#### provide solution for RD Sharma maths class 12 chapter Differentiation exercise  10.4 question 30

$\frac{d y}{d x}=8\left[\frac{4}{16+\pi^{2}}-\frac{1}{\log 2}\right]_{\mathrm{At}} x=\frac{\pi}{4}$

Hint:

Use chain rule and differentiation of inverse trigonometric function

Given:

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

Solution:

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

$y=\left[\log _{\cos x} \sin x\right]\left[\log _{\cos x} \sin x\right]+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$                $\left[\because \log _{b} a=\left(\log _{a} b\right)^{-1}\right]$

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

$y=\left(\frac{\log \sin x}{\log \cos x}\right)^{2}+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$                        $\left[\because \log _{a} b=\frac{\log b}{\log a}\right]$

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

Differentiate w.r.t x

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

$=\left[\frac{d\left(\frac{\log \sin x}{\log \cos x}\right)^{2}}{d\left(\frac{\log \sin x}{\log \cos x}\right)} \times \frac{d\left(\frac{\log \sin x}{\log \cos x}\right)}{d x}\right]+\frac{d \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)}{d\left(\frac{2 x}{1+x^{2}}\right)} \times \frac{d\left(\frac{2 x}{1+x^{2}}\right)}{d x}$     [Using chain rule]

$\frac{d y}{d x}=2\left(\frac{\log \sin x}{\log \cos x}\right) \times \frac{d}{d x}\left(\frac{\log \sin x}{\log \cos x}\right)+\frac{1}{\sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^{2}}} \times \frac{d}{d x}\left(\frac{2 x}{1+x^{2}}\right)$

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

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

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

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

$\frac{d y}{d x}=\frac{2 \log \sin x}{(\log \cos x)^{3}} \times \frac{[\cot x \times(\log \cos x)+\tan x \times(\log \sin x)]}{1}+\frac{2}{1+x^{2}}$

$\frac{d y}{d x}=\frac{2 \log \sin x}{(\log \cos x)^{3}} \times(\cot x \log \cos x+\tan x \log \sin x)+\frac{2}{1+x^{2}}$

Put $x=\frac{\pi}{4}$

$\frac{d y}{d x}=2\left(\frac{\log \left(\sin \frac{\pi}{4}\right)}{\left(\log \left(\cos \frac{\pi}{4}\right)\right)^{3}}\right) \times\left(\cot \frac{\pi}{4} \times \log \cos \frac{\pi}{4}+\tan \frac{\pi}{4} \cdot \log \sin \frac{\pi}{4}\right)+\frac{2}{\left[1+\left(\frac{\pi}{4}\right)^{2}\right]}$                $\left[\because \sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right] ;\left[\tan \frac{\pi}{4}=\cot \frac{\pi}{4}=1\right]$

$\frac{d y}{d x}=2\left[\frac{\log \left(\frac{1}{\sqrt{2}}\right)}{\left(\log \left(\frac{1}{\sqrt{2}}\right)\right)^{3}}\right]\left[1 \times \log \frac{1}{\sqrt{2}}+1 \times \log \frac{1}{\sqrt{2}}\right]+\frac{2}{\left(1+\frac{\pi^{2}}{16}\right)}$$\frac{d y}{d x}=\frac{2}{\left(\log \frac{1}{\sqrt{2}}\right)^{2}}\left[2 \log \frac{1}{\sqrt{2}}\right]+\frac{32}{16+\pi^{2}}$

$=\frac{4}{\left(\log \frac{1}{\sqrt{2}}\right)}+\frac{32}{16+\pi^{2}}$

$=\frac{4}{\log (2)^{\frac{-1}{2}}}+\frac{32}{16+\pi^{2}}$

$=\frac{4}{\frac{-1}{2} \log 2}+\frac{32}{16+\pi^{2}}$

$=\frac{-8}{\log 2}+\frac{32}{16+\pi^{2}}$

$\therefore\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{4}}=8\left[\frac{4}{16+\pi^{2}}-\frac{1}{\log 2}\right]$