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

$\Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x}$

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

$\Rightarrow \frac{d y}{d x}=u \cdot v \cdot w\left(\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x}\right)$

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

EXERCISE 5.5 ( Differentiation )

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