Question

Differentiate

${\tan }^{-1}\left(\frac{2a^x}{1-a^{2x}}\right),a>1,-\infty <x<0$

Answer 1

$$\begin{gathered} \mathrm{Let},y={\tan }^{-1}\left\{\frac{2a^x}{1-a^{2x}}\right\} \\ \mathrm{put}{a}^x=\tan \theta \\ \Rightarrow y={\tan }^{-1}\left\{\frac{2\times a^x}{1-{\left(a^x\right)}^2}\right\} \\ \Rightarrow y={\tan }^{-1}\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right) \\ \Rightarrow y={\tan }^{-1}\left(\tan 2\theta \right)...\left(\mathrm{i}\right) \\ \mathrm{Here},-\infty <x<0 \\ \Rightarrow a^{-\infty }<a^x<2° \\ \Rightarrow 0<\tan \theta <1 \\ \Rightarrow 0<\theta <\frac{\mathrm{\pi }}{4} \\ \Rightarrow 0<2\theta <\frac{\mathrm{\pi }}{2} \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ y=2\theta \left[\mathrm{Since},{\tan }^{-1}\left(\tan \theta \right)=\theta ,\mathrm{if}\theta \in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)\right] \\ \Rightarrow y=2{\tan }^{-1}\left(a^x\right) \\ \mathrm{Differentiating}\mathrm{it}\mathrm{with}\mathrm{respect}\mathrm{to}x, \\ \frac{dy}{dx}=\frac{2}{1+{\left(a^x\right)}^2}\frac{d}{dx}\left(a^x\right) \\ \Rightarrow \frac{dy}{dx}=\frac{2\times a^x{\log }_{e}a}{1+a^{2x}} \\ \therefore \frac{dy}{dx}=\frac{2a^x{\log }_{e}a}{1+a^{2x}} \end{gathered}$$