question_answer

If \[y={{\tan }^{-1}}\left( \frac{a+x}{1-ax} \right),\] find \[\frac{dy}{dx}.\]

Answer:

We have, \[y={{\tan }^{-1}}\left( \frac{a+x}{1-ax} \right)={{\tan }^{-1}}a+{{\tan }^{-1}}x\] \[\therefore \]      \[\frac{dy}{dx}=0+\frac{1}{1+{{x}^{2}}}=\frac{1}{1+{{x}^{2}}}\]