How to show by the chain rule and the FTC?

Question

How can I apply the FTC and the chain rule to prove this: $$\frac{d}{dx}\displaystyle\int_{a}^{x}\int_{c}^{d}f(x,y,z)dzdy=\int_{c}^{d}f(x,y,z)dz+\int_{a}^{x}\int_{c}^{d}f_{x}(x,y,z)dzdy.$$

Answer 1

I forget about the inner integral because it does not play any role.

Define the function $$ F(t,x)=\int_a^t f(x,y) \,d y\,. $$ What you want to compute is nothing but $$ \frac{d}{d x} (F\circ i)(x) $$ where $i$ is the map $x\mapsto(x,x)$. By the chain rule, you have $$ \frac{d}{d x} (F\circ i)(x)=(\nabla F)(i(x))\cdot \frac{d}{d x}i(x)=(\partial_t F)(x,x)+(\partial_x F)(x,x) $$ because $\frac{d}{d x}i(x)=(1,1)$. To conclude, observe that $$ (\partial_t F)(t,x)= f(x,t) \qquad (\partial_x F)(t,x)=\int_a^t (\partial_x f)(x,y) \,d y. $$