Solve this


If $y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)$, then write the value of $\frac{d y}{d x}$.


$y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)$

$=\cos ^{-1}\left(\frac{x-1}{x+1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)$

Which exists for $-1 \leq \frac{x-1}{x+1} \leq 1$ and is equal to $\frac{\pi}{2}$

Now, $\frac{x-1}{x+1} \leq 1$

$\Rightarrow \frac{x-1}{x+1}-1 \leq 0$

$\Rightarrow \frac{x-1}{x+1}-\frac{x+1}{x+1} \leq 0$

$\Rightarrow-\frac{2}{x+1} \leq 0$

$\Rightarrow \frac{2}{x+1} \geq 0$

$\Rightarrow x+1>0$

$\Longrightarrow x>-1 \ldots$ (i)

Also, $\frac{x-1}{x+1} \geq-1$

$\Rightarrow \frac{x-1}{x+1}+1 \geq 0$

$\Rightarrow \frac{x-1}{x+1}+\frac{x+1}{x+1} \geq 0$

$\Rightarrow \frac{2 x}{x+1} \geq 0$

$\Rightarrow x \geq 0$ or $x<-1 \ldots$ (ii)

Comparing equations (i) and (ii), we understand that the condition satisfying both inequalities is $x \geq 0$.

So, for $x \geq 0$,

$y=\cos ^{-1}\left(\frac{x-1}{x+1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)=\frac{\pi}{2}$, which is a constant

So, $\frac{d y}{d x}=\left\{\begin{array}{c}0, x \geq 0 \\ \text { does not exist for } x<0\end{array}\right.$ (Ans)

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