Limit Continuity And Differentiability Ques 26
- Find the derivative with respect to $x$ of the function $y=\left\{\left(\log _{\cos x} \sin x\right)\left(\log _{\sin x} \cos x\right)^{-1}+\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right\}$ at $x=\frac{\pi}{4}$.
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Answer:
Correct Answer: 26.$(\frac{-8}{\log e^2}+\frac{32}{16+\pi^2})$
Solution: Given, $\mathrm{y}=\left\{\left(\log _{\cos x} \sin x\right) \cdot\left(\log _{\sin x} \cos x\right)^{-1}+\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right\}$
$\therefore \quad y=\left(\frac{\log _e(\sin x)}{\log _e(\cos x)}\right)^2+\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$
$\Rightarrow \quad \frac{d y}{d x}=2\left\{\frac{\log _e(\sin x)}{\log _e(\cos x)} \cdot \frac{\left(\log _e(\cos x) \cdot \cot x\right.}{\left\{\log _e(\cos x)\right\}^2}\right\}+\frac{2}{1+x^2}$
$\Rightarrow \quad \left(\frac{d y}{d x}\right)_{\left(x-\frac{\pi}{4}\right)}=2\left\{1 \cdot \frac{2 \cdot \log \left(\frac{1}{\sqrt{2}}\right)}{\left(\log \frac{1}{\sqrt{2}}\right)^2}\right\}+\frac{2}{1+\frac{\pi^2}{16}}$
$ = \quad -\frac{8}{\log _e 2}+\frac{32}{16+\pi^2} $
BETA
