Answer
$ \displaystyle \frac{1+2x}{\sqrt{1-x^{2}}}$
Work Step by Step
f(x) is a composite function$, f(x)=u(v(x))$, where
$u(x)=\sqrt{x}=x^{1/2},\quad v(x)=x+x^{2}$
Apply the Chain rule: $\displaystyle \frac{df}{dx}=\frac{du}{dv}\frac{dv}{dx}$
$\displaystyle \frac{du}{dv}=\frac{1}{2}v^{-1/2}=\frac{1}{2\sqrt{v}}=\frac{1}{2\sqrt{x+x^{2}}}$
$\displaystyle \frac{dv}{dx}=1+2x$
$\displaystyle \frac{df}{dx}=\frac{du}{dv}\frac{dv}{dx}=\frac{1}{2\sqrt{x+x^{2}}}\cdot(1+2x)$
$=\displaystyle \frac{1+2x}{\sqrt{1-x^{2}}}$