A) \[{{\tan }^{-1}}x\]
B) \[2x{{\tan }^{-1}}x\]
C) \[2x{{\tan }^{-1}}x\]
D) \[\frac{2x}{{{\tan }^{-1}}x}\]
Correct Answer: B
Solution :
\[y=(1+{{x}^{2}}){{\tan }^{-1}}x-x\] \[\Rightarrow \]\[y={{\tan }^{-1}}x+{{x}^{2}}{{\tan }^{-1}}x-x\] \[\Rightarrow \]\[\frac{dy}{dx}=\frac{1}{1+{{x}^{2}}}+{{x}^{2}}.\frac{1}{1+{{x}^{2}}}+{{\tan }^{-1}}x.2x-1\] \[\Rightarrow \]\[\frac{dy}{dx}=1+2x{{\tan }^{-1}}x-1\Rightarrow \frac{dy}{dx}=2x{{\tan }^{-1}}x\]
You need to login to perform this action.
You will be redirected in
3 sec
