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