ becomes continuous on $[0,\pi]$
AnswerGiven, f is continuous on $[0,\pi]$
$\therefore$ f is continuous at $\text{x }\frac{\pi}{4}$ and $\frac{\pi}{2}$
At $\text{x}=\frac{\pi}{4},$ we have
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{4}-\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[\Big(\frac{\pi}{4}-\text{h}\Big)+\text{a}\sqrt{2}\sin\Big(\frac{\pi}{4}-\text{h}\Big)\Big]\\=\Big[\frac{\pi}{4}+\text{a}\sqrt{2}\sin\Big(\frac{\pi}{4}\Big)\Big]=\Big[\frac{\pi}{4}+\text{a}\Big]$
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{4}+\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[2\Big(\frac{\pi}{4}+\text{h}\Big)\cot\Big(\frac{\pi}{4}+\text{h}\Big)+\text{b}\Big]\\=\Big[\frac{\pi}{2}\cot\Big(\frac{\pi}{4}\Big)+\text{b}\Big]=\Big[\frac{\pi}{2}+\text{b}\Big]$
At $\text{x}=\frac{\pi}{2},$ we have
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{2}-\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[2\Big(\frac{\pi}{2}-\text{h}\Big)\cot\Big(\frac{\pi}{2}-\text{h}\Big)+\text{b}\Big]=\text{b}$
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{2}+\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[\text{a}\cos2\Big(\frac{\pi}{2}+\text{h}\Big)-\text{b}\sin\Big(\frac{\pi}{2}+\text{h}\Big)\Big]=-\text{a}-\text{b}$
Since f is continuous at $\text{x}=\frac{\pi}{4}$ and $\text{x}=\frac{\pi}{2},$ we have
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}$ and $\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^+}\text{f(x)}$
$\Rightarrow-\text{b}-\text{a}=\text{b}$ and $\frac{\pi}{4}+\text{a}=\frac{\pi}{2}+\text{b}$
$\Rightarrow\text{b}=\frac{-\text{a}}{2}\ ...(\text{i})$ and $\frac{-\pi}{4}=\text{b}-\text{a}\ ...(\text{ii})$
$\Rightarrow\frac{-\pi}{4}=\frac{-3\text{a}}{2}$ [Substituting the value of b in eq. (ii)]
$\Rightarrow\text{a}=\frac{\pi}{6}$
$\Rightarrow\text{b}=\frac{-\pi}{12}$ [From eq. (i)]
View full question & answer→Question 5365 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{4\text{x}}{1-4\text{x}^2}\Big),-\frac{1}{2}<\text{x}<\frac{1}{2}$
AnswerLet $\text{y}=\tan^{-1}\Big\{\frac{4\text{x}}{1-4\text{x}^2}\Big\}$
Put $2\text{x}=\tan\theta,\text{ so}$
$\text{y}=\tan^{-1}\Big\{\frac{2\tan\theta}{1-\tan^2\theta}\Big\}$
$\text{y}=\tan^{-1}\{\tan2\theta\}\ .....(\text{i})$
Here, $-\frac{1}{2}<\text{x}<\frac{1}{2}$
$\Rightarrow -1<2\text{x}<1$
$\Rightarrow -1<\tan\theta<1$
$\Rightarrow -\frac{\pi}{4}<\theta<\frac{\pi}{4}$
$\Rightarrow -\frac{\pi}{2}<(2\theta)<\frac{\pi}{2}$
So, from equation (i),
$\text{y}=\theta\Big[\text{Since}, \tan^{-1}(\tan\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=2\tan^{-1}(2\text{x})\ \big[\text{Since}, 2\text{x}=\tan\theta\big]$
Differentiating ti with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=2\Big(\frac{1}{1+(2\text{x})^2}\Big)\frac{\text{d}}{\text{dx}}(2\text{x})$
$\frac{\text{dy}}{\text{dx}}=\frac{4}{1+4\text{x}^2}$
View full question & answer→Question 5375 Marks
If $\text{x}\sin(\text{a}+\text{y})+\sin\text{a}\cos(\text{a}+\text{y})=0,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
AnswerWe have, $\text{x}\sin(\text{a}+\text{y})+\sin\text{a}\cos(\text{a}+\text{y})=0$
Differentiate with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}\big[\text{x}\sin(\text{a}+\text{y})\big]+\frac{\text{d}}{\text{dx}}\big[\sin\text{a}\cos(\text{a}+\text{y})\big]=0$
$\Rightarrow\Big[\text{x}\frac{\text{d}}{\text{dx}}\sin(\text{a}+\text{y})+\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})\Big]+\sin\text{a}\frac{\text{d}}{\text{dx}}\cos(\text{a}+\text{y})=0$
$\Rightarrow\Big[\text{x}\cos(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{a}+\text{y})+\sin(\text{a}+\text{y})(1)\Big] \\ +\sin\text{a}\Big[-\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{a}+\text{y})\Big]=0$
$\Rightarrow\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}(\text{a}+\text{y})+\sin(\text{a}+\text{y})-\sin\text{a}\sin(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[\text{x}\cos(\text{a}+\text{y})-\sin\text{a}\sin(\text{a}+\text{y})\big] \\ =-\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big[-\sin\text{a}\frac{\cos^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}-\sin\text{a}\sin(\text{a}+\text{y})\Big] \\ =-\sin(\text{a}+\text{y})$
$\Big[\because\ \text{x}=-\sin\text{a}\frac{\cos(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}\Big]$
$ \Rightarrow-\frac{\text{dy}}{\text{dx}}\Big[\frac{\sin\text{a}\cos^2(\text{a}+\text{y})+\sin\text{a}\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}\Big] \\ =-\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})\Big[\frac{\sin(\text{a}+\text{y})}{\sin\text{a}\{\cos^2(\text{a}+\text{y})+\sin^2(\text{a}+\text{y})\}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
View full question & answer→Question 5385 Marks
Discuss the continuity and differentiability of $\text{f(x)}=\text{e}^{|\text{x}|}.$
AnswerGiven:
$\text{f(x)}=\text{e}^{|\text{x}|}$
$\Rightarrow\text{f(x)}=\begin{cases}\text{e}^\text{x},&\text{x}\geq0\\\text{e}^{-\text{x}},&\text{x}<0\end{cases}$
f is Continuity:
(LHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0{^-}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{-(0-\text{h})}$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{-\text{h}}$
$=1$
(RHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0{^{+}}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{(0+\text{h})}$
$=1$
and f(0)
$=\text{e}^0=1$
Thus, $\lim_\limits{\text{x}\rightarrow0^{-}}-\text{f(x)}=\lim_\limits{\text{h}\rightarrow0^{+}}-\text{f(x)}=\text{f(0)}$
Hence, function is continuous at x = 0.
Differentiability at x = 0.
(LHL at x = 0)
$=\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{0-\text{h}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{-(0-\text{h})}-1}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{-\text{h}}=-1\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\Big]$
(RHL at x = 0)
$=\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{0+\text{h}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{-(0-\text{h})}-1}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\Big]$
LHL at (x = 0) $\neq$ RHL at (x = 0)
Hence the function is not differentiable at x = 0.
View full question & answer→Question 5395 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\},0<\text{x}<1$
AnswerLet $\text{y}=\sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\}$
Put $\text{x}=\cos2\theta,\text{ So}$
$=\sin^{-1}\Big\{\frac{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}{2}\Big\}$
$=\sin^{-1}\Big\{\frac{\sqrt{2\cos^2\theta}+\sqrt{2\sin^2\theta}}{2}\Big\}$
$=\sin^{-1}\Big\{\frac{\sqrt{2}\cos\theta+\sqrt{2}\sin\theta}{2}\Big\}$
$=\sin^{-1}\Big\{\cos\theta\Big(\frac{1}{\sqrt{2}}\Big)+\Big(\frac{1}{\sqrt{2}}\Big)\sin\theta\Big\}$
$=\sin^{-1}\Big\{\cos\theta\sin\Big(\frac{\pi}{4}\Big)+\cos\frac{\pi}{4}\sin\theta\Big\}$
$\text{y}=\sin^{-1}\Big\{\sin\big(\theta+\frac{\pi}{4}\Big)\Big\}\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow 0<\cos2\theta<1$
$\Rightarrow 0<2\theta<\frac{\pi}{2}$
$\Rightarrow 0 < \theta < \frac{\pi}{4}$
$\Rightarrow \frac{\pi}{4}<\Big(\theta+\frac{\pi}{4}\Big)<\frac{\pi}{2}$
So from eqaution (i),
$\text{y}=\theta+\frac{\pi}{4}\ \Big[\text{Since}, \sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\text{y}=\frac{1}{2}\cos^{-1}\text{x}+\frac{\pi}{4}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)+0$
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\sqrt{1-\text{x}^2}}$
View full question & answer→Question 5405 Marks
Show that the function defined by $f(x) = \cos (x^2 )$ is a continuous function.
AnswerIt is given function is $\text{f(x)} = \cos(\text{x}^{2})$
This function f is defined for every real number and f can be written as the composition of two function as,
f = goh, where, $\text{g(x}) = \cos\text{x}\ \text {and}\ \text{h(x)} = \text{x}^{2}$
First we have to prove that $\text{g(x}) = \cos\text{x}\ \text {and}\ \text{h(x)} = \text{x}^{2}$ are continuous functions.
We know that g is defined for every real number.
Let k be a real number.
Then, g(k) = cos k
Now, put $x = k + h$
lf $x \rightarrow k$, then $h \rightarrow 0$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\cos\text{x}$
$= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{-0}}\cos (\text{k}+\text{h})$
$= ^{\text{lim}}_{\text{h}\rightarrow\text{0}}-[\cos\text{k}\cos\text{h} - \sin\text{k}.\sin\text{h}]$
$= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{-0}}\cos\text{k}\cos\text{h} -^{\ \ \text{lim}}_{\text{h}\rightarrow\text{-0}} \sin\text{k}\sin\text{h}$
$= \cos\text{k}\cos0 - \sin\text{k}\sin0$
$= \cos\text{k} \times 1 - \sin \times\ 0$
$= \cos\text{k}$
$\therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} =\text{g(k)}$
Thus, g(x) = cosx is continuous function.
Now, $h(x) = x^2$
So, h is defined for every real number.
Let c be a real number, then $h(c) = c^2$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{c}}\text{h(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{c}}\text{x}^{2}$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{c}}\text{h(x)} =\text{h(c)}$
Therefore, h is a continuous function.
We know that for real valued functions g and h, Such that (fog) is continuous at c.
Therefore, $f(x) = (goh)(x) = \cos(x^2)$ is a continuous function.
View full question & answer→Question 5415 Marks
If $\text{y}=\log\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}+\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{\sqrt{3}\text{x}}{1-\text{x}^2}\Big),$ find $\frac{\text{dy}}{\text{dx}}$
AnswerHere,
$\text{y}=\log\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}+\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{\sqrt{3}\text{x}}{1-\text{x}^2}\Big)$
Differentiating it with respect to x using chain rule and quotient rule,
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)+\frac{2}{\sqrt{3}}\frac{\text{d}}{\text{dx}}\tan^{-1}\Big(\frac{\sqrt{3}\text{x}}{1-\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)}\frac{\text{d}}{\text{dx}}{\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)}+\frac{2}{\sqrt{3}}\Bigg\{\frac{1}{1+\Big(\frac{\sqrt{3}\text{x}}{1-\text{x}^2}\Big)}\Bigg\}\frac{\text{d}}{\text{dx}}\Big(\frac{\sqrt{3}\text{x}}{1-\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)\bigg(\frac{(\text{x}^2-\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)-(\text{x}^2+\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{x}+1)}{(\text{x}^2-\text{x}+1)^2}\bigg) \\ +\frac{2}{\sqrt{3}}\Big\{\frac{(1-\text{x})^2}{1+\text{x}^4-2\text{x}^2+3\text{x}^2}\Big\}\bigg\{\frac{(1-\text{x}^2)^2\frac{\text{d}}{\text{dx}}(\sqrt{3\text{x}})-\sqrt{3}\text{x}\frac{\text{d}}{\text{dx}}(1-\text{x})^2}{(1-\text{x}^2)^2}\bigg\}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{1}{\text{x}^2-\text{x}+1}\Big)\Big(\frac{(\text{x}^2-\text{x}+1)(2\text{x}+1)-(\text{x}^2+\text{x}+1)(2\text{x}-1)}{(\text{x}^2-\text{x}+1)}\Big) \\ +\frac{2}{\sqrt{3}}\Big(\frac{(1-\text{x}^2)^2}{1+\text{x}^2+\text{x}^4}\Big)\Big(\frac{(1-\text{x}^2)(\sqrt{3})-\sqrt{3}\text{x}(-2\text{x})}{(1-\text{x}^2)^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{2\text{x}^3-2\text{x}^2+2\text{x}+\text{x}^2-\text{x}+1-2\text{x}^3-2\text{x}^2-2\text{x}+\text{x}^2+\text{x}+1}{\text{x}^4+2\text{x}^2+1-\text{x}^2}\Big) \\ +\frac{2}{\sqrt{3}}\Big(\frac{\sqrt{3}-\sqrt{3}\text{x}^2\sqrt{3}\text{x}^2}{1+\text{x}^2+\text{x}^4}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{-2\text{x}^2+2}{\text{x}^4+\text{x}^2+1}\Big)+\frac{2\sqrt{3}(\text{x}^2+1)}{\sqrt{3}(1+\text{x}^2+\text{x}^4)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2(1-\text{x}^2)}{(\text{x}^4+\text{x}^2+1)}+\frac{2(\text{x}^2+1)}{1+\text{x}^2+\text{x}^4}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2(1-\text{x}^2+\text{x}^2+1)}{1+\text{x}^2+\text{x}^4}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2+\text{x}^4}$
View full question & answer→Question 5425 Marks
Differentiate the following functions from first principles:
$\sin^{-1}(2\text{x}+3)$
AnswerLet $\text{f(x)}=\sin^{-1}(2\text{x}+3)$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\sin^{-1}(2(\text{x}+\text{h})+3)$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\sin^{-1}(2\text{x}+2\text{h}+3)$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}(2\text{x}+2\text{h}+3)-\sin^{-1}(2\text{x}+3)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}\Big[(2\text{x}+2\text{h}+3)\sqrt{1+(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\Big]}{\text{h}}$
$\Big[\text{Since}, \sin^{-1}\text{x}-\sin^{-1}\text{y}=\sin^{-1}\big[\text{x}\sqrt{1-\text{y}^2}-\text{y}\sqrt{1-\text{x}^2}\big]\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}\text{z}}{\text{z}}\times\frac{\text{z}}{\text{h}}$
Where, $\text{z}=(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}$
$\text{and }\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}\text{h}}{\text{h}}=1$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{z}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2\text{x}+2\text{h}+3)^2-(2\text{x}+3)^2-(2\text{x}+3)^2\big(1-(2\text{x}+2\text{h}+3)^2\big)}{\big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\big\}}$
[Since, rationalizing numerator]
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big[(2\text{x}+3)^2+4\text{h}^2+4\text{h}(2\text{x}+3)\big]\big(1-(2\text{x}+3)^2\big)-(2\text{x}+3)^2\big[1-(2\text{x}+3)^2-4\text{h}^2-4\text{h}(2\text{x}+3)\big]}{\text{h}\big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\big\}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\Big[(2\text{x}+3)^2+4\text{h}^2+4\text{h}(2\text{x}+3)-(2\text{x}+3)^4-4\text{h}^2(2\text{x}+3)^2-4\text{h}(2\text{x}+3)^3 -(2\text{x}+3)^2+(2\text{x}+3)^3+4\text{h}^2(2\text{x}+3)^2+4\text{h}(2\text{x}+3)^3\Big]}{\text{h}\big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\big\}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{4\text{h}\big[\text{h}+(2\text{x}+3)\big]}{\text{h}\Big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\Big\}}$
$=\frac{4\text{h}\big[\text{h}+(2\text{x}+3)\big]}{(2\text{x}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+3)^2}}$
$=\frac{4(2\text{x}+3)}{2(2\text{x}+3)\sqrt{1-(2\text{x}+3)^2}}$
$=\frac{2}{\sqrt{1-(2\text{x}+3)^2}}$
So,
$\frac{\text{d}}{\text{dx}}\big(\sin^{-1}(2\text{x}+3)\big)=\frac{2}{\sqrt{1-(2\text{x}+3)^2}}$
View full question & answer→Question 5435 Marks
Differentiate $\tan^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$ with respect to $\sin^{-1}\big(3\text{x}-4\text{x}^3\big),$ if $-\frac{1}{2}<\text{x}<\frac{1}{2}$
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$Put $\text{x}=\tan\theta$
$\Rightarrow \text{u}=\tan^{-1}\Big(\frac{\tan\theta-1}{\tan\theta+1}\Big)$
$\Rightarrow \text{u}=\tan^{-1}\bigg(\frac{\tan\theta-\tan\frac{\pi}{4}}{1+\tan\theta\tan\frac{\pi}{4}}\bigg)$
$\Rightarrow\text{u}\tan^{-1}\Big[\tan\big(\theta-\frac{\pi}{4}\big)\Big]\ .....(\text{i})$
Here, $-\frac{1}{2}<\text{x}<\frac{1}{2}$
$\Rightarrow-\frac{1}2{}<\tan\theta<\frac{1}{2}$
$\Rightarrow-\tan^{-1}\big(\frac{1}{2}\big)<\theta<\tan^{-1}\big(\frac{1}{2}\big)$
So, from equation (i)
$\text{u}=\theta-\frac{\pi}{4}$
$\Big[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\tan^{-1}\text{x}-\frac{\pi}{4} \ [\text{Since,x}= \tan\theta]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{1}{1+\text{x}^2}-0$
$\Rightarrow \frac{\text{du}}{\text{dx}}=\frac{1}{1+\text{x}^2}\ ..... \text{(ii)}$
And,
Let, $\text{v}=\sin^{-1}(3\text{x}-4\text{x}^3)$
Put $\text{x}=\sin\theta$
$\Rightarrow\text{v}=\sin^{-1}(3\sin\theta-4\sin^3\theta)$
$\Rightarrow\text{v}=\sin^{-1}(\sin3\theta)\ .....\text{(iii)}$
Now, $-\frac{1}{2}<\text{x}<\frac{1}{2}$
$\Rightarrow-\frac{1}{2}<\sin\theta<\frac{1}{2}$
$\Rightarrow-\frac{1}{6}<\theta<\frac{\pi}{6}$
So, from equation (iii),
$\text{v}=3\theta\Big[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\text{v}=3\sin^{-1}\text{x}[\text{Since, x}=\sin\theta]$
Dirrerentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{3}{\sqrt{1-\text{x}^2}}\ .....\text{(iv)}$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{1}{1+\text{x}^2}\times\frac{\sqrt{1-\text{x}^2}}{3}$
$\therefore\frac{\text{du}}{\text{dv}}=\frac{\sqrt{1-\text{x}^2}}{3(1+\text{x}^2)}$
View full question & answer→Question 5445 Marks
Differentiate $\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$ with respect to $\sqrt{1-4\text{x}^2},$ if:
$\text{x}\in\Big(-\frac{1}{2\sqrt{2}},\frac{1}{\sqrt{2\sqrt{2}}}\Big)$
AnswerLet $\text{u}=\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$
Put $2\text{x}=\cos\theta$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\times\cos\theta\sqrt{1-\cos^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\cos\theta\sin\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let, $\text{v}=\sqrt{1-4\text{x}^2}\ .....(\text{ii})$
Here,
$\text{x}\in\Big(-\frac{1}{2\sqrt{2}},\frac{1}{2\sqrt{2}}\Big)$
$\Rightarrow2\text{x}\in\Big(-\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}\Big)$
$\Rightarrow\theta\in\Big(\frac{\pi}{4},\frac{3\pi}{4}\Big)$
So, from equation (i),
$\text{u}=\pi=2\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\pi-\theta,\text{ if }\theta\in\Big(\frac{\pi}{2},\pi\Big)\Big]$
$\Rightarrow\text{u}=\pi-2\cos^{-1}(2\text{x})\big[\text{Since},2\text{x}=\cos\theta\big]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=0-2\bigg(\frac{-1}{\sqrt{1-(2\text{x})^2}}\bigg)\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2}{\sqrt{1-4\text{x}^2}}(2)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{4}{\sqrt{1-4\text{x}^2}}\ .....(\text{iii})$
From equation (ii)
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4\text{x}}{\sqrt{1-4\text{x}^2}}$
But, $\text{x}\in\Big(-\frac{1}{2},\frac{1}{2\sqrt{2}}\Big)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4(-\text{x})}{\sqrt{1-4(-\text{x})^2}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{4\text{x}}{\sqrt{1-4\text{x}^2}}\ .....(\text{iv})$
Diferentiating equation (ii) with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-4\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-4\text{x}^2)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-4\text{x}^2}}(-8\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4\text{x}}{\sqrt{1-4\text{x}^2}}\ .....(\text{v})$
Dividing equation (iii) by (v)
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{4}{\sqrt{1-4\text{x}^2}}\times\frac{\sqrt{1-4\text{x}^2}}{-4\text{x}}$
$\therefore\frac{\text{du}}{\text{dv}}=-\frac{1}{\text{x}}$
View full question & answer→Question 5455 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\log\text{x}}+(\log\text{x}^\text{x})$
AnswerLet $\text{y}=\text{x}^{\log\text{x}}+(\log\text{x}^\text{x})$
Also, let $\text{u}=(\log\text{x})^\text{x}\text{ and v}=\text{x}^{\log\text{x}}$
$\therefore\ \text{y}=\text{v}+\text{u}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{dv}}{\text{dx}}+\frac{\text{du}}{\text{dx}}\ .....(\text{i})$
Now, $\text{u}=(\log\text{x})^\text{x}$
$\Rightarrow\log\text{u}=\log\big[(\log\text{x})^\text{x}\big]$
$\Rightarrow\log\text{u}=\text{x}\log(\log\text{x})$
Differentiating both sides with respect to x,
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log(\log\text{x})\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{d}}{\text{dx}}\big[\log(\log\text{x})\big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[\log(\log\text{x})+\text{x}\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\log(\log\text{x})+\frac{\text{x}}{\log\text{x}}\times\frac{1}{\text{x}}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\log(\log\text{x})+\frac{1}{\log\text{x}}\Big]\ .....(\text{ii})$
Also, $\text{v}=\text{x}^{\log\text{x}}$
$\Rightarrow\log\text{v}=\log\text{x}^{\log\text{x}}$
$\Rightarrow\log\text{v}=\log\text{x}\log\text{x}=(\log\text{x})^2$
Differentiating both sides with respect to x,
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[(\log)^2\big]$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=2(\log\text{x})\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=2\text{v}(\log\text{x})\frac{1}{\text{x}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=2\text{x}^{\log\text{x}}\frac{\log\text{x}}{\text{x}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=2\text{x}^{\log\text{x}}\frac{\log\text{x}}{\text{x}}\ .....(\text{iii})$
From (i), (ii) and (iii), we obtain
$\frac{\text{dy}}{\text{dx}}=2\text{x}^{\log\text{x}}\frac{\log\text{x}}{\text{x}}+(\log\text{x})^\text{x}\Big[\log(\log\text{x})+\frac{1}{\log\text{x}}\Big]$
View full question & answer→Question 5465 Marks
If $\text{x}=\frac{\sin^3\text{t}}{\sqrt{\cos^2\text{t}}},\text{y}=\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}},$ find $\frac{\text{dy}}{\text{dx}}$
AnswerWe have, $\text{x}=\frac{\sin^{3}\text{t}}{\sqrt{\cos2\text{t}}}$ and $\text{y}=\frac{\cos^{3}\text{t}}{\sqrt{\cos2\text{t}}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\frac{\sin^{3}\text{t}}{\sqrt{\cos2\text{t}}}\Big] $
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}}\frac{\text{d}}{\text{dt}}(\sin^{3})-\sin^{3}\text{t}\frac{\text{d}}{\text{dt}}\sqrt{\cos2\text{t}}}{\cos2\text{t}}$
[Using quotient rule]
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}}(3\sin^{3}\text{t})\frac{\text{d}}{\text{dt}}(\sin\text{t})-\sin^{3}\times\frac{1}{2\sqrt{\cos2\text{t}}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{3\sqrt{\cos2\text{t}}(\sin^{2}\text{t}\cos\text{t})-\frac{\sin^{3}\text{t}}{2\sqrt{\cos2\text{t}}}(-2\sin2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{3\cos2\text{t}\sin^{\text{2}}\text{t}\cos\text{t}+\sin^{3}\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}$
Now, $\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\frac{\cos^{3}}{\sqrt{\cos2\text{t}}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}} \frac{\text{d}}{\text{dt}}(\cos^{3}\text{t})-\cos^{3}\text{t}\frac{\text{d}}{\text{dt}}\sqrt{\cos2\text{t}}}{\cos2\text{t}}$
[Using quotient rule]
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}}(3\cos^{2}\text{t})\frac{\text{d}}{\text{dt}}(\cos\text{t})-\cos^{3}\text{t}\times\frac{1}{2\sqrt{\cos2\text{t}}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{3\sqrt{\cos2\text{t}}\cos^{2}\text{t}-(\sin\text{t})-\frac{\cos^{3}\text{t}}{2\sqrt{\cos2\text{t}}}(-2\sin2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-3\cos2\text{t}\cos^{2}\text{t}+\cos^{3}\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}$
$\therefore\frac{\text{dy}}{\text{dt}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-3\cos2\text{t}\cos^{2}\text{t}\sin\text{t}+\cos^{3}\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}\times\frac{\cos2\text{t}\sqrt{\cos2\text{t}}}{3\cos2\text{t}\sin^{2}\text{t}\cos\text{t}+\sin^{3}\text{t}\sin2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{\sin\text{t}\cos{\text{t}}[-3\cos2\text{t}\cos\text{t}+2\cos^{3}\text{t}]}{\sin\text{t}\cos\text{t}[3\cos2\text{t}\sin{\text{t}}+2\sin^{3}\text{t}]}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{[-3(2\cos^{2}\text{t}-1)\cos\text{t}+2\cos^{3}\text{t}]}{[3(1-2\sin^{2}\text{t})\sin\text{t}+2\sin^{3}\text{t}]}$
$\begin{bmatrix} \cos2\text{t}=2\cos^2\text{t}-1 \\ \cos2\text{t}=1-2\sin^2\text{t} \end{bmatrix}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-4\cos^{3}\text{t}+3\cos\text{t}}{3\sin\text{t}-4\sin^{3}\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-\cos3\text{t}}{\sin3\text{t}}$
$\begin{bmatrix} \cos3\text{t}=4\cos^3\text{t}-3\cos\text{t} \\ \sin3\text{t}=3\sin^2\text{t}-4\sin^3\text{t} \end{bmatrix}$
$\therefore\frac{\text{dy}}{\text{dx}}=-\cos3\text{t}$
View full question & answer→