Question

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