Question

If \[y = \left\{ \log_{\cos x} \sin x \right\} \left\{ \log_{\sin x} \cos x \right\}^{- 1} + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right), \text{ find } \frac{dy}{dx} \text{ at }x = \frac{\pi}{4}\] ?

Answer 1

\[\text{ We have, y } = \left[ \log_{\cos x} \sin x \right] \left[ \log_{\sin x} \cos x \right]^{- 1} + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\]

\[ \Rightarrow y = \left[ \log_{\cos x }\sin x \right]\left[ \log_{\cos x }\sin x \right] + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) ..........\left[ \because \log_a b = \left( \log_b a \right)^{- 1} \right]\]

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

Differentiating with respect to x,

\[\frac{dy}{dx} = \frac{d}{dx} \left[ \frac{\log \sin x}{\log \cos x} \right]^2 + \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\}\]

\[ \Rightarrow \frac{dy}{dx} = 2\left[ \frac{\log \sin x}{\log \cos x} \right]\frac{d}{dx}\left( \frac{\log \sin x}{\log \cos x} \right) + \frac{1}{\sqrt{1 - \left( \frac{2x}{1 + x^2} \right)^2}} \times \frac{d}{dx}\left[ \frac{2x}{1 + x^2} \right] \]

\[ \Rightarrow \frac{dy}{dx} = 2\left[ \frac{\log \sin x}{\log \cos x} \right]\left[ \frac{\left( \log \cos x \right)\frac{d}{dx}\left( \log \sin x \right) - \log \sin x\frac{d}{dx}\left( \log \cos x \right)}{\left( \log \cos x \right)^2} \right] + \left[ \frac{\left( 1 + x^2 \right)}{\sqrt{1 + x^4 - 2 x^2}} \right]\left[ \frac{\left( 1 + x^2 \right)\left( 2 \right) - \left( 2x \right)\left( 2x \right)}{\left( 1 + x^2 \right)^2} \right] \]

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

\[ \Rightarrow \frac{dy}{dx} = 2\left[ \frac{\log \sin x}{\log \cos x} \right]\left[ \frac{\log \cos x \times \left( \frac{\cos x}{\sin x} \right) + \log \sin x \times \left( \frac{\sin x}{\cos x} \right)}{\left( \log \cos x \right)^2} \right] + \left[ \frac{1 + x^2}{\sqrt{\left( 1 - x^2 \right)^2}} \right]\left[ \frac{2 + 2 x^2 - 4 x^2}{\left( 1 + x^2 \right)^2} \right]\]

\[ \Rightarrow \frac{dy}{dx} = 2\frac{\log \sin x}{\left( \log \cos x \right)^3}\left( \cot x \log \cos x + \tan x \log \sin x \right) + \frac{2}{1 + x^2}\]

\[\text{ put x } = \frac{\pi}{4}\]

\[ \Rightarrow \frac{dy}{dx} = 2\left\{ \frac{\log \sin\frac{\pi}{4}}{\left( \log \cos\frac{\pi}{4} \right)^3} \right\} \left( \cot\frac{\pi}{4} \log \cos\frac{\pi}{4} + \tan\frac{\pi}{4} \log \sin\frac{\pi}{4} \right) + 2\left\{ \frac{1}{1 + \left( \frac{\pi}{4} \right)^2} \right\}\]

\[ \Rightarrow \frac{dy}{dx} = 2\left\{ \frac{1}{\left( \log\frac{1}{\sqrt{2}} \right)^2} \right\}\left( 1 \times \log\frac{1}{\sqrt{2}} + 1 \times \log\frac{1}{\sqrt{2}} \right) + 2\left( \frac{16}{16 + \pi^2} \right) \]

\[ \Rightarrow \frac{dy}{dx} = 2 \times \frac{2\log\left( \frac{1}{\sqrt{2}} \right)}{\left\{ \log\left( \frac{1}{\sqrt{2}} \right) \right\}^2} + \frac{32}{16 + \pi^2}\]

\[ \Rightarrow \frac{dy}{dx} = 4\frac{1}{\log\left( \frac{1}{\sqrt{2}} \right)} + \frac{32}{16 + \pi^2}\]

\[ \Rightarrow \frac{dy}{dx} = 4\frac{1}{- \frac{1}{2}\log2} + \frac{32}{16 + \pi^2}\]

\[ \Rightarrow \frac{dy}{dx} = - \frac{8}{\log2} + \frac{32}{16 + \pi^2}\]

\[So, \left( \frac{dy}{dx} \right)_{x = \frac{\pi}{4}} = 8\left[ \frac{4}{16 + \pi^2} - \frac{1}{\log2} \right]\]