Question 13 Marks
If $\text{y}=\sqrt{\cos\text{x}+\sqrt{\cos\text{x}+\sqrt{\cos\text{x}+\ .... \text{to }\infty}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sin\text{x}}{1-2\text{y}}$
AnswerHere,
$\text{y}=\sqrt{\cos\text{x}+\sqrt{\cos\text{x}+\sqrt{\cos\text{x}+\ .... \text{to }\infty}}}$
$\text{y}=\sqrt{\cos\text{x}+\text{y}}$
Squaring both the sides,
$\text{y}^2=\cos\text{x}+\text{y}$
Differentiating it with respect to x,
$2\text{y}\frac{\text{dy}}{\text{dx}}=-\sin\text{x}+\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}(2\text{y}-1)=-\sin\text{x}$
$\frac{\text{dy}}{\text{dx}}=\frac{-\sin\text{x}}{(2\text{y}-1)}$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin\text{x}}{1-2\text{y}}$
View full question & answer→Question 23 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$
AnswerGiven,
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$
Differentiating with resepct to x,
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}\Big)=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2}{\text{a}^2}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{y}^2}{\text{b}^2}\Big)=0$
$\Rightarrow\frac{1}{\text{a}^2}(2{\text{x}})+\frac{1}{\text{b}^2}(2\text{y})\frac{\text{d}}{\text{dx}}=0$
$\Rightarrow\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=-\frac{2{\text{x}}}{\text{a}^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\Big(\frac{2{\text{x}}}{\text{a}^2}\Big)\Big(\frac{\text{b}^2}{2\text{y}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{{\text{b}^2\text{x}}}{\text{a}^2\text{y}}$
View full question & answer→Question 33 Marks
Differentiate the following functions with respect to x:
$3^{\text{x}^2+2\text{x}}$
AnswerConsider $\text{y}=3^{\text{x}^2+2\text{x}}$
Differentiate it with respect to x,
$\frac{\text{d}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(3^{\text{x}^2+2\text{x}}\Big)$
$=3^{\text{x}^2+2\text{x}}\times\log3\frac{\text{d}}{\text{dx}}(\text{x}^2+2\text{x})$
[Using chain rule]
$=(2\text{x}+2)\log3\times3^{\text{x}^2+2\text{x}}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}\big(3\text{x}^2+2\text{x}\big)=(2\text{x}+2)\log3\times3^{\text{x}^2+2\text{x}}$
View full question & answer→Question 43 Marks
If f(x) is an even function, then write whether f'(x) is even of odd.
AnswerHere,
f(x) is even function, so
f(-x) = f(x)
Differentiating it with respect to x,
$\frac{\text{d}}{\text{dx}}(\text{f}(-\text{x}))=\frac{\text{d}}{\text{dx}}(\text{f}(\text{x}))$
$\text{f}'(-\text{x})\frac{\text{d}}{\text{dx}}(-\text{x})=\text{f}'\text{(x)}$
$\text{f}'(-\text{x})\times(-1)=\text{f}'(\text{x})$
$-\text{f}'(-\text{x})=\text{f}'(\text{x})$
$\text{f}'(-\text{x})=-\text{f}'(\text{x})$
So,
f'(x) is odd function.
View full question & answer→Question 53 Marks
If $\text{y}=\text{x}\sin\text{y},$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\sin^2\text{y}}{(1-\text{x}\cos\text{y})}$
AnswerHere,
$\text{y}=\text{x}\sin\text{y}$
Differentiate with respect to x,
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{y})+\sin\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
[Using product rule]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\cos\frac{\text{dy}}{\text{dx}}+\sin\text{y}(1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(1-\text{x}\cos\text{y})=\sin\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin\text{y}}{1-\text{x}\cos\text{y}}$
View full question & answer→Question 63 Marks
If $e^{x+y} - x = 0$, prove that $\frac{\text{dy}}{\text{dx}}=\frac{1-\text{x}}{\text{x}}$
AnswerHere,
$e^{x+y} - x = 0$
$e^{x+y} = x$ .....(i)
Differentiating it with respect to x using chain rule,
$\frac{\text{d}}{\text{dx}}\big(\text{x}^{\text{x}+\text{y}}\big)=\frac{\text{d}}{\text{dx}}(\text{x})$
$\text{e}^{\text{x}+\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})=1$
$\text{x}\Big[1+\frac{\text{dy}}{\text{dx}}\Big]=1$
[Using euqation (i)]
$1+\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}-1$
$\frac{\text{dy}}{\text{dx}}=\frac{1-\text{x}}{\text{x}}$
View full question & answer→Question 73 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+\sec^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big),\text{x}\in\text{R}$
AnswerLet $\text{y}=\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+\sec^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
$\Rightarrow\text{y}=\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
$\Big[\text{Since},\sec^{-1}\text{x}=\cos^{-1}\big(\frac{1}{\text{x}}\big)\Big]$
$\Rightarrow \text{y}=\frac{\pi}{2}\Big[\text{Since}, \sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\Big]$
Differentiate it with respect to x,
$\therefore \frac{\text{dy}}{\text{dx}}=0$
View full question & answer→Question 83 Marks
If $y = x^x$, find $\frac{\text{dy}}{\text{dx}}\text{at x}=\text{e}$
AnswerWe have, $y = x^x$ .....(i)
Taking log on both sides,
$\log\text{y}=\log\text{x}^\text{x}$
$\Rightarrow\log\text{y}=\text{x}\log\text{x}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1+\log\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}(1+\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})$
[Using equation (i)]
Putting x = e, we get,
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{e}(1+\log_\text{e}\text{e})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^\text{e}(1+1)\big[\because\log_\text{e}\text{e}=1\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\text{e}^\text{e}$
View full question & answer→Question 93 Marks
Differentiate the following functions with respect to x:
$\cos(\log\text{ x})^2$
AnswerConsider $\text{y}=\cos(\log\text{ x})^2$
Differentiate it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\cos(\log\text{ x})^2$
$=-\sin(\log\text{x})^2\frac{\text{d}}{\text{dx}}(\log\text{ x})^2$
$=-\sin(\log\text{x})^2\frac{2\log\text{x}}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\frac{-2\log\text{x}\sin(\log\text{x})^2}{\text{x}}$
So, The solution is $\frac{\text{dy}}{\text{dx}}=\frac{-2\log\text{x}\sin(\log\text{x})^2}{\text{x}}$
View full question & answer→Question 103 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big)$
AnswerLet $\text{y}=\sin^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big)$
Put $\text{x}=\cot\theta$
$\text{y}=\sin^{-1}\Big(\frac{1}{\sqrt{1+\cot^2\theta}}\Big)$
$=\sin^{-1}\Big(\frac{1}{\sqrt{\text{cosec}^2\theta}}\Big)$
$=\sin^{-1}(\sin\theta)$
$=\theta$
$\text{y}=\cot^{-1}\text{x}\ [\text{Since}, \cot\theta=\text{x}]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{(1+\text{x}^2)}$
View full question & answer→Question 113 Marks
Differentiate the following functions with respect to x:
$10^{(10^\text{x})}$
AnswerLet $\text{y}=10^{(10^\text{x})}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log_\text{e}10^{(10^\text{x})}$
$\log\text{y}=10^{\text{x}}\log_\text{e}10$
Differentiating with respect to x,
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log_\text{e}10\times10^\text{x}\log_\text{e}10$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=10^\text{x}\times(\log_\text{e}10)^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\big[10^{\text{x}}\times(\log_\text{e}10)^2\big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=10^{(10\text{x})}\times10^\text{x}\times(\log_\text{e}10)^2$
[Using equation (i)]
View full question & answer→Question 123 Marks
If $f(0)=f(1)=0, f^{\prime}(1)=1$ and $y=f\left(e^x\right) e^{f(x)}$, write the value of $\frac{\text{dy}}{\text{dx}}\text{ at x} = 0.$
AnswerHere,
$f(0) = f(1) = 0, f'(1) = 2$
And, $y = f(e^x)d^{f(x)}$
Differentiating ti with respect to x using product rule, chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\text{f}(\text{e}^\text{x})\times\text{e}^{\text{f(x)}}\big]$
$=\text{f}(\text{e}^\text{x})\frac{\text{d}}{\text{dx}}\text{e}^{\text{f(x)}}+\text{e}^{\text{f(x)}}\frac{\text{d}}{\text{dx}}\text{f}(\text{e}^\text{x})$
$=\text{f}(\text{e}^\text{x})\text{e}^{\text{f(x)}}\frac{\text{d}}{\text{dx}}\text{f(x)}+\text{e}^{\text{f(x)}}\times\text{f}'(\text{e}^\text{x})\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})$
$=\text{f}(\text{e}^\text{x})\times\text{e}^{\text{f(x)}}\times\text{f}'\text{(x)}+\text{e}^{\text{f(x)}}+\text{f}'(\text{e}^\text{x})\times\text{e}^\text{x}$
Put x = 0
$=\text{f}(\text{e}^0)\text{e}^{\text{f}(0)}\text{f}'(0)+\text{e}^{\text{f}(0)}\text{f}^{1}(\text{e}^0)\times\text{e}^0$
$=\text{f}(1)\text{e}^{\text{f}(0)}\times\text{f}'(0)+\text{e}^{\text{f}(0)}\times\text{f}'(1)\times1$
$=0\times\text{e}^0\times\text{f}'(0)+\text{e}^02\times1$
$\big[\text{Since},\text{f}(0)=\text{f}(1)=0,\text{f}'(1)=2\big]$
$=0+1\times2\times1$
$=2$
So,
$\frac{\text{dy}}{\text{dx}}=2$
View full question & answer→Question 133 Marks
If $\text{y}=\log_\text{a}\text{x},$, find $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=\log_\text{a}\text{x},$
$\Rightarrow\text{y}=\frac{\log\text{x}}{\log\text{a}} \Big[\because\log_\text{a}\text{b}=\frac{\log\text{b}}{\log\text{a}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\log\text{a}}\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\log\text{a}}\Big(\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}\log\text{a}}$
View full question & answer→Question 143 Marks
If $\text{y}=\log|3\text{x}|,\text{x}\neq0,$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=\log|3\text{x}|$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\log|3\text{x}|)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{3\text{x}}\frac{\text{d}}{\text{dx}}(3\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{3\text{x}}(3)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
View full question & answer→Question 153 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big),\pi<\text{x}<\pi$
AnswerLet $\text{f(x)}=\tan^{-1}\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
This function is defined for all real numbers where $\cos\text{x}\neq1$
$\text{f(x)}=\tan^{-1}\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
$\Rightarrow\ \text{f(x)}=\tan^{-1}\Bigg[\frac{2\sin\big(\frac{\text{x}}{2}\big)\cos\big(\frac{\text{x}}{2}\big)}{2\cos^2\big(\frac{\text{x}}{2}\big)}\Bigg]$
$\Rightarrow\ \text{f(x)}=\tan^{-1}\big[\tan\big(\frac{\text{x}}{2}\big)\big]=\frac{\text{x}}{2}$
Thus, $\text{f'(x)}=\frac{\text{d}}{\text{dx}}\big(\frac{\text{x}}{2}\big)=\frac{1}{2}$
View full question & answer→Question 163 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\sin^{-1}2\text{x}}$
AnswerConsider $\text{y}=\text{e}^{\sin^{-1}2\text{x}}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sin^{-1}2\text{x}}\big)$
$=\text{e}^{\sin^{-1}2\text{x}}\times\frac{\text{d}}{\text{dx}}\big(\sin^{-1}2\text{x}\big)$
[Using chain rule]
$=\text{e}^{\sin^{-1}2\text{x}}\times\frac{1}{\sqrt{1-(2\text{x})^2}}\frac{\text{d}}{\text{dx}}(2\text{x})$
$=\frac{2\text{e}^{\sin^{-1}}2\text{x}}{\sqrt{1-4\text{x}^2}}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sin^{-1}}2\text{x}\big)=\frac{2\text{e}^{\sin^{-1}}2\text{x}}{\sqrt{1-4\text{x}^2}}$
View full question & answer→Question 173 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\text{e}^{\text{x}-\text{y}}=\log\Big(\frac{\text{x}}{\text{y}}\Big)$
AnswerWe have, $\tan^{-1}\big(\text{x}^2+\text{y}^2\big)=\text{a}$
Differentiating with respect to x, we get
$\frac{\text{d}}{\text{dx}}\big[\tan^{-1}\big(\text{x}^2+\text{y}^2\big)\big]=\frac{\text{d}}{\text{dx}}(\text{a})$
$\Rightarrow\frac{1}{1+(\text{x}^2+\text{y}^2)^2}\times\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{y}^2\big)=0$
$\Rightarrow\Big[\frac{1}{1+(\text{x}^2+\text{y}^2)^2}\Big]\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}}$
View full question & answer→Question 183 Marks
Differentiate the following functions with respect to x:
$\tan^2\text{x}$
AnswerLet,
$\text{y}=\tan^2\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=2\tan\text{x }\frac{\text{d}}{\text{dx}}(\tan\text{x})\ \big[\text{using chain rule}\big]$
$=2\tan\text{x}\times\sec^2\text{x}$
So,
$\frac{\text{d}}{\text{dx}}=\big(\tan^2\text{x}\big)=2\tan\text{x }\sec^2\text{x}.$
View full question & answer→Question 193 Marks
Differentiate the following functions with respect to x:
$\log(\tan^{-1}\text{x})$
AnswerConsider $\text{y}=\log\big(\tan^{-1}\text{x}\big)$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\big(\tan^{-1}\text{x}\big)$
$=\frac{1}{\tan^{-1}\text{x}}\times\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{x}\big)$
[Using chain rule]
$=\frac{1}{\big(1+\text{x}^2\big)\tan^{-1}\text{x}}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}\big(\log\tan^{-1}\text{x}\big)=\frac{1}{\big(1+\text{x}^2\big)\tan^{-1}\text{x}}$
View full question & answer→Question 203 Marks
Differentiate the following functions with respect to x:
$\log_7(2\text{x}-3)$
AnswerLet, $\text{y}=\log_7(2\text{x}-3)$
$\Rightarrow\ \text{y}=\frac{\log(2\text{x}-3)}{\log_7}\ \Big[\text{Since}, \log^\text{b}_\text{a}=\frac{\log\text{b}}{\log\text{a}}\Big]$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\log7}\frac{\text{d}}{\text{dx}}\big(\log(2\text{x}-3)\big)$
$=\frac{1}{\log7}\times\frac{1}{(2\text{x}-3)}\frac{\text{d}}{\text{dx}}(2\text{x}-3)$
[Using chain rule]
$=\frac{2}{(2\text{x}-3)\log7}$
Hence, $\frac{\text{d}}{\text{dx}}(\log_7(2\text{x}-3))=\frac{2}{(2\text{x}-3)\log7}$
View full question & answer→Question 213 Marks
Differentiate the following functions with respect to x:
$3^{\text{x}\log\text{x}}$
AnswerLet $\text{y}=3^{\text{x}\log\text{x}}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(3^{\text{x}\log\text{x}}\big)$
$=3^{\text{x}\log\text{x}}\times\log_\text{e}3\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})$
[Using chain rule]
$=3^{\text{x}\log\text{x}}\times\log_\text{e}3\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=3^{\text{x}\log\text{x}}\times\log_\text{e}3\Big[\frac{\text{x}}{\text{x}}+\log\text{x}\Big]$
$=3^{\text{x}\log\text{x}}\big[1+\log\text{x}\big]\times\log_\text{e}3$
So,
$\frac{\text{d}}{\text{dx}}\big(3^{\text{x}\log\text{x}}\big)=3^{\text{x}\log\text{x}}\big[1+\log\text{x}\big]\log_\text{e}3$
View full question & answer→Question 223 Marks
If $\text{y}=\sqrt{\text{x}+\sqrt{\text{x}+\sqrt{\text{x}+\ .... \text{to }\infty}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{1}{2\text{y}-1}$
AnswerWe have, $\text{y}=\sqrt{\text{x}+\sqrt{\text{x}+\sqrt{\text{x}+\ .... \text{to }\infty}}}$
Squaring both sides, we get,
$y^2 = x + y$
$\Rightarrow 2\text{y}\frac{\text{dy}}{\text{dx}}=1+\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(2\text{y}-1)=1$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\text{y}-1}$
View full question & answer→Question 233 Marks
Differentiate the following functions with respect to x:
$2^{\text{x}^3}$
AnswerConsider $\text{y}=2^{\text{x}^3}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(2^{\text{x}^3}\Big)$
$=2^{\text{x}^3}\times\log_2\frac{\text{d}}{\text{dx}}(\text{x}^3)$
[Using chain rule]
$=3\text{x}^2\times2^{\text{x}^3}\times\log_2$
Hence, the solution is $\frac{\text{d}}{\text{dx}}\big(2^{\text{x}^3}\big)=3\text{x}^2\times2^{\text{x}^3}\log_2$
View full question & answer→Question 243 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\tan3\text{x}}$
AnswerLet, $\text{y}=\text{e}^{\tan3\text{x}}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan3\text{x}}\big)$
$=\text{e}^{\tan3\text{x}}\frac{\text{d}}{\text{dx}}(\tan3\text{x})$
[Using chain rule]
$\text{e}^{\tan3\text{x}}\times\sec^23\text{x}\times\frac{\text{d}}{\text{dx}}(3\text{x})$
So,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan3\text{x}}\big)=3\text{e}^{\tan3\text{x}}\times\sec^2 3\text{x}$
View full question & answer→Question 253 Marks
If $\text{y}=\sqrt{\tan\text{x}+\sqrt{\tan\text{x}+\sqrt{\tan\text{x}+\ .... \text{to }\infty}}}$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sec^2\text{x}}{2\text{y}-1}$
AnswerWe have, $\text{y}=\sqrt{\tan\text{x}+\sqrt{\tan\text{x}+\sqrt{\tan\text{x}+\ .... \text{to }\infty}}}$
$\Rightarrow\text{y}=\sqrt{\tan\text{x}+\text{y}}$
Squaring both sides, we get,
$\text{y}^2=\tan\text{x}+\text{y}$
$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}=\sec^2\text{x}+\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(2\text{y}-1)=\sec^2\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sec^2\text{x}}{2\text{y}-1}$
View full question & answer→Question 263 Marks
If $\text{y}=\cos^{-1}(2\text{x})+2\cos^{-1}\sqrt{1-4\text{x}^2}, 0 <\text{x}<\frac{1}{2},$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\cos^{-1}(2\text{x})+2\cos^{-1}\sqrt{1-4\text{x}^2}$
Put $2\text{x}=\cos\theta$
$\therefore\ \text{y}=\cos^{-1}(\cos\theta)+2\cos^{-1}\sqrt{1-\cos^2\theta}$
$\Rightarrow \text{y}=\cos^{-1}(\cos\theta)+2\cos^{-1}(\sin\theta)$
$\therefore\ \text{y}=\cos^{-1}(\cos\theta)+2\cos^{-1}\Big[\cos\big(\frac{\pi}{2}\big)-\theta\Big]\ .....(\text{i})$
Here, $0<\text{x}<\frac{1}{2}$
$\Rightarrow 0<2\text{x}<1$
$\Rightarrow 0<\cos\theta<1$
$\Rightarrow 0<\theta<\frac{\pi}{2}$
And
$\Rightarrow 0> -\theta>-\frac{\pi}{2}$
$\Rightarrow\ \frac{\pi}{2}>\big(\frac{\pi}{2}-\theta\big)>0$
View full question & answer→Question 273 Marks
Differentiate the following functions with respect to x:
$\log(\cos\text{x}^2)$
AnswerConsider $\text{y}=\log(\cos\text{x}^2)$
Differentiate it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log(\cos\text{x}^2)$
$=\frac{-2\text{x}\sin\text{x}^2}{\cos\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=-2\text{x}\tan\text{x}^2$
Therefore,
$\frac{\text{dy}}{\text{dx}}=-2\text{x}\tan\text{x}^2$
View full question & answer→Question 283 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$xy = c^2$
AnswerWe have, $xy = c^2$
Differentiating with respect to x, we get
$\frac{\text{d}}{\text{dx}}(\text{xy})=\frac{\text{d}}{\text{dx}}(\text{c}^2)$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})=0$
[Using product rule]
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=-\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}$
View full question & answer→Question 293 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\tan\text{x}}$
AnswerLet $\text{y}=\text{e}^{\tan \text{x}}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\text{e}^{\tan\text{x}}\Big)$
$=\text{e}^{\tan\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x})$
[Using chain rule]
$=\text{e}^{\tan\text{x}}\times\sec^2\text{x} $
So, $\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan\text{x}}\big)=\sec^2\text{xe}^{\tan\text{x}}$
View full question & answer→Question 303 Marks
If $\text{x}=3\sin\text{t}-\sin3\text{t},\text{y}=3\cos3\text{t}-\cos3\text{t}$ find $\frac{\text{dy}}{\text{dx}}\text{ at t}=\frac{\pi}{3}$
Answer$\text{x}=3\sin\text{t}-\sin3\text{t and } \text{y}=3\cos3\text{t}-\cos3\text{t}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=3\cos\text{t}-3\cos3\text{t}\text{ and} \\ \frac{\text{dy}}{\text{dt}}=-3\sin\text{t}-3\sin3\text{t}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-3\sin\text{t}+3\sin3\text{t}}{3\cos\text{t}-3\cos3\text{t}}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{t}=\frac{\pi}{3}}=\frac{-3\sin\frac{\pi}{3}+3\sin\pi}{3\cos\frac{\pi}{3}-3\cos\pi}$
$=\frac{3\times\frac{\sqrt{3}}{2}+0}{3\times\frac{1}{2}+3}$
$=\frac{\frac{-3\sqrt{3}}{2}}{\frac{9}{2}}$
$=-\frac{1}{\sqrt{3}}$
View full question & answer→Question 313 Marks
Write the derivative of $\sin\text{x}$ with respect to $\cos\text{x}$.
AnswerLet $\text{u}=\sin\text{x}\text{ and v}=\cos\text{x}$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\cos\text{x and }\frac{\text{dv}}{\text{dx}}=-\sin\text{x}$
$\therefore\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{\cos\text{x}}{-\sin\text{x}}$
$\Rightarrow\frac{\text{du}}{\text{dv}}=-\cot\text{x}$
View full question & answer→Question 323 Marks
If $\text{y}=\sec^{-1}\Big(\frac{\text{x}+1}{\text{x}-1}\Big)+\sin^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big),\text{x}>0.$ Find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\sec^{-1}\Big(\frac{\text{x}+1}{\text{x}-1}\Big)+\sin^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$
$\text{y}=\cos^{-1}\Big(\frac{\text{x}+1}{\text{x}-1}\Big)+\sin^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$
$\Big[\text{Since, } \sec^{-1}(\text{x})=\cos^{-1}\Big(\frac{1}{\text{x}}\Big)\Big]$
$\text{y}=\frac{\pi}{2} \Big[\text{Since}, \cos^{-1}\text{x}+\sin^{-1}=\frac{\pi}{2}\Big]$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=0$
View full question & answer→Question 333 Marks
If $\text{x}=\frac{1+\log\text{t}}{\text{t}^2},\text{y}=\frac{3+2\log\text{t}}{\text{t}},$ find $\frac{\text{dy}}{\text{dx}}$
Answer$\text{x}=\frac{1+\log\text{t}}{\text{t}^2},\text{y}=\frac{3+2\log\text{t}}{\text{t}}$
$\frac{\text{dy}}{\text{dt}}=\frac{\text{t}^2\big(\frac{1}{\text{t}}\big)-(1+\log\text{t})(2\text{t})}{\text{t}^4} \\ =\frac{\text{t}-2\text{t}-2\text{t}\log\text{t}}{\text{t}^4}=\frac{-2\log\text{t}-1}{\text{t}^3}$
$\frac{\text{dy}}{\text{dt}}=\frac{\text{t}\big(\frac{2}{\text{t}}\big)-(3+2\log\text{t})(1)}{\text{t}^2} \\ =\frac{2-3-2\log\text{t}}{\text{t}^2}=\frac{-2\log\text{t}-1}{\text{t}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\frac{-2\log\text{t}-1}{\text{t}^2}}{\frac{-2\log\text{t}-1}{\text{t}^3}}=\text{t}$
View full question & answer→Question 343 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\tan^{-1}\big(\text{x}^2+\text{y}^2\big)=\text{a}$
AnswerWe have, $\tan^{-1}\big(\text{x}^2+\text{y}^2\big)=\text{a}$Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}\big[\tan^{-1}\big(\text{x}^2+\text{y}^2\big)\big]=\frac{\text{d}}{\text{dx}}(\text{a})$
$\Rightarrow\frac{1}{1+(\text{x}^2+\text{y}^2)^2}\times\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{y}^2\big)=0$
$\Rightarrow\Big[\frac{1}{1+(\text{x}^2+\text{y}^2)^2}\Big]\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}}$
View full question & answer→Question 353 Marks
Let $g(x)$ be the inverse of an invertible function $f(x)$ which is derivable at $x = 3$. If $f(3) = 3$ and $f(3) = 9$, write the value of $g'(9)$.
AnswerWe have,$ f(3) = 9, f'(3) = 9$
and $g(x) = f^{-1} (x)$
$\Rightarrow(\text{gof})(\text{x})=\text{x}$
$\Rightarrow\text{g}\{\text{f(x)}\}=\text{x}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big[\text{g}\{\text{f}\text{(x)}\}\big] = 1$
$\Rightarrow\text{g}'\big\{\text{f}\text{(x)}\big\}\frac{\text{d}}{\text{dx}}\big\{\text{f}(\text{x})\big\}=1$
$\Rightarrow\text{g}'\big\{\text{f}\text{(x)}\big\}\times\text{f}'\text{(x)}=1$
Putting x = 3, we get,
$\text{g}'\big\{\text{f}(3)\big\}\times\text{f}'(3)=1$
$\Rightarrow\text{g}'(9)\times9=1\big[\because\text{f}(3)=9,\text{f}'(3)=9\big]$
$\Rightarrow\text{g}'(9)=\frac{1}{9}$
View full question & answer→Question 363 Marks
If f(x) is an odd function, then write whether f'(x) is even of odd.
AnswerWe have, f(x) is an odd function.
$\Rightarrow\text{f}(-\text{x})=-\text{f}(\text{x})$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big\{\text{f}(-\text{x})\big\}=-\frac{\text{d}}{\text{dx}}\big\{\text{f}(\text{x})\big\}$
$\Rightarrow\text{f}'(-\text{x})\frac{\text{d}}{\text{dx}}(-\text{x})=-\text{f}'\text{(x)}$
$\Rightarrow\text{f}'(-\text{x})\times(-1)=-\text{f}'\text{(x)}$
$\Rightarrow\text{f}'(-\text{x})=-\text{f}'\text{(x)}$
$\Rightarrow\text{f}'(-\text{x})=\text{f}'\text{(x)}$
Thus, f'(x) is an even function.
View full question & answer→Question 373 Marks
Differentiate the following functions with respect to x:
$\sin(\log\text{x})$
AnswerConsider $\text{y}=\sin(\log\text{x})$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin(\log\text{x})$
$=\cos(\log\text{x})\frac{\text{d}}{\text{dx}}(\log\text{x})$
[Using chain rule]
$=\frac{1}{\text{x}}\cos(\log\text{x})$
Hence, the solution is $\frac{\text{d}}{\text{dx}}=(\sin(\log\text{x}))=\frac{1}{\text{x}}\cos(\log\text{x})$
View full question & answer→Question 383 Marks
If $\text{y}=\log\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{2\text{x}(\text{x}+1)}$
AnswerWe have, $\text{y}=\log\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
$=\frac{1}{\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}}\frac{\text{d}}{\text{dx}}\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
$=\frac{\sqrt{\text{x}}}{\text{x}+1}\Big(\frac{1}{2\sqrt{\text{x}}}-\frac{1}{2\text{x}\sqrt{\text{x}}}\Big)$
$=\frac{1}{2}\frac{\sqrt{\text{x}}}{\text{x}+1}\Big(\frac{\text{x}-1}{\text{x}\sqrt{\text{x}}}\Big)$
$=\frac{\text{x}-1}{2\text{x}(\text{x}+1)}$
So,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{2\text{x}(\text{x}+1)}$
View full question & answer→Question 393 Marks
Differentiate the following functions from first principles:
$e^{-x}$.
AnswerConsider $f(x) = e^{-x}$
$\Rightarrow f(x + h) = e^{-(x+h)}$
$\frac{\text{d}}{\text{dx}}(\text{f}(\text{x}))=\frac{\lim}{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\frac{\lim}{\text{h}\rightarrow0}\frac{\text{e}^{-(\text{x}+\text{h})}\text{e}^{-\text{x}}}{\text{h}}$
$=\frac{\lim}{\text{h}\rightarrow0}\frac{\text{e}^{-\text{x}}\times\text{e}^{-\text{h}}-\text{e}^{-\text{x}}}{\text{h}}$
$=\frac{\lim}{\text{h}\rightarrow0}\text{e}^{-\text{x}}\left\{\Big(\frac{\text{e}^{-\text{h}}-1}{-\text{h}}\Big)\right\}\times(-1)$
$\Big[\text{Since, }\frac{\lim}{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\Big]$
$=-\text{e}^{-\text{x}}$
So,
$\frac{\text{d}}{\text{dx}}(\text{e}^{-\text{x}})=-\text{e}^{-\text{x}}$
View full question & answer→Question 403 Marks
Differentiate $\log(1+\text{x}^2)$ with respect to $\tan^{-1}\text{x}$
AnswerLet $\text{u}=\log(1+\text{x}^2)$
Differentiating it with respect to x using chain rule,
$\frac{\text{du}}{\text{dx}}=\frac{1}{(1+\text{x}^2)}\frac{\text{d}}{\text{dx}}(1+\text{x}^2)$
$=\frac{1}{(1+\text{x}^2)}(2\text{x})$
$\frac{\text{du}}{\text{dx}}=\frac{2\text{x}}{(1+\text{x}^2)}\ .....(\text{i})$
Let $\text{v}=\tan^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{1+\text{x}^2}\ .....(\text{ii})$
Dividing equation (i) by (ii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2\text{x}}{(1+\text{x}^2)}\times\frac{(1+\text{x}^2)}{1}$
$\frac{\text{du}}{\text{dx}}=2\text{x}$
View full question & answer→Question 413 Marks
Differentiate the following functions with respect to x:
$\log\Big\{\cot\Big(\frac{\pi}{4}+\frac{\pi}{2}\Big)\Big\}$
Answer$\frac{\text{d}}{\text{dx}}\Big[\log\Big\{\cot\Big(\frac{\Pi}{4}+\frac{\pi}{2}\Big)\Big\}\Big]$
$\frac{1}{\cot\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)}\times\Big(-\text{cosec}^2\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big)\times\frac{1}{2}$
$\frac{-1}{2\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\sin\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)}=-\frac{1}{\sin2\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)}$
$=-\frac{1}{\sin2\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)}=-\frac{1}{\cos\text{x}}=-\sec\text{x}$
View full question & answer→Question 423 Marks
Differentiate the following functions with respect to x:
$\tan(\text{e}^{\sin\text{x}})$
AnswerConsider $\text{y}=\tan(\text{e}^{\sin\text{x}})$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\tan\text{e}^{\sin\text{x}}\big]$
$=\sec^2\big(\text{e}^{\sin\text{e}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sin\text{x}}\big)$
[Using chain rule]
$=\sec^2\big(\text{e}^{\sin\text{x}}\big)\times\text{e}^{\sin\text{x}}\times\frac{\text{d}}{\text{dx}}(\sin\text{ x})$
View full question & answer→Question 433 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}(\text{e}^{\text{x}})$
AnswerConsider $\text{y}=\tan^{-1}(\text{e}^{\text{x}})$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{e}^{\text{x}}\big)$
$=\frac{1}{1+\big(\text{e}^{2\text{x}}\big)^2}=\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\big)$
[using chain rule]
$=\frac{1}{1+\text{e}^{2\text{x}}}\times\text{e}^\text{x}$
$=\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{e}^\text{x}\big)=\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}$
View full question & answer→Question 443 Marks
Differentiate:$\tan(\text{x}^\circ+45^\circ)$
AnswerLet, $\text{y}=\tan(\text{x}^\circ+45^\circ)$
$\Rightarrow \text{y}=\tan\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\tan\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}$
$=\sec^2\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}\times\frac{\text{d}}{\text{dx}}(\text{x}+45)\frac{\pi}{180}$
[Using chain rule]
$=\frac{\pi}{180}\sec^2(\text{x}^\circ+45^\circ)$
So,
$\frac{\text{d}}{\text{dx}}\big\{\tan(\text{x}^\circ+45^\circ)\big\}=\frac{\pi}{180}\sec^2(\text{x}^\circ+45^\circ)$
View full question & answer→Question 453 Marks
Differentiate the following functions with respect to x:
$3^{\text{e}^{\text{x}}}$
AnswerLet, $\text{y}=3^{\text{e}^{\text{x}}}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(3^{\text{e}^\text{x}}\Big)$
$=3^{\text{e}^\text{x}}\log3\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\big)$
[Using chain rule]
$=\text{e}^\text{x}\times3^{\text{e}^\text{x}}\log3 $
So,
$\frac{\text{d}}{\text{dx}}\Big(3^{\text{e}^\text{x}}\Big)=\text{e}^\text{x}\times3^{\text{e}^\text{x}}\log3$
View full question & answer→Question 463 Marks
Differentiate the following functions with respect to x:
$\tan(\text{x}^\circ+45^\circ)$
AnswerLet, $\text{y}=\tan(\text{x}^\circ+45^\circ)$
$\Rightarrow\text{y}=\tan\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}$
Differentiating it with respect to x we get,
$\frac{\text{dx}}{\text{dy}}=\frac{\text{d}}{\text{dx}}\tan\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}$
$=\sec^2\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}\times\frac{\text{d}}{\text{dx}}(\text{x}+45)\frac{\pi}{180}$
[Using chain rule]
$=\frac{\pi}{180}\sec^2(\text{x}^\circ+45^\circ)$
So,
$=\frac{\text{d}}{\text{dx}}\Big\{\tan(\text{x}^\circ+45^\circ)\Big\}=\frac{\pi}{180}\sec^2(\text{x}^\circ+45^\circ)$
View full question & answer→Question 473 Marks
Differentiate the following functions with respect to x:
$\sin^2(2\text{x}+1)$
AnswerCobnsider $\text{y}=\sin^2(2\text{x}+1)$
Differentiate it with respect to x,
$\frac{\text{d}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\sin^2(2\text{x}+1)\big]$
$=2\sin(2\text{x}+1)\frac{\text{d}}{\text{dx}}\sin(2\text{x}+1)$
[Using chain rule]
$=2\sin(2\text{x}+1)\cos(2\text{x}+1)\frac{\text{d}}{\text{dx}}\sin(2\text{x}+1)$
[Using chain rule]
$=4\sin(2\text{x}+1)\cos(2\text{x}+1)$
$=2\sin(2\text{x}+1)$
$\Big[\text{Since}, \sin^2\text{A}=2\sin\text{A}\cos\text{A}\Big]$
$2\sin(4\text{x}+2)$
Hence, the solution is $\frac{\text{d}}{\text{dx}}\big(\sin^2(2\text{x}+1)\big)=2\sin(4\text{x}+2)$
View full question & answer→Question 483 Marks
If $\text{y}=\log\sqrt{\tan\text{x}},$ write $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=\log\sqrt{\tan\text{x}}$
$\Rightarrow\text{y}=\log(\tan\text{x})^\frac{1}{2}$
$\Rightarrow\text{y}=\frac{1}{2}\log(\tan\text{x})\big[\because\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\times\frac{1}{\tan\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\times\frac{1}{\tan\text{x}}(\sec^2\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\frac{\sin\text{x}}{\cos\text{x}}}\times\cos^2\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sin\text{x}\cos\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\sin2\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{cosec }2\text{x}$
View full question & answer→Question 493 Marks
Differentiate the following functions with respect to x:
$\tan5\text{x}^\circ$
AnswerLet, $\text{y}=\tan5\text{x}^\circ$
$\Rightarrow\ \text{y}=\tan\Big(5\text{x}\times\frac{\pi}{180}\Big)$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\tan\Big(5\text{x}\times\frac{\pi}{180}\Big)$
$=\sec^2\Big(5\text{x}\times\frac{\pi}{180}\Big)\frac{\text{d}}{\text{dx}}\Big(5\text{x}\times\frac{\pi}{108}\Big) $
[Using chain rule]
$=\Big(\frac{5\text{x}}{180}\Big)\sec^2\Big(5\text{x}\times\frac{\pi}{180}\Big)$
$=\frac{5\pi}{180}\sec^2(5\text{x}^\circ)$
Hence, $\frac{\text{d}}{\text{dx}}(\tan5\text{x}^\circ)=\frac{5\pi}{180}\sec^2(5\text{x}^\circ)$
View full question & answer→Question 503 Marks
Differentiate the following functions from first principles:
$e^{3x}$.
AnswerLet $f(x) = e^{3x}$
$\Rightarrow f(x + h) = e^{3(x + h)}$
$\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{3(\text{x}+\text{h})}-\text{e}^{3\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{3\text{x}}\text{e}^{3\text{h}}-\text{e}^{3\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{e}^{3\text{x}}\left\{\frac{(\text{e}^{3\text{h}}-1)}{3\text{h}}\right\}\times3$
$=3\text{e}^{3\text{x}}\Big[\text{Since, }\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\Big]$
Hence,
$\frac{\text{d}}{\text{dx}}(\text{e}^{3\text{x}})=3\text{e}^{3\text{x}}$
View full question & answer→Question 513 Marks
If $\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big),$ write the value of $\frac{\text{dy}}{\text{dx}}\text{ for x}>1.$
AnswerWe have, $\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
Putting $\text{x}=\tan\theta$
$\Rightarrow 1 <\tan\theta<\infty$
$\Rightarrow\frac{\pi}{4}<\theta<\frac{\pi}{2}$
$\frac{\pi}{2}<2\theta<\pi$
$\therefore\text{y}=\sin^{-1}(\sin2\theta)$
$\Rightarrow\text{y}=\sin^{-1}\big\{\sin(\pi-2\theta)\big\}$
$\Rightarrow\text{y}=\pi-2\theta$
$\Rightarrow\text{y}=\pi-2\tan^{-1}\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=0-\frac{2}{1+\text{x}^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{2}{1+\text{x}^2}$
View full question & answer→Question 523 Marks
Differentiate the following functions with respect to x:
$\sin(3\text{x}+5)$
AnswerConsider $\text{y}=\sin(3\text{x}+5)$
Differentiate y with the respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\sin(3\text{x}+5)\big)$
$=\cos(3\text{x}+5)\frac{\text{d}}{\text{dx}}(3\text{x}+5)$
[using chain rule]
$=\cos(3\text{x}+5)\times[3(1)+0]$
$=3\cos(3\text{x}+5)$
Hence, the solution is $\frac{\text{d}}{\text{dx}}(\sin(3\text{x}+5))=3\cos(3\text{x}+5)$
View full question & answer→Question 533 Marks
If $\text{x}=\text{a}(\theta-\sin\theta)\text{ and},\text{y}=\text{a}(1+\cos\theta),$ find $\frac{\text{dy}}{\text{dx}}\text{ at }\theta=\frac{\pi}{3}$
AnswerHere,
$\text{x}=\text{a}(\theta-\sin\theta)\text{ and y}=\text{a}(1+\cos\theta)$
Then,
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\big[\text{a}(\theta-\sin\theta)\big]=\text{a}(1-\cos\theta)$
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\big[\text{a}(1+\sin\theta)\big]=\text{a}(1-\sin\theta)$
$\therefore\frac{\text{dy}}{\text{dx}}=\Bigg[\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\text{a}\sin\theta}{\text{a}(1-\cos\theta)}\Bigg]_{\theta=\frac{\pi}{3}}$
$=-\frac{\sin\frac{\pi}{2}}{1-\cos\frac{\pi}{3}}=\frac{\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}=-\sqrt{3}$
View full question & answer→Question 543 Marks
If $\text{y}=\sqrt{\log\text{x}+\sqrt{\log\text{x}+\sqrt{\log\text{x}+\ .... \text{to }\infty}}},$ prove that $(2\text{y}-1)\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
AnswerWe have, $\text{y}=\sqrt{\log\text{x}+\sqrt{\log\text{x}+\sqrt{\log\text{x}+\ .... \text{to }\infty}}}$
$\Rightarrow\text{y}=\sqrt{\log\text{x}+\text{y}}$
Squaring both sides, we get,
$\text{y}^2=\log\text{x}+\text{y}$
$=2\text{y}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}+\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(2\text{y}-1)=\frac{1}{\text{x}}$
View full question & answer→Question 553 Marks
Differentiate the following functions with respect to x:
$(\log\sin\text{x})^2$
AnswerLet $\text{y}=(\log\sin\text{x})^2$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\log\sin\text{x})^2$
$=2(\log\sin\text{x})\frac{\text{d}}{\text{dx}}(\log\sin\text{x})$
$=2(\log\sin\text{x})\times\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})$
$=2(\log\sin\text{x})\times\frac{1}{\sin\text{x}}\cos\text{x}$
$=2(\log\sin\text{x})\cot\text{x}$
So,
$\frac{\text{d}}{\text{dx}}(\log\sin\text{x})^2=2(\log\sin\text{x})\cot\text{x}$
View full question & answer→Question 563 Marks
Differentiate the following functions with respect to x:
$\sin(\log\sin\text{x})$
AnswerConsider $\text{y}=\sin(\log\sin\text{x})$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin(\log\sin\text{x})$
$=\cos(\log\sin\text{x})\frac{\text{d}}{\text{dx}}(\log\sin\text{x})$
[Using chain rule]
$=\cos(\log\sin\text{x})\times\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}0\sin\text{x}$
$=\cos(\log\sin\text{x})\frac{\cos\text{x}}{\sin\text{x}}$
$=\cos(\log\sin\text{x})\times\cot\text{x}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}(\sin(\log\sin\text{x}))=\cos(\log\sin\text{x})\text{x}\cot\text{x}$
View full question & answer→Question 573 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{a}+\text{x}}{1-\text{ax}}\Big)$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\text{a}+\text{x}}{1-\text{ax}}\Big)$
$\text{y}=\tan^{-1}\text{a}+\tan^{-1}\text{x}$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{a})+\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{x})$
$=0+\frac{1}{1+\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$
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Differentiate the following functions with respect to x:
$\text{e}^{\sin\sqrt{\text{x}}}$
AnswerLet, $\text{y}=\text{e}^{\sin\sqrt{\text{x}}}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sin\sqrt{\text{x}}}\big)$
$=\text{e}^{\sin\sqrt{\text{x}}}\frac{\text{d}}{\text{dx}}\big(\sin\sqrt{\text{x}}\big)$
[Using chain rule]
$=\text{e}^{\sin\sqrt{\text{x}}}\times\cos\sqrt{\text{x}}\frac{\text{d}}{\text{dx}}\sqrt{\text{x}}$
[Using chain rule]
$=\text{e}^{\sin\sqrt{\text{x}}}\times\cos\sqrt{\text{x}}\times\frac{1}{2\sqrt{\text{x}}}$
$=\frac{1}{2\sqrt{\text{x}}}\times\cos\sqrt{\text{x}}\times\text{e}^{\sin\sqrt{\text{x}}}$
So,
$\frac{\text{d}}{\text{dx}}=\big(\text{e}^{\sin^\sqrt{\text{x}}}\big)=\frac{1}{2\sqrt{\text{x}}}\cos\sqrt{\text{x}}\times\text{e}^{\sin\sqrt{\text{x}}}$
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