Question

The derivative of $co{s}^{-1}\left(\frac{x^{-1}-x}{x^{-1}+x}\right)atx=-1$ is

• A. - 2
• B. - 1
• C. 1

Answer 1

The correct option is C 1

We have, $y={\cos }^{-1}\left(\frac{x^{-1}-x}{x^{-1}+x}\right)\Rightarrow y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
Now, Put $x=\tan \theta$
We get $y={\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)\Rightarrow y={\cos }^{-1}\left(\cos 2\theta \right)\Rightarrow y=2\theta \Rightarrow y=2{\tan }^{-1}x\therefore \frac{dy}{dx}=\frac{2}{1+x^2}\therefore \frac{dy}{dx}{|}_{x=-1}=\frac{2}{1+(-1{)}^2}=2$