How do I differentiate this integral?

That is:

$$\left(\int_{a(x)}^{b(x)}\!f(x,t)\,dt\right)'$$

I don't know how to differentiate a integral if functions of $x$ are at its limits.

Can you guys show me how to do this?

$$\frac{d}{dx}\int_{a(x)}^{b(x)} f(x,t)dt=f(x,b(x))\frac{d}{dx}b(x)-f(x,a(x))\frac{d}{dx}a(x)+\int_{a(x)}^{b(x)}\frac{\partial}{\partial{x}}f(x,t)dt,$$

where I have used Leibniz's Rule.

Note that if $a(x)$ and $b(x)$ are constants, then we have a special case of Leibniz's Rule.

There is an intermediate function of three variables involved, namely $$F(u,v,w):=\int_u^v f(w,t)\ dt\ .$$ One has $$F_u(u,v,w)=-f(w,u)\ ,\quad F_v(u,v,w)=f(w,v)\ ,\quad F_w(u,v,w)=\int_u^v f_w(w,t)\ dt\ ,$$ where the last formula is Leibniz' Rule "without extras".

When the variables $u$, $v$, $w$ become functions of $x$: $$u(x):=a(x)\ ,\quad v(x):=b(x)\ ,\quad w(x):=x\ ,$$ then the composition with $F$ defines a function $\phi(x):=F\bigl(a(x),b(x),x\bigr)$. In order to compute the derivative $\phi'$ we have to use the chain rule and obtain the formulas given in Nana's answer.