$\frac{d}{d x}\left\{\left(1+x^2 ight) an ^{-1}(x) ight\}=$

Question

$\frac{d}{d x}\left\{\left(1+x^2ight) 	an ^{-1}(x)ight\}=$

Aicharya · Accepted Answer

We have, $ \begin{gathered} \frac{d}{d x}(u \cdot v)=u \cdot \frac{d v}{d x}+v \cdot \frac{d v}{d x} \ \frac{d}{d x}\left[(1+x)^2 \cdot 	an ^{-1}(x)ight]=\left(1+x^2ight) \ \frac{d}{d x} 	an ^{-1} x+\frac{d}{d x}\left(1+x^2ight) \cdot 	an ^{-1} x \ =\left(1+x^2ight) \cdot \frac{1}{1+x^2}+(2 x) \cdot 	an ^{-1} x \ =1+2 x 	an ^{-1} x=2 x 	an ^{-1} x+1 \end{gathered} $ Hence, option (3) is correct

$\frac{d}{d x}\left\{\left(1+x^2\right) \tan ^{-1}(x)\right\}=$

Solution