Question

Differentiate ${\tan }^{-1}\left(\frac{1+ax}{1-ax}\right)$ with respect to $\sqrt{1+a^2{x}^2}$

Answer 1

$$\begin{gathered} \mathrm{Let},u={\tan }^{-1}\left(\frac{1+ax}{1-ax}\right) \\ \mathrm{Put}ax=\tan \theta \\ \Rightarrow u={\tan }^{-1}\left(\frac{1+\tan \theta }{1-\tan \theta }\right) \\ \Rightarrow u={\tan }^{-1}\left(\frac{\tan {\displaystyle \frac{\pi }{4}}+\tan \theta }{1-{\displaystyle \tan \frac{{\displaystyle \pi }}{4}}\tan \theta }\right) \\ \Rightarrow u={\tan }^{-1}\left[\tan \left(\frac{\mathrm{\pi }}{4}+\theta \right)\right] \\ \Rightarrow u=\frac{\mathrm{\pi }}{4}+\theta \\ \Rightarrow u=\frac{\mathrm{\pi }}{4}+{\tan }^{-1}\left(ax\right)\left[\mathrm{Since},\tan \theta =ax\right] \end{gathered}$$

Differentiating it with respect to x,

$$\begin{gathered} \frac{du}{dx}=0+\frac{1}{1+{\left(ax\right)}^2}\frac{d}{dx}\left(ax\right) \\ \Rightarrow \frac{du}{dx}=\frac{a}{1+a^2{x}^2}...\left(\mathrm{i}\right) \\ \mathrm{Now}, \\ \mathrm{Let},v=\sqrt{1+a^2{x}^2} \end{gathered}$$

Differentiating it with respect to x,

$$\begin{gathered} \frac{dv}{dx}=\frac{1}{2\sqrt{1+a^2{x}^2}}\frac{d}{dx}\left(1+a^2{x}^2\right) \\ \Rightarrow \frac{dv}{dx}=\frac{1}{2\sqrt{1+a^2{x}^2}}\left(2a^{2}x\right) \\ \Rightarrow \frac{dv}{dx}=\frac{a^{2}x}{\sqrt{1+a^2{x}^2}}...\left(\mathrm{ii}\right) \\ \mathrm{Dividing}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{by}\left(\mathrm{ii}\right), \\ \frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}=\frac{a}{1+a^2{x}^2}\times \frac{\sqrt{1+a^2{x}^2}}{a^{2}x} \\ \frac{du}{dv}=\frac{1}{ax\sqrt{1+a^2{x}^2}} \end{gathered}$$