Question

Differentiate ${\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right)$ with respect to ${\sec }^{-1}x$

Answer 1

$$\begin{gathered} \mathrm{Let},u={\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right) \\ \Rightarrow u={\tan }^{-1}\left[\tan \left(\frac{\mathrm{\pi }}{4}-\frac{x}{2}\right)\right] \\ \Rightarrow u=\frac{\mathrm{\pi }}{4}-\frac{x}{2} \end{gathered}$$

Differentiating it with respect to x,

$$\begin{gathered} \frac{du}{dx}=0-\left(\frac{1}{2}\right) \\ \frac{du}{dx}=-\frac{1}{2}...\left(\mathrm{i}\right) \\ \mathrm{Let},v=se{c}^{-1}x \end{gathered}$$

Differentiating it with respect to x,

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