#### Please solve RD Sharma class 12 chapter 10 Differentiation exercise Very short answers question 21 maths textbook solution

$\frac{d y}{d x}=\{0, x \geq 0\} \text { does not exist for } x<0$

Hint:

changing   $\sec ^{-1}\left(\frac{x+1}{x-1}\right) \text { into } \cos ^{-1}\left(\frac{x-1}{x+1}\right)$

Given:

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

\begin{aligned} &y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x+1}{x-1}\right) \\\\ &y=\cos ^{-1}\left(\frac{x-1}{x+1}\right)+\sin ^{-1}\left(\frac{x+1}{x-1}\right) \end{aligned}

Which exists for

$-1 \leq\left(\frac{x-1}{x+1}\right) \leq 1\text { and is equal to } \frac{\pi}{2}$

Now,

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

\begin{aligned} &\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 \end{aligned}

\begin{aligned} &\Rightarrow x+1>0 \\\\ &\Rightarrow x>1 \ldots \ldots \ldots . . \text { (i) } \end{aligned}

Also,

\begin{aligned} &\frac{x-1}{x+1} \geq-1 \\\\ &\Rightarrow \frac{x-1}{x+1}+1 \geq 0 \end{aligned}

\begin{aligned} &\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 \text { or } x<-1 \ldots \ldots \ldots \ldots \ldots \ldots(ii) \end{aligned}

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,

$\frac{d y}{d x}=\{0, x \geq 0\} \text { does not exist for } x<0$