Show $\frac{d}{dx} \int\limits^{a(x)}_{0} f(x,y)dy = \int\limits^{a}_{0} \frac{\partial f}{\partial x}dy + a'(x)f(x,a)$ I am trying to show that 
$$
\frac{d}{dx} \int\limits^{a(x)}_{0} f(x,y)dy = \int\limits^{a}_{0}  \frac{\partial f}{\partial x}dy + a'(x)f(x,a)
$$
I know this has something to do with the fundamental theorem of calculus but am having trouble making any sort of progress. If someone could point me in the right direction, it would be much appreciated.
 A: We set
$$
F(x,y)=\int_0^yf(x,s)\,ds,
$$
and assume that $\partial_1F$ and $\partial_2F$ exist. This is the case, e.g. when  $f$ is continuous in $y$ and differentiable in $x$. Then
$$
\partial_1F(x,y)=\int_0^y\frac{\partial f}{\partial x}(x,s)\,ds,\quad
\partial_2F(x,y)=f(x,y).
$$
(See, e.g. Prove that $h'(t)=\int_{a}^{b}\frac{\partial\phi}{\partial t}ds$. for the first identity). 
It follows that
\begin{eqnarray}
\frac{d}{dx}\int_0^{a(x)}f(x,y)\,dy&=&\frac{d}{dx}(F(x,a(x)))=\partial_1F(x,a(x))+a'(x)\partial_2F(x,a(x)\\
&=&\int_0^{a(x)}\frac{\partial f}{\partial x}(x,y)\,dy+a'(x)f(x,a(x)).
\end{eqnarray}
A: This is Leibniz's integration rule. The proof is by partial differentiation. Define the function $F(x,z):=\int_0^z f(x,y)dy$. Then your integral becomes $F(x,a(x))$. By partial differentiation, we get:
$$\frac{dF}{dx}=\frac{\partial F}{\partial x}+\frac{\partial F}{\partial z}\frac{dz}{dx}$$
Now, $\frac{\partial F}{\partial x}=\int_0^z \frac{\partial f}{\partial x}$. As well, $\frac{\partial F}{\partial z}=f(x,y)dy$ by fundamental theorem of calculus. Finally, writing $z=a(x)$, we have $\frac{dz}{dx}=a'(x)$ giving the desired result.
A: Hint: Define $H(x,z)=\int_0^{z} f(x,y)dy$, hence you want to calculate $\displaystyle\frac{d H(x,z(x))}{dx}$ where $z(x)=a(x)$
A: Let a $C^1$ function $f$ and 
$$
F(x,u,v,t)=\int_{u}^{v}f(x,t) dx.
$$
Use the implicit differentiation ruler in function 
$$
\varphi(x)=F(x,u(x),v(x),t(x))= \int_{u(x)}^{v(x)}f(x,t(x)) dx .
$$
Here, $u=u(x)$, $v=v(x)$, $t=t(x)$ are $C^1$ functions. 
Implicit differentiation rule: For all fix $x_0$  we have 
 $u_0=u(x_0)$, $v_0=v(x_0)$, $t_0=t(x_0)$ and 
\begin{align}
\left.\frac{d\varphi(x)}{dx}\right|_{x=x_0}=
&
\left.\frac{\partial F(x,u_0,v_0,t_0)}{\partial x}\right|_{x=x_0}
\\
+
&
\left.\frac{\partial F (x_0,u(x),v_0,t_0)}{\partial x}\right|_{x=x_0}
\\
+
&
\left.\frac{\partial F (x_0,u(x),v_0,t_0)}{\partial x}\right|_{x=x_0}
\\
+
&
\left.\frac{\partial F (x_0,u_0,v(x),t_0)}{\partial x}\right|_{x=x_0}
\\
+
&
\left.\frac{\partial F (x_0,u_0,v_0,t(x))}{\partial x}\right|_{x=x_0}
\\
\end{align}
By  chain rule, we get:
\begin{align}
\left.\frac{d\varphi(x)}{dx}\right|_{x=x_0}=
&
\left(\left.\frac{\partial F(x,u_0,v_0,t_0)}{\partial x}\right|_{x=x_0}\right)
\cdot
\left(\left.\frac{dx}{dx}\right|_{x=x_0}\right)
\\
+
&
\left(\left.\frac{\partial F (x_0,u,v_0,t_0)}{\partial u}\right|_{u=u_0}\right)
\cdot
\left(\left.\frac{du}{dx}\right|_{x=x_0}\right)
\\
+
&
\left(\left.\frac{\partial F (x_0,u_0,v,t_0)}{\partial v}\right|_{v=v_0}\right)
\cdot
\left(\left.\frac{dv}{dx}\right|_{x=x_0}\right)
\\
+
&
\left(\left.\frac{\partial F (x_0,u_0,v_0,t)}{\partial t}\right|_{t=t_0}\right)
\cdot
\left(\left.\frac{dt}{dx}\right|_{x=x_0}\right)
\end{align}
Then the result fellowing by Leibniz rule and fundametal teorem of calculus.
