A) \[-\frac{1}{1+{{x}^{2}}}\]
B) \[\frac{1}{1+{{a}^{2}}}-\frac{1}{1+{{x}^{2}}}\]
C) \[\frac{1}{1+{{\left( \frac{a-x}{1+ax} \right)}^{2}}}\]
D) \[\frac{-1}{\sqrt{1-{{\left( \frac{a-x}{1+ax} \right)}^{2}}}}\]
Correct Answer: A
Solution :
\[\frac{d}{dx}\left[ {{\tan }^{-1}}\left( \frac{a-x}{1+ax} \right) \right].\] = \[\frac{d}{dx}[{{\tan }^{-1}}a-{{\tan }^{-1}}x]=0-\frac{1}{1+{{x}^{2}}}=-\frac{1}{1+{{x}^{2}}}\].
You need to login to perform this action.
You will be redirected in
3 sec
