Differentiation of Function Defined as Integral of Multiple Variables I am stuck on a question regarding finding the derivative $f'(0)$ of a function $f(x)$ defined as $f(x) = \int_{-\infty}^{e^{2x}} e^{\frac{-1}{2}(y-sin(x))^2}\, dy $ for $x \epsilon {\Bbb R}$. 

I am vaguely aware that Leibniz' Integral Rule is the starting point for this question, but I don't know how to deal with the $-\infty$ in the lower bound of the integral. (https://en.wikipedia.org/wiki/Leibniz_integral_rule) 

How do I compute this integral? Many thanks for any help given.
We have that $$f(x) = \int_{-\infty}^{e^{2x}} e^{-\frac{1}{2}\left(y-\sin x\right)^2}\mathrm d y.$$ Thus, $$f'(x) = \left(e^{2x}\right)'\left[e^{-\frac{1}{2}\left(e^{2x}-\sin x\right)^2}\right]+\int_{-\infty}^{e^{2x}} e^{-\frac{1}{2}\left(y-\sin x\right)^2}[-\left(y-\sin x\right)](-\cos x)\mathrm d y=2e^{2x}\left[e^{-\frac{1}{2}\left(e^{2x}-\sin x\right)^2}\right]+\int_{-\infty}^{e^{2x}} e^{-\frac{1}{2}\left(y-\sin x\right)^2}\left(y-\sin x\right)(\cos x)\mathrm d y.$$

Now set $x=0$ to obtain $f'(0).$
We have that $$f(x) = \int_{-\infty}^{e^{2x}} e^{-\frac{1}{2}\left(y-\sin x\right)^2}\mathrm d y.$$ Thus, $$f'(x) = \left(e^{2x}\right)'\left[e^{-\frac{1}{2}\left(e^{2x}-\sin x\right)^2}\right]+\int_{-\infty}^{e^{2x}} e^{-\frac{1}{2}\left(y-\sin x\right)^2}[-\left(y-\sin x\right)](-\cos x)\mathrm d y=2e^{2x}\left[e^{-\frac{1}{2}\left(e^{2x}-\sin x\right)^2}\right]+\int_{-\infty}^{e^{2x}} e^{-\frac{1}{2}\left(y-\sin x\right)^2}\left(y-\sin x\right)(\cos x)\mathrm d y.$$

Now set $x=0$ to obtain $f'(0).$