Question

The derivative of $cose{c}^{-1}\left(\frac{1}{2x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ is

• A. $-4$
• B. $4$
• C. $-1$
• D. none of these

Answer 1

Answer: (A)

$-4$

Let $y=cose{c}^{-1}\left(\frac{1}{2x^2-1}\right)$ and $z=\sqrt{1-x^2}$
We have,
$y=cose{c}^{-1}\left(\frac{1}{2x^2-1}\right)$
$\Rightarrow y={\sin }^{-1}(2x^2-1)=\frac{\pi }{2}-{\cos }^{-1}(2x^2-1)$
$\Rightarrow y=\{\begin{array}{ccc}\frac{\pi }{2}-2{\cos }^{-1}x& & ,\text{if}0\le x\le 1\\ -\frac{3\pi }{2}+2{\cos }^{-1}x& & ,\text{if}-1\le x\le 0\end{array}$
$\Rightarrow \frac{dy}{dx}=\{\begin{array}{ccc}\frac{2}{\sqrt{1-x^2}}& & ,\text{if}0<x<1\\ \frac{-2}{\sqrt{1-x^2}}& & ,\text{if}-1<x<0\end{array}$
and,
$z=\sqrt{1-x^2}\Rightarrow \frac{dz}{dx}=\frac{-x}{\sqrt{1-x^2}}\text{for all}x\in (-1,1)$
$\therefore \frac{dy}{dz}=\frac{dy/dx}{dz/dx}=\{\begin{array}{ccc}-\frac{2}{x}& & ,\text{if}0<x<1\\ \frac{2}{x}& & ,\text{if}-1<x<0\end{array}$
$\Rightarrow {\left(\frac{dy}{dz}\right)}_{x=1/2}=-4$