Answer
$f'(x)=\frac{2x\cdot e^{-4x^{2}}-e^{-(\ln x)^{2}}}{x}$
Work Step by Step
$$f(x)=\int_{\ln x}^{2x}e^{-t^{2}}dt$$
Use the rule $\frac{d}{dx}\left(\int_{h(x)}^{p(x)}g(t)dt\right)=p'(x)g(p(x))-h'(x)g(h(x))$ to find the first derivative of $f$ so it follows: $f'(x)=(2x)'\cdot e^{-(2x)^{2}}-(\ln x)'\cdot e^{-(\ln x)^{2}}$ $f'(x)=2\cdot e^{-4x^{2}}-\frac{1}{x}\cdot e^{-(\ln x)^{2}}$
$f'(x)=\frac{2x\cdot e^{-4x^{2}}-e^{-(\ln x)^{2}}}{x}$
