$ F(x)= \int_{\sin x}^{\cos x} e^{t^2+xt}\,dt$. What is $F'(0)=$? 
$$\begin{align*} F(x)&= \int_{\sin x}^{\cos x} e^{t^2+xt}\,\mathrm dt\\F'(0)&=\,?\end{align*}$$

I only know how to deal with the kind like $$F(x)=\int h(x)g(t)\,\mathrm dt $$ for which I can treat $h(x)$ as a constant relative to $t$ and move it out of the integral and differentiate. But in this problem I cannot separate $x$ from $t$.
 A: Let $F(x)$ be given by
$$F(x)=\int_{\sin(x)}^{\cos(x)}e^{t^2+xt}\,dt$$
Using Leibniz's Rule for Differentiating Under the Integral, we have
$$\begin{align}
F'(x)&=\left.\left(e^{t^2+tx}\right)\right|_{t=\cos(x)}\left(\frac{d\cos(x)}{dx}\right)-\left.\left(e^{t^2+tx}\right)\right|_{t=\sin(x)}\left(\frac{d\sin(x)}{dx}\right)\\\\
&+\int_{\sin(x)}^{\cos(x)}\frac{\partial \left(e^{t^2+xt}\right)}{\partial x}\,dt\\\\
&=-\sin(x)e^{\cos^2(x)+x\cos(x)}-\cos(x)e^{\sin^2(x)+x\sin(x)}+\int_{\sin(x)}^{\cos(x)}te^{t^2+xt}\,dt\tag 1
\end{align}$$
Setting $x=0$ in $(1)$ reveals
$$\begin{align}
F'(0)&=-1+\int_0^1te^{t^2}\,dt\\\\
&=-1+\frac12(e^1-1)\\\\
&=\frac12(e^1-3)
\end{align}$$
A: I apologize if this is just Dr. MV's idea, but I hope this is a simpler exposition.
Using $v=\cos(x)$, $u=\sin(x)$, and $y=t^2$, we have
$$
\begin{align}
&\frac{\mathrm{d}}{\mathrm{d}x}\int_{\sin(x)}^{\cos(x)}e^{t^2+xt}\mathrm{d}t\\[3pt]
&=\frac{\mathrm{d}v}{\mathrm{d}x}\frac{\partial}{\partial v}\int_u^ve^{t^2+xt}\mathrm{d}t
-\frac{\mathrm{d}u}{\mathrm{d}x}\frac{\partial}{\partial u}\int_u^ve^{t^2+xt}\mathrm{d}t
+\frac{\partial}{\partial x}\int_u^ve^{t^2+xt}\mathrm{d}t\\[3pt]
&=\frac{\mathrm{d}v}{\mathrm{d}x}e^{v^2+xv}
-\frac{\mathrm{d}u}{\mathrm{d}x}e^{u^2+xu}
+\int_u^vte^{t^2+xt}\mathrm{d}t\\
&=-\sin(x)e^{\cos^2(x)+x\cos(x)}
-\cos(x)e^{\sin^2(x)+x\sin(x)}
+\int_{\sin(x)}^{\cos(x)}te^{t^2+xt}\mathrm{d}t\\
&=-0-1+\int_0^1te^{t^2}\mathrm{d}t\\
&=-1+\frac12\int_0^1e^y\mathrm{d}y\\[3pt]
&=\frac12e-\frac32
\end{align}
$$
A: Consider the function of three variables
$$g(x,y,z)=\int_z^y e^{t^2+xt}\,dt.$$
By the fundamental theorem of calculus, you know that $\frac{\partial g}{\partial y}(x,y,z)=e^{y^2+xy},$ while $\frac{\partial g}{\partial z}(x,y,z)=-e^{z^2+xz}$. Also  $\frac{\partial g}{\partial x}(x,y,z)=\int_z^y  \frac{\partial}{\partial x}(e^{t^2+xt})dt=\int_z^yt (e^{t^2+xt})\,dt.$
Now your function is the composition $g(x,\cos x,\sin x)$ and so you can use the chain rule to find that
$$F'(x)=\frac{\partial g}{\partial x}(x,\cos x,\sin x)+\frac{\partial g}{\partial y}(x,\cos x,\sin x)(-\sin x)+\frac{\partial g}{\partial z}(x,\cos x,\sin x)  (\cos x).$$
