## University Calculus: Early Transcendentals (3rd Edition)

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