Answer
$\frac{dy}{dx}={(2x)}e^{\tan^{-1}{x}}+(e^{\tan^{-1}{x}})$
Work Step by Step
We are given that:
$y=(1+x^2)e^{\tan^{-1}{x}}$
$\frac{dy}{dx}=\frac{d((1+x^2))}{dx}e^{\tan^{-1}{x}}+(1+x^2)\frac{d(e^{\tan^{-1}{x}})}{dx}$
$\frac{dy}{dx}={(2x)}e^{\tan^{-1}{x}}+(1+x^2)(e^{\tan^{-1}{x}})\frac{d({\tan^{-1}{x}})}{dx}$
$\frac{dy}{dx}={(2x)}e^{\tan^{-1}{x}}+(1+x^2)(e^{\tan^{-1}{x}})\frac{{{1}}}{1+x^2}$
$\frac{dy}{dx}={(2x)}e^{\tan^{-1}{x}}+(e^{\tan^{-1}{x}})$
