Question

Let $f(x)={\cos }^{-1}\left(\frac{2x}{1+x^2}\right)$ and $g(x)={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, then derivative of $f(x)$ with respect to $g(x)$ at $x=\frac{1}{2}$ is equal to

• A. 0
• B. 1
• C. -1
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

Given, $f(x)={\cos }^{-1}\left(\frac{2x}{1+x^2}\right)=\frac{\pi }{2}-{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=\frac{\pi }{2}-2{\tan }^{-1}x,-1\le x\le 1$
${\therefore f^{\prime }(x)]}_k=\frac{1}{2}=\frac{-2}{1+x^2}=\frac{-2}{1+\frac{1}{4}}=\frac{-8}{5}$
Also, $g(x)={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\frac{\pi }{2}-{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\frac{\pi }{2}-2{\tan }^{-1}x,x\ge 0$
${\therefore g^{\prime }(x)]}_{x=\frac{1}{2}}=\frac{-2}{1+x^2}=\frac{-2}{1+\frac{1}{4}}=\frac{-8}{5}$
So, derivative of $f(x)$ with respect to $g(x)$ at $x=\frac{1}{2}\Rightarrow \frac{f^{\prime }\left(\frac{1}{2}\right)}{g^{\prime }\left(\frac{1}{2}\right)}=1$