Question

If $y=\left\{{\log }_{\cos x}\sin x\right\}{\left\{{\log }_{\sin x}\cos x\right\}}^{-1}+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right),\mathrm{find}\frac{dy}{dx}atx=\frac{\pi }{4}$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y=\left[{\log }_{\cos x}\sin x\right]{\left[{\log }_{\sin x}\cos x\right]}^{-1}+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right) \\ \Rightarrow y=\left[{\log }_{\cos x}\sin x\right]\left[{\log }_{\cos x}\sin x\right]+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\left[\because {\log }_{a}b={\left({\log }_{b}a\right)}^{-1}\right] \\ \Rightarrow y={\left[\frac{\log \sin x}{\log \cos x}\right]}^2+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\left[\because {\log }_{a}b=\frac{\log b}{\log a}\right] \end{gathered}$$

Differentiating with respect to x,

$$\begin{gathered} \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({\displaystyle \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){\displaystyle \frac{d}{dx}}\left(\log \sin x\right)-\log \sin x{\displaystyle \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^4-2x^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 {\displaystyle \frac{1}{\sin x}}{\displaystyle \frac{d}{dx}}\left(\sin x\right)-\log \sin x\times {\displaystyle \frac{1}{\cos x}}{\displaystyle \frac{d}{dx}}\left(\cos x\right)}{{\left(\log \cos x\right)}^2}\right]+\left[\frac{\left(1+x^2\right)}{\sqrt{1+x^4-2x^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({\displaystyle \frac{\cos x}{\sin x}}\right)+\log \sin x\times \left({\displaystyle \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+2x^2-4x^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} \\ \mathrm{put}x=\frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{dy}{dx}=2\left\{\frac{\log \sin {\displaystyle \frac{\mathrm{\pi }}{4}}}{{\left(\log \cos {\displaystyle \frac{\mathrm{\pi }}{4}}\right)}^3}\right\}\left(\cot \frac{\mathrm{\pi }}{4}\log \cos \frac{\mathrm{\pi }}{4}+\tan \frac{\mathrm{\pi }}{4}\log \sin \frac{\mathrm{\pi }}{4}\right)+2\left\{\frac{1}{1+{\left({\displaystyle \frac{\mathrm{\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+{\mathrm{\pi }}^2}\right) \\ \Rightarrow \frac{dy}{dx}=2\times \frac{2\log \left({\displaystyle \frac{1}{\sqrt{2}}}\right)}{{\left\{\log \left(\frac{1}{\sqrt{2}}\right)\right\}}^2}+\frac{32}{16+{\mathrm{\pi }}^2} \\ \Rightarrow \frac{dy}{dx}=4\frac{1}{\log \left({\displaystyle \frac{1}{\sqrt{2}}}\right)}+\frac{32}{16+{\mathrm{\pi }}^2} \\ \Rightarrow \frac{dy}{dx}=4\frac{1}{-{\displaystyle \frac{1}{2}}\log 2}+\frac{32}{16+{\mathrm{\pi }}^2} \\ \Rightarrow \frac{dy}{dx}=-\frac{8}{\log 2}+\frac{32}{16+{\mathrm{\pi }}^2} \\ \mathrm{So},{\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{4}}=8\left[\frac{4}{16+{\mathrm{\pi }}^2}-\frac{1}{\log 2}\right] \end{gathered}$$