How to differentiate $F(x,y)=\int_x^y \sqrt{e^{tx}+3y}dt$ I want to compute $D_1f$ and $D_2f$, two partial derivatives.
The only tool I have now is the fundamental theorem of calculus and chain rule. Maybe I can write $F(x,y)$ as some composition functions and then apply the chain rule?
The answer is:
$D_1f(x,y)=-\sqrt{e^{x^2}+3y}+\int_x^y \frac{te^{tx}}{2\sqrt{e^{tx}+3y}}dt$
$D_2f(x,y)=\sqrt{e^{xy}+3y}+\int_x^y\frac{3}{2\sqrt{e^{tx}+3y}}dt$
 A: Hint
Just to make it more general, consider the more complex case where$$G(x,y)=\int_{a(x,y)}^{b(x,y)} F[x,y,t] \, dt$$ Then, the fundamental theorem of calculus leads to $$\frac{dG(x,y)}{dx}=\int_{a(x,y)}^{b(x,y)} \frac{dF[x,y,t]}{dx} \, dt+F[x,y,b(x,y)]\frac{db(x,y)}{dx}-F[x,y,a(x,y)]\frac{da(x,y)}{dx}$$ $$\frac{dG(x,y)}{dy}=\int_{a(x,y)}^{b(x,y)} \frac{dF[x,y,t]}{dy} \, dt+F[x,y,b(x,y)]\frac{db(x,y)}{dy}-F[x,y,a(x,y)]\frac{da(x,y)}{dy}$$
The case of the post is much simpler since  $$a(x,y)=x~,~ \frac{da(x,y)}{dx}=1~,~\frac{da(x,y)}{dy}=0$$ $$b(x,y)=y~,~\frac{db(x,y)}{dx}=0~,~\frac{db(x,y)}{dy}=1$$ and so $$\frac{dG(x,y)}{dx}=-F[x,y,x]+\int_{x}^{y} \frac{dF[x,y,t]}{dx} \, dt$$ $$\frac{dG(x,y)}{dy}=F[x,y,y]+\int_{x}^{y} \frac{dF[x,y,t]}{dy} \, dt$$
A: Hint: $F(x,y)=\int_x^y \sqrt{e^{tx}+3y}dt = \int_x^a \sqrt{e^{tx}+3y}dt +\int_a^y \sqrt{e^{tx}+3y}dt = - \int_a^x \sqrt{e^{tx}+3y}dt +\int_a^y \sqrt{e^{tx}+3y}dt$ for any constant $a$ between $x$ and $y$. Now, apply fundamental theorem of calculus and chain rule to both terms. 
A: Your idea to use composition and the chain rule is good. Consider $G(u,v,w,z)=\int_u^v \sqrt{e^{tw}+3z}\,dt$. Note that your $F$ is $G$ with compositions $u(x,y)=x$, $v(x,y)=y$, $w(x,y)=x$, and $z(x,y)=y$.
So 
$$\begin{align}
\frac{\partial F}{\partial x}&=\frac{\partial G}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial G}{\partial v}\frac{\partial v}{\partial x}+\frac{\partial G}{\partial w}\frac{\partial w}{\partial x}+\frac{\partial G}{\partial z}\frac{\partial z}{\partial x}\\
\end{align}$$
Can you find all these simpler pieces? And remember that once you have the right side evaluated, any $u,v,w$, or $z$ should be rewritten in terms of $x$ and $y$.
