Generalizing the Leibniz rule Given the Leibniz rule:
$$\frac{d}{dy}\int^a_b f(x,y) dx = \int_b^a \frac{\partial f}{\partial y} (x,y) dx$$
How do I prove a more general case using the chain rule and the above:
$$\frac{d}{dy} \int_{g_1(y)}^{g_2(y)} f(x,y) dx =?$$
From the fundamental theorem of calculus we have that: $$\frac{d}{dt} \int_{f_2(t)}^{f_1(t)} g(s) ds = g(f_1(t))f'_1(t) - g(f_2(t))f'_2(t)$$
But when I just apply the fundamental theorem of calculus I get an incorrect answer (because I also need to use Leibniz rule itself...). 
 A: Do the linear change of variables $$x = g_1(y)+(g_2(y)-g_1(y))u.$$
The integral becomes $$\int_{g_1(y)}^{g_2(y)} f(x,y) dx =(g_2(y)-g_1(y)) \int_0^1 f(g_1(y)+(g_2(y)-g_1(y))u,y) du$$ and now you can apply the traditionnal Leibniz Rule.
A: You're very, very close. The last term is the Leibnitz rule as you are familiar with it itself. That is,
$$\dfrac{d}{dt}\int_a^bf(x,t)dx=f(b,t)b'-f(a,t)a'+\int_a^b\dfrac{\partial f(x,t)}{\partial t}dx.$$ This is because our integral
$$=F(b,t)-F(a,t)$$ which when differentiated gives
$$\dfrac{\partial F(b,t)}{\partial b}b'+\dfrac{\partial F(b,t)}{\partial t}-\dfrac{\partial F(a,t)}{\partial b}a'-\dfrac{\partial F(a,t)}{\partial t}$$ which afterwards it becomes obvious that
$$\dfrac{\partial F(b,t)}{\partial b}b'-\dfrac{\partial F(a,t)}{\partial a}a'=f(b,t)b'-f(a,t)a'$$ and that because we took the partial derivatives of F as if $b$ and $a$ were independent of $t$, we obtain that the last term is
$$\dfrac{\partial F(b,t)}{\partial t}-\dfrac{\partial F(a,t)}{\partial t}=\dfrac{\partial F(x,t)}{\partial t}\Bigg|_{x=a}^{x=b}=\dfrac{\partial}{\partial t}\int_a^b f(x,t)dx=\int_a^b \dfrac{\partial}{\partial t}f(x,t)dx$$ Done!
