Question

If $y=se{c}^{-1}\left(\frac{x+1}{x-1}\right)+{\sin }^{-1}\left(\frac{x-1}{x+1}\right),x>0.\mathrm{Find}\frac{dy}{dx}$.

Answer 1

$$\begin{gathered} \mathrm{Here},y=se{c}^{-1}\left(\frac{x+1}{x-1}\right)+{\sin }^{-1}\left(\frac{x-1}{x+1}\right) \\ \Rightarrow y={\cos }^{-1}\left(\frac{x-1}{x+1}\right)+{\sin }^{-1}\left(\frac{x-1}{x+1}\right)\left[\mathrm{Since},se{c}^{-1}\left(x\right)={\cos }^{-1}\left(\frac{1}{x}\right)\right] \\ \Rightarrow y=\frac{\mathrm{\pi }}{2}\left[\mathrm{Since},{\cos }^{-1}\alpha +{\sin }^{-1}\alpha =\frac{\mathrm{\pi }}{2}\right] \end{gathered}$$

Differentiate it with respect to x,

$\therefore \frac{dy}{dx}=0$