Class 12 NCERT solution math exercise 5.5 differentiation

The Class 12 NCERT solution math exercise 5.5 differentiation prepared by expert Mathematics teacher at gmath.in as per CBSE guidelines. See our Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Questions with Solutions to help you to revise complete Syllabus and Score More marks in your Board and School exams.

EXERCISE 5.5 ( Differentiation )

Differentiate the function with respect to (Class 12 ncert solution math exercise 5.5 differentiation)

Question 1: $x: \cos x \cdot \cos 2 x \cdot \cos 3 x$

Solution: Let $y=\cos x \cdot \cos 2 x \cdot \cos 3 x$

Taking logarithm on both the sides, we obtain

$\log y=\log (\cos x \cdot \cos 2 x \cdot \cos 3 x)$

$\Rightarrow \log y=\log (\cos x)+\log (\cos 2 x)+\log (\cos 3 x)$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{y} \frac{d y}{d x}=\frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)+\frac{1}{\cos 2 x} \cdot \frac{d}{d x}(\cos 2 x)+\frac{1}{\cos 3 x} \cdot \frac{d}{d x}(\cos 3 x)$

$\Rightarrow \frac{d y}{d x}=y\left[-\frac{\sin x}{\cos x}-\frac{\sin 2 x}{\cos 2 x} \cdot \frac{d}{d x}(2 x)-\frac{\sin 3 x}{\cos 3 x} \cdot \frac{d}{d x}(3 x)\right]$

$\therefore \frac{d y}{d x}=-\cos x \cdot \cos 2 x \cdot \cos 3 x[\tan x+2 \tan 2 x+3 \tan 3 x]$

Question 2:Differentiate the function with respect to $x: \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

Solution: Let $y=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

Taking logarithm on both the sides, we obtain

$\log y=\log \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

$\Rightarrow \log y=\frac{1}{2} \log \left[\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right]$

$\Rightarrow \log y=\frac{1}{2}[\log \{(x-1)(x-2)\}-\log \{(x-3)(x-4)(x-5)\}]$

$\Rightarrow \log y=\frac{1}{2}[\log (x-1)+\log (x-2)-\log (x-3)-\log (x-4)-\log (x-5)]$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{y} \frac{d y}{d x}=\frac{1}{2}\left[\begin{array}{l} \frac{1}{x-1} \cdot \frac{d}{d x}(x-1)+\frac{1}{x-2} \cdot \frac{d}{d x}(x-2)-\frac{1}{x-3} \cdot \frac{d}{d x}(x-3)-\frac{1}{x-4} \cdot \frac{d}{d x}(x-4)-\frac{1}{x-5} \cdot \frac{d}{d x}(x-5) \end{array}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{y}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]$

$\therefore \frac{d y}{d x}=\frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]$

Question 3: Differentiate the function with respect to $x:(\log x)^{\cos x}$

Solution: Let $y=(\log x)^{\cos x}$

Taking logarithm on both the sides, we obtain

$\log y=\cos x \cdot \log (\log x)$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x}(\cos x) \cdot \log (\log x)+\cos x \cdot \frac{d}{d x}[\log (\log x)]$

$\Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=-\sin x \log (\log x)+\cos x \cdot \frac{1}{\log x} \cdot \frac{d}{d x}(\log x)$
$\Rightarrow \frac{d}{d x}=y\left[-\sin x \log (\log x)+\frac{\cos x}{\log x} \cdot \frac{1}{x}\right]$

$\therefore \frac{d y}{d x}=(\log x)^{\cos x}\left[\frac{\cos x}{x \log x}-\sin x \log (\log x)\right]$

Question 4:Differentiate the function with respect to $x: x^x-2^{\sin x}$

Solution: Let $y=x^x-2^{\sin x}$
Also, let $x^x=u$ and $2^{\sin x}=v$
$\therefore y=u-v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-\frac{d v}{d x}$

$u=x^x$

Taking logarithm on both the sides, we obtain

$\log u=x \log x$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{u} \cdot \frac{d u}{d x}=\left[\frac{d}{d x}(x) \times \log x+x \times \frac{d}{d x}(\log x)\right]$

$\Rightarrow \frac{d u}{d x}=u\left[1 \times \log x+x \times \frac{1}{x}\right]$

$\Rightarrow \frac{d u}{d x}=x^x(\log x+1)$

$\Rightarrow \frac{d u}{d x}=x^x(1+\log x)$

$v=2^{\sin x}$

Taking logarithm on both the sides, we obtain

$\log v=\sin x \cdot \log 2$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{v} \cdot \frac{d v}{d x}=\log 2 \cdot \frac{d}{d x}(\sin x)$

$\Rightarrow \frac{d v}{d x}=v \log 2 \cos x$

$\Rightarrow \frac{d v}{d x}=2^{\sin x} \cos x \log 2$

$\therefore \frac{d y}{d x}=x^x(1+\log x)-2^{\sin x} \cos x \log 2$

Question 5: Differentiate the function with respect to $x:(x+3)^2 \cdot(x+4)^3 \cdot(x+5)^4$

Solution: Let $y=(x+3)^2 \cdot(x+4)^3 \cdot(x+5)^4$

Taking logarithm on both the sides, we obtain

$\log y=\log (x+3)^2+\log (x+4)^3+\log (x+5)^4$

$\Rightarrow \log y=2 \log (x+3)+3 \log (x+4)+4 \log (x+5)$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{y} \cdot \frac{d y}{d x}=2 \cdot \frac{1}{x+3} \cdot \frac{d}{d x}(x+3)+3 \cdot \frac{1}{x+4} \cdot \frac{d}{d x}(x+4)+4 \cdot \frac{1}{x+5} \cdot \frac{d}{d x}(x+5)$

$\Rightarrow \frac{d y}{d x}=y\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]$

$\Rightarrow \frac{d y}{d x}=(x+3)^2(x+4)^3(x+5)^4 \cdot\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]$

$\Rightarrow \frac{d y}{d x}=(x+3)^2(x+4)^3(x+5)^4\left[\frac{ 2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)}\right]$

$\Rightarrow \frac{d y}{d x}=(x+3)(x+4)^2(x+5)^3 \cdot\left[2\left(x^2+9 x+20\right)+3\left(x^2+8 x+15\right) \\ +4\left(x^2+7 x+12\right)\right]$

$\therefore \frac{d y}{d x}=(x+3)(x+4)^2(x+5)^3\left(9 x^2+70 x+133\right)$

Question 6: Differentiate the function with respect to $x:\left(x+\frac{1}{x}\right)^x+x^{\left(1+\frac{1}{x}\right)}$

Solution: Let $y=\left(x+\frac{1}{x}\right)^x+x^{\left(1+\frac{1}{x}\right)}$

Also, let $u=\left(x+\frac{1}{x}\right)^x$ and $v=x^{\left(1+\frac{1}{x}\right)}$

$\therefore y=u+v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}---(1)$

Then, $u=\left(x+\frac{1}{x}\right)^x$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log u=\log \left(x+\frac{1}{x}\right)^x$

$\Rightarrow \log u=x \log \left(x+\frac{1}{x}\right)$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x) \times \log \left(x+\frac{1}{x}\right)+x \times \frac{d}{d x}\left[\log \left(x+\frac{1}{x}\right)\right]$

$\Rightarrow \frac{1}{u} \cdot \frac{d u}{d x}=1 \times \log \left(x+\frac{1}{x}\right)+x \times \frac{1}{\left(x+\frac{1}{x}\right)} \cdot \frac{d}{d x}\left(x+\frac{1}{x}\right)$

$\Rightarrow \frac{d u}{d x}=u\left(\log \left(x+\frac{1}{x}\right)+\frac{x}{\left(x+\frac{1}{x}\right) \times\left(1-\frac{1}{x^2}\right)}\right]$

$\Rightarrow \frac{d u}{d x}=\left(x+\frac{1}{x}\right)^x\left[\log \left(x+\frac{1}{x}\right)+\frac{\left(x-\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)}\right]$

$\Rightarrow \frac{d u}{d x}=\left(x+\frac{1}{x}\right)^x\left[\log \left(x+\frac{1}{x}\right)+\frac{x^2-1}{x^2+1}\right]$

$\Rightarrow \frac{d u}{d x}=\left(x+\frac{1}{x}\right)^x\left[\frac{x^2-1}{x^2+1}+\log \left(x+\frac{1}{x}\right)\right]---(2)$

$\text { Now, } v=x^{\left(1+\frac{1}{x}\right)}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log v=\log \left[x^{\left(1+\frac{1}{x}\right)}\right]$

$\Rightarrow \log v=\left(1+\frac{1}{x}\right) \log x$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{v} \cdot \frac{d v}{d x}=\left[\frac{d}{d x}\left(1+\frac{1}{x}\right)\right] \times \log x+\left(1+\frac{1}{x}\right) \cdot \frac{d}{d x} \log x$

$\Rightarrow \frac{1}{v} \cdot \frac{d v}{d x}=\left(-\frac{1}{x^2}\right) \log x+\left(1+\frac{1}{x}\right) \cdot \frac{1}{x}$

$\Rightarrow \frac{1}{v} \cdot \frac{d v}{d x}=-\frac{\log x}{x^2}+\frac{1}{x}+\frac{1}{x^2}$

$\Rightarrow \frac{d v}{d x}=v\left[\frac{-\log x+x+1}{x^2}\right]$

$\Rightarrow \frac{d v}{d x}=x^{\left(1+\frac{1}{x}\right)}\left[\frac{-\log x+x+1}{x^2}\right]--(3)$

Therefore, from (1), (2) and (3);

$\frac{d y}{d x}=\left(x+\frac{1}{x}\right)^x\left[\frac{x^2-1}{x^2+1}+\log \left(x+\frac{1}{x}\right)\right]+x^{\left(1+\frac{1}{x}\right)}\left[\frac{x+1-\log x}{x^2}\right]$

Question 7: Differentiate the function with respect to $x:(\log x)^x+x^{\log x}$

Solution: Let $y=(\log x)^x+x^{\log x}$

$\therefore y=u+v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

Then, $u=(\log x)^x$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log u=\log \left[(\log x)^x\right]$

$\Rightarrow \log u=x \log (\log x)$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x) \times \log (\log x)+x \cdot \frac{d}{d x}[\log (\log x)]$

$\Rightarrow \frac{d u}{d x}=u\left[1 \times \log (\log x)+x \cdot \frac{1}{(\log x)} \cdot \frac{d}{d x}(\log x)\right]$

$\Rightarrow \frac{d u}{d x}=(\log x)^x\left[\log (\log x)+\frac{x}{(\log x)} \cdot \frac{1}{x}\right]$

$\Rightarrow \frac{d u}{d x}=(\log x)^x\left[\log (\log x)+\frac{1}{(\log x)}\right]$

$\Rightarrow \frac{d u}{d x}=(\log x)^x\left[\frac{\log (\log x) \cdot \log x+1}{\log x}\right]$

$\Rightarrow \frac{d u}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]$

$v=x^{\log x}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log v=\log \left(x^{\log x}\right)$

$\Rightarrow \log v=\log x \log x=(\log x)^2$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{v} \cdot \frac{d v}{d x}=\frac{d}{d x}\left[(\log x)^2\right]$

$\Rightarrow \frac{1}{v} \cdot \frac{d v}{d x}=2(\log x) \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d v}{d x}=2 v(\log x) \cdot \frac{1}{x}$

$\Rightarrow \frac{d v}{d x}=2 x^{\log x} \frac{\log x}{x}$

& $\Rightarrow \frac{d v}{d x}=2 x^{\log x-1} \cdot \log x$

Therefore, from (1), (2) and (3);

$\frac{d y}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]+2 x^{\log x-1} \cdot \log x$

Question 8:Differentiate the function with respect to $x:(\sin x)^x+\sin ^{-1} \sqrt{x}$

Solution: Let $y=(\sin x)^x+\sin ^{-1} \sqrt{x}$

Also, let $u=(\sin x)^x$ and $v=\sin ^{-1} \sqrt{x}$

$\therefore y=u+v ---(1)$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

Then, $u=(\sin x)^x$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log u=\log (\sin x)^x$

$\Rightarrow \log u=x \log (\sin x)$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x) \times \log (\sin x)+x \cdot \frac{d}{d x}[\log (\sin x)]$

$\Rightarrow \frac{d u}{d x}=u\left[1 \times \log (\sin x)+x \cdot \frac{1}{(\sin x)} \cdot \frac{d}{d x}(\sin x)\right]$

$\Rightarrow \frac{d u}{d x}=(\sin x)^x\left[\log (\sin x)+\frac{x}{(\sin x)} \cdot \cos x\right]$

$\Rightarrow \frac{d u}{d x}=(\sin x)^x[x \cot x+\log \sin x]---(2)$

$v=\sin ^{-1} \sqrt{x}$

Differentiating both sides with respect to $x$ , we obtain

$\frac{d v}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{d}{d x}(\sqrt{x})$

$\Rightarrow \frac{d v}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}}$

$\Rightarrow \frac{d v}{d x}=\frac{1}{2 \sqrt{x-x^2}}---(3)$

Therefore, from (1), (2) and (3);

$\frac{d y}{d x}=(\sin x)^x[x \cot x+\log \sin x]+\frac{1}{2 \sqrt{x-x^2}}$

Question 9:Differentiate the function with respect to $x: x^{\sin x}+(\sin x)^{\cos x}$

Solution: Let $y=x^{\sin x}+(\sin x)^{\cos x}$

Also, let $u=x^{\sin x}$ and $v=(\sin x)^{\cos x}$

$\therefore y=u+v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} --(1)$

Then, $u=x^{\sin x}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log u=\log \left(x^{\sin x}\right)$

$\Rightarrow \log u=\sin x \log x$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(\sin x) \cdot \log x+\sin x \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d u}{d x}=u\left[\cos x \log x+\sin x \cdot \frac{1}{x}\right]$

$\Rightarrow \frac{d u}{d x}=x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}\right]---(2)$

$v=(\sin x)^{\cos x}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log v=\log (\sin x)^{\cos x}$

$\Rightarrow \log v=\cos x \log (\sin x)$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{v} \frac{d v}{d x}=\frac{d}{d x}(\cos x) \times \log (\sin x)+\cos x \times \frac{d}{d x}[\log (\sin x)]$

$\Rightarrow \frac{d v}{d x}=v\left[-\sin x \cdot \log (\sin x)+\cos x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)\right]$

$\Rightarrow \frac{d v}{d x}=(\sin x)^{\cos x}\left[-\sin x \log (\sin x)+\frac{\cos x}{\sin x} \cos x\right]$

$\Rightarrow \frac{d v}{d x}=(\sin x)^{\cos x}[-\sin x \log (\sin x)+\cot x \cos x]---(3)$

$\Rightarrow \frac{d v}{d x}=(\sin x)^{\cos x}[\cot x \cos x-\sin x \log (\sin x)]$

Therefore, from $(1),(2)$ and $(3)$

$\frac{d y}{d x}=x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}\right]+(\sin x)^{\cos x}[\cot x \cos x-\sin x \log (\sin x)]$

Question 10:Differentiate the function with respect to $x: x^{x^{x \cos x}+\frac{x^2+1}{x^2-1}}$

Solution:Let $y=x^{x \cos x}+\frac{x^2+1}{x^2-1}$

Also, let $u=x^{x \cos x}$ and $v=\frac{x^2+1}{x^2-1}$

$\therefore y=u+v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

Then, $u=x^{x \cos x}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log u=\log \left(x^{x \cos x}\right)$

$\Rightarrow \log u=x \cos x \log x$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x) \cdot \cos x \cdot \log x+x \cdot \frac{d}{d x}(\cos x) \cdot \log x+x \cos x \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d u}{d x}=u\left[1 \cdot \cos x \cdot \log x+x \cdot(-\sin x) \log x+x \cos x \cdot \frac{1}{x}\right]$

$\Rightarrow \frac{d u}{d x}=x^{x \cos x}[\cos x \log x-x \cdot \sin x \log x+\cos x]$

$\Rightarrow \frac{d u}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \cdot \sin x \log x]---(2)$

$v=\frac{x^2+1}{x^2-1}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log v=\log \left(x^2+1\right)-\log \left(x^2-1\right)$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{v} \frac{d v}{d x}=\frac{2 x}{x^2+1}-\frac{2 x}{x^2-1}$

$\Rightarrow \frac{d v}{d x}=v\left[\frac{2 x\left(x^2-1\right)-2 x\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2-1\right)}\right]$

& $\Rightarrow \frac{d v}{d x}=\frac{x^2+1}{x^2-1} \times\left[\frac{-4 x}{\left(x^2+1\right)\left(x^2-1\right)}\right]$

$\Rightarrow \frac{d v}{d x}=\frac{-4 x}{\left(x^2-1\right)^2}---(3)$

Therefore, from $(1),(2)$ and (3);

$\frac{d y}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \cdot \sin x \log x]-\frac{4 x}{\left(x^2-1\right)^2}$

Question 11: Differentiate the function with respect to $x:(x \cos x)^x+(x \sin x)^{\frac{1}{x}}$

Solution: Let $y=(x \cos x)^x+(x \sin x)^{\frac{1}{x}}$
Also, let $u=(x \cos x)^x$ and $v=(x \sin x)^{\frac{1}{x}}$

$\therefore y=u+v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}---(1)$

Then, $u=(x \cos x)^x$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log u=(x \cos x)^x$

$\Rightarrow \log u=x \log (x \cos x)$

$\Rightarrow \log u=x[\log x+\log \cos x]$

$\Rightarrow \log u=x \log x+x \log \cos x$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x \log x)+\frac{d}{d x}(x \log \cos x)$

$\Rightarrow \frac{d u}{d x}=u\left[\left\{\log x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log x)\right\}+\left\{\log \cos x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log \cos x)\right\}\right]$

$\Rightarrow \frac{d u}{d x}=(x \cos x)^x\left[\left(\log x \cdot 1+x \cdot \frac{1}{x}\right)+\left\{\log \cos x \cdot 1+x \cdot \frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)\right\}\right]$

$\Rightarrow \frac{d u}{d x}=(x \cos x)^x\left[(\log x+1)+\left\{\log \cos x+\frac{x}{\cos x} \cdot(-\sin x)\right\}\right]$

$\Rightarrow \frac{d u}{d x}=(x \cos x)^x[(1+\log x)+(\log \cos x-x \tan x)]$

$\Rightarrow \frac{d u}{d x}=(x \cos x)^x[(1-x \tan x)+(\log x+\log \cos x)]$

$\Rightarrow \frac{d u}{d x}=(x \cos x)^x[1-x \tan x+\log (x \cos x)]---(2)$

$v=(x \sin x)^{\frac{1}{x}}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log v=\log (x \sin x)^{\frac{1}{x}}$

$\Rightarrow \log v=\frac{1}{x} \log (x \sin x)$

$\Rightarrow \log v=\frac{1}{x}(\log x+\log \sin x)$

$\Rightarrow \log v=\frac{1}{x} \log x+\frac{1}{x} \log \sin x$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{v} \frac{d v}{d x}=\frac{d}{d x}\left(\frac{1}{x} \log x\right)+\frac{d}{d x}\left[\frac{1}{x} \log (\sin x)\right]$

$\Rightarrow \frac{1}{v} \frac{d v}{d x}=\left[\log x \cdot \frac{d}{d x}\left(\frac{1}{x}\right)+\frac{1}{x} \cdot \frac{d}{d x}(\log x)\right]+\left[\log (\sin x) \cdot \frac{d}{d x}\left(\frac{1}{x}\right)+\frac{1}{x} \cdot \frac{d}{d x}\{\log (\sin x)\}\right]$

$\Rightarrow \frac{1}{v} \frac{d v}{d x}=\left[\log x \cdot\left(-\frac{1}{x^2}\right)+\frac{1}{x} \cdot \frac{1}{x}\right]+\left[\log (\sin x) \cdot\left(-\frac{1}{x^2}\right)+\frac{1}{x} \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)\right]$

$\Rightarrow \frac{1}{v} \frac{d v}{d x}=\frac{1}{x^2}(1-\log x)+\left[-\frac{\log (\sin x)}{x^2}+\frac{1}{x \sin x} \cdot \cos x\right]$

$\Rightarrow \frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1-\log x}{x^2}+\frac{-\log (\sin x)+x \cot x}{x^2}\right]$

$\Rightarrow \frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1-\log x-\log (\sin x)+x \cot x}{x^2}\right]$

$\Rightarrow \frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1-\log (x \sin x)+x \cot x}{x^2}\right]---(3)$

Therefore, from (1), (2) and (3);

$\frac{d y}{d x}=(x \cos x)^x[1-x \tan x+\log (x \cos x)]+(x \sin x)^{\frac{1}{x}}\left[\frac{x \cot x+1-\log (x \sin x)}{x^2}\right]$

Question 12: Find $\frac{d y}{d x}$ of the function $x^y+y^x=1$

Solution:The given function is $x^y+y^x=1$

Let, $x^y=u$ and $y^x=v$

$\therefore u+v=1$

$\Rightarrow \frac{d u}{d x}+\frac{d v}{d x}=0 ---(1)$

Then, $u=x^y$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log u=\log \left(x^y\right)$

$\Rightarrow \log u=y \log x$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{u} \cdot \frac{d u}{d x}=\log x \frac{d y}{d x}+y \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d u}{d x}=u\left[\log x \frac{d y}{d x}+y \cdot \frac{1}{x}\right]$

$\Rightarrow \frac{d u}{d x}=x^y\left[\log x \frac{d y}{d x}+\frac{y}{x}\right]---(2)$

Now, $v=y^x$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log v=\log \left(y^x\right)$

$\Rightarrow \log v=x \log y$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{v} \frac{d v}{d x}=\log y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log y)$

$\Rightarrow \frac{d v}{d x}=v\left[\log y \cdot 1+x \cdot \frac{1}{y} \cdot \frac{d y}{d x}\right]$

$\Rightarrow \frac{d v}{d x}=y^x\left[\log y+\frac{x}{y} \cdot \frac{d y}{d x}\right]---(3)$

Therefore, from (1), (2) and (3);

$x^y\left[\log x \frac{d y}{d x}+\frac{y}{x}\right]+y^x\left[\log y+\frac{x}{y} \cdot \frac{d y}{d x}\right]=0$

$\Rightarrow\left(x^y \log x+x y^{x-1}\right) \frac{d y}{d x}=-\left(y x^{y-1}+y^x \log y\right)$

$\therefore \frac{d y}{d x}=\frac{-\left(y x^{y-1}+y^x \log y\right)}{\left(x^y \log x+x y^{x-1}\right)}$

Question 13: Find $\frac{d y}{d x}$ of the function $y^x=x^y$

Solution: The given function is $y^x=x^y$

Taking logarithm on both the sides, we obtain

$x \log y=y \log x$

Differentiating both sides with respect to $x$ , we obtain

$\log y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log y)=\log x \cdot \frac{d}{d x}(y)+y \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \log y \cdot 1+x \cdot \frac{1}{y} \cdot \frac{d y}{d x}=\log x \cdot \frac{d y}{d x}+y \cdot \frac{1}{x}$

$\Rightarrow \log y+\frac{x}{y} \cdot \frac{d y}{d x}=\log x \cdot \frac{d y}{d x}+\frac{y}{x}$

$\Rightarrow\left(\frac{x}{y}-\log x\right) \frac{d y}{d x}=\frac{y}{x}-\log y$

$\Rightarrow\left(\frac{x-y \log x}{y}\right) \frac{d y}{d x}=\frac{y-x \log y}{x}$

$\therefore \frac{d y}{d x}=\frac{y}{x}\left(\frac{y-x \log y}{x-y \log x}\right)$

Question 14: Find $\frac{d y}{d x}$ of the function $(\cos x)^y=(\cos y)^x$

Solution: The given function is $(\cos x)^y=(\cos y)^x$

Taking logarithm on both the sides, we obtain

$y \log \cos x=x \log \cos y$

Differentiating both sides with respect to $x$ , we obtain

$\log \cos x \cdot \frac{d y}{d x}+y \cdot \frac{d}{d x}(\log \cos x)=\log \cos y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log \cos y)$

$\Rightarrow \log \cos x \cdot \frac{d y}{d x}+y \cdot \frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)=\log \cos y \cdot 1+x \cdot \frac{1}{\cos y} \cdot \frac{d}{d x}(\cos y)$

$\Rightarrow \log \cos x \cdot \frac{d y}{d x}+\frac{y}{\cos x} \cdot(-\sin x)=\log \cos y+\frac{x}{\cos y} \cdot(-\sin y) \cdot \frac{d y}{d x}$

$\Rightarrow \log \cos x \cdot \frac{d y}{d x}-y \tan x=\log \cos y-x \tan y \frac{d y}{d x}$

$\Rightarrow(\log \cos x+x \tan y) \frac{d y}{d x}=y \tan x+\log \cos y$

$\therefore \frac{d y}{d x}=\frac{y \tan x+\log \cos y}{x \tan y+\log \cos x}$

Question 15: Find $\frac{d y}{d x}$ of the function $x y=e^{(x-y)}$

Solution: The given function is $x y=e^{(x-y)}$

Taking logarithm on both the sides, we obtain

$\log (x y)=\log \left(e^{x-y}\right)$

$\Rightarrow \log x+\log y=(x-y) \log e$

$\Rightarrow \log x+\log y=(x-y) \times 1$

$\Rightarrow \log x+\log y=(x-y)$

Differentiating both sides with respect to $x$ , we obtain

$\frac{d}{d x}(\log x)+\frac{d}{d x}(\log y)=\frac{d}{d x}(x)-\frac{d y}{d x}$

$\Rightarrow \frac{1}{x}+\frac{1}{y} \frac{d y}{d x}=1-\frac{d y}{d x}$

$\Rightarrow\left(1+\frac{1}{y}\right) \frac{d y}{d x}=1-\frac{1}{x}$

$\Rightarrow\left(\frac{y+1}{y}\right) \frac{d y}{d x}=\frac{x-1}{x}$

$\therefore \frac{d y}{d x}=\frac{y(x-1)}{x(y+1)}$

Question 16: Find the derivative of the function given by $f(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)$ and hence find $f^{\prime}(1)$

Solution: The given function is $f(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)$

Taking logarithm on both the sides, we obtain

$\log f(x)=\log (1+x)+\log \left(1+x^2\right)+\log \left(1+x^4\right)+\log \left(1+x^8\right)$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{f(x)} \cdot \frac{d}{d x}[f(x)]=\frac{d}{d x} \log (1+x)+\frac{d}{d x} \log \left(1+x^2\right)$

$+\frac{d}{d x} \log \left(1+x^4\right)+\frac{d}{d x} \log \left(1+x^8\right)$

$\Rightarrow \frac{1}{f(x)} \cdot f^{\prime}(x)= \frac{1}{1+x} \cdot \frac{d}{d x}(1+x)+\frac{1}{1+x^2} \cdot \frac{d}{d x}\left(1+x^2\right)$

$+\frac{1}{1+x^4} \cdot \frac{d}{d x}\left(1+x^4\right)+\frac{1}{1+x^8} \cdot \frac{d}{d x}\left(1+x^8\right)$

$\Rightarrow f^{\prime}(x)=f(x)\left[\frac{1}{1+x}+\frac{1}{1+x^2} \cdot 2 x+\frac{1}{1+x^4} \cdot 4 x^3+\frac{1}{1+x^8} \cdot 8 x^7\right]$

$\therefore f^{\prime}(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left[\frac{1}{1+x}+\frac{2 x}{1+x^2}+\frac{4 x^3}{1+x^4}+\frac{8 x^7}{1+x^8}\right]$

Hence,

$f^{\prime}(1) =(1+1)\left(1+1^2\right)\left(1+1^4\right)\left(1+1^8\right)\left[\frac{1}{1+1}+\frac{2(1)}{1+1^2}+\frac{4(1)^3}{1+1^4}+\frac{8(1)^7}{1+1^8}\right]$

$=2 \times 2 \times 2 \times 2\left[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right]$

$=16\left(\frac{15}{2}\right)$

$=120$

Question 17:Differentiate $\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)$ in three ways mentioned below.
(i) By using product rule
(ii) By expanding the product to obtain a single polynomial.
(iii) By logarithmic differentiation.
Do they all give the same answer?

Solution: Let $y=\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)$

(i) By using product rule

Let $u=\left(x^2-5 x+8\right)$ and $v=x^3+7 x+9$

$\therefore y=u v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x} \cdot v+u \cdot \frac{d v}{d x}$

$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(x^2-5 x+8\right) \cdot\left(x^3+7 x+9\right)+\left(x^2-5 x+8\right) \cdot \frac{d}{d x}\left(x^3+7 x+9\right)$

$\Rightarrow \frac{d y}{d x}=(2 x-5)\left(x^3+7 x+9\right)+\left(x^2-5 x+8\right)\left(3 x^2+7\right)$

$\Rightarrow \frac{d y}{d x}=2 x\left(x^3+7 x+9\right)-5\left(x^3+7 x+9\right)+x^2\left(3 x^2+7\right)$

$-5 x\left(3 x^2+7\right)+8\left(3 x^2+7\right)$

$\Rightarrow \frac{d y}{d x}=\left(2 x^4+14 x^2+18 x\right)-5 x^3-35 x-45+\left(3 x^4+7 x^2\right)$

$\Rightarrow \frac{d y}{d x}=-15 x^3-35 x+24 x^2+56$

$\therefore \frac{d y}{d x}=5 x^4-20 x^3+45 x^2-52 x+11$

(ii) By expanding the product to obtain a single polynomial.

$y=\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)$

$=x^2\left(x^3+7 x+9\right)-5 x\left(x^3+7 x+9\right)+8\left(x^3+7 x+9\right)$

$=x^5+7 x^3+9 x^2-5 x^4-35 x^2-45 x+8 x^3+56 x+72$

$=x^5-5 x^4+15 x^3-26 x^2+11 x+72$

Therefore,

$\frac{d y}{d x}=\frac{d}{d x}\left(x^5-5 x^4+15 x^3-26 x^2+11 x+72\right)$

$=\frac{d}{d x}\left(x^5\right)-5 \frac{d}{d x}\left(x^4\right)+15 \frac{d}{d x}\left(x^3\right)-26 \frac{d}{d x}\left(x^2\right)+11 \frac{d}{d x}(x)+\frac{d}{d x}(72)$

$=5 x^4-5\left(4 x^3\right)+15\left(3 x^2\right)-26(2 x)+11(1)+0$

$=5 x^4-20 x^3+45 x^2-52 x+11$

(iii) By logarithmic differentiation.

$y=\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)$

Taking logarithm on both the sides, we obtain

$\log y=\log \left(x^2-5 x+8\right)+\log \left(x^3+7 x+9\right)$

Differentiating both sides with respect to $x$ , we obtain

$\frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x} \log \left(x^2-5 x+8\right)+\frac{d}{d x} \log \left(x^3+7 x+9\right)$

$\Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{x^2-5 x+8} \cdot \frac{d}{d x}\left(x^2-5 x+8\right)+\frac{1}{x^3+7 x+9} \cdot \frac{d}{d x}\left(x^3+7 x+9\right)$

$\Rightarrow \frac{d y}{d x}=y\left[\frac{1}{x^2-5 x+8} \cdot(2 x-5)+\frac{1}{x^3+7 x+9} \cdot\left(3 x^2+7\right)\right]$

$\Rightarrow \frac{d y}{d x}=\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)\left[\frac{2 x-5}{x^2-5 x+8}+\frac{3 x^2+7}{x^3+7 x+9}\right]$

$\Rightarrow \frac{d y}{d x}=\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)\left[\frac{(2 x-5)\left(x^3+7 x+9\right)+\left(3 x^2+7\right)\left(x^2-5 x+8\right)}{\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)}\right]$

$\Rightarrow \frac{d y}{d x}=2 x\left(x^3+7 x+9\right)-5\left(x^3+7 x+9\right)+3 x^2\left(x^2-5 x+8\right)+7\left(x^2-5 x+8\right)$

$\Rightarrow \frac{d y}{d x}=2 x^4+14 x^2+18 x-5 x^3-35 x-45+3 x^5-15 x^3+24 x^2+7 x^2-35 x+56$

$\Rightarrow \frac{d y}{d x}=5 x^4-20 x^3+45 x^2-52 x+11$

From the above three observations, it can be concluded that

all the results of $\frac{d y}{d x}$ are same.

Question 18:If $u, v$ and $w$ are functions of $x$ , then show that

$\frac{d}{d x}(u \cdot v \cdot w)=\frac{d u}{d x} v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot \frac{d w}{d x}$

in two ways – first by repeated application of product rule, second by logarithmic differentiation.

Solution: Let $y=u . v . w=u .(v . w)$

By applying product rule, we get

$\frac{d y}{d x}=\frac{d u}{d x} \cdot(v \cdot w)+u \cdot \frac{d}{d x}(v \cdot w)$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x} \cdot(v \cdot w)+u\left[\frac{d v}{d x} \cdot w+v \cdot \frac{d w}{d x}\right] \quad \text { (Again applying product rule) }$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x} \cdot v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot \frac{d w}{d x}$

Taking logarithm on both the sides of the equation $y=u . v . w$ ,
we obtain, $\log y=\log u+\log v+\log w$

Differentiating both sides with respect to $x$ , we obtain
$\frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x}(\log u)+\frac{d}{d x}(\log v)+\frac{d}{d x}(\log w)$