If $f(x)=x \tan ^{-1} x$, then f'(1) is equal to
Answer & explanation
Correct answer: option 1
We have,
$f(x)=x \tan ^{-1} x$
$\Rightarrow f'(x)=\tan ^{-1} x+\frac{x}{1+x^2}$
$\Rightarrow f'(1)=\tan ^{-1} 1+\frac{1}{2}=\frac{\pi}{4}+\frac{1}{2}$
If $f(x)=x \tan ^{-1} x$, then f'(1) is equal to
Correct answer: option 1
We have,
$f(x)=x \tan ^{-1} x$
$\Rightarrow f'(x)=\tan ^{-1} x+\frac{x}{1+x^2}$
$\Rightarrow f'(1)=\tan ^{-1} 1+\frac{1}{2}=\frac{\pi}{4}+\frac{1}{2}$