TITLE: Leibniz rule derivation
QUESTION [11 upvotes]: How is Leibniz Integral rule derived?
$$\frac {\mathrm{d}}{\mathrm{d}x}\left(\int_{a(x)}^{b(x)}f(x, t) \,\mathrm{d}t\right)= f(x,b(x))\frac{\mathrm{d}}{\mathrm{d}x}b(x)- f(x, a(x))\dfrac{\mathrm{d}}{\mathrm{d}x}a(x)+ \displaystyle\int_{a(x)}^{b(x)}\dfrac{\partial f(x,t)}{\partial x} \,\mathrm{d}t.$$
Also, what is the intuition behind this formula?

REPLY [6 votes]: We have a function $\Phi$of three variables, namely
$$\Phi(u,v,w):=\int_u^v f(w,t)\>dt\ ,$$
with the necessary continuity assumptions when $(u,v,w)$ range in some three-dimensional domain $\Omega$. With these givens the function$$g(x):=\Phi\bigl(a(x),b(x),x\bigr)$$
is defined, and we are told to compute its derivative $g'(x)$. By the chain rule we have
$$g'(x)=\Phi_{.1}\bigl(a(x),b(x),x\bigr)a'(x)+\Phi_{.2}\bigl(a(x),b(x),x\bigr)b'(x)+\Phi_{.3}\bigl(a(x),b(x),x\bigr)x'(x)\ .$$
Since
$$\Phi_{.1}\bigl(u,v,w\bigr)=-f(w,u),\quad \Phi_{.2}\bigl(u,v,w\bigr)=f(w,v),\quad \Phi_{.3}\bigl(u,v,w\bigr)=\int_u^v f_{.1}(w,t)\>dt\ ,$$
whereby only the third part really needs some work, we arrive at the stated formula by plugging everything in. 
It remains to prove the "vanilla" Leibniz rule
$${d\over dw}\int_u^v f(w,t)\>dt=\int_u^v f_{.1}(w,t)\>dt\ .$$
For this we need that under suitable continuity assumptions the convergence
$$\lim_{h\to0}{f(w+h,t)-f(w,t)\over h}=f_{.1}(w,t)$$
is uniform in $t\in[u,v]$.