Chain rule for integrals, how?

Question

Can you please give some hints how to solve such a task:

Given 3 smooth functions: $f: \mathbb R^2 \rightarrow \mathbb R$, $a,b: \mathbb R \rightarrow \mathbb R$. I should determin the derivative of:

$$x \rightarrow \int_{a(x)}^{b(x)}f(t,x)dt$$

I have really no idea. Even with these hints: $$\frac{\partial}{\partial x}\int_{a}^{b}f(t,x)dt = \int_{a}^{b}\left(\frac{\partial}{\partial x}f\right)(t,x)dt$$

Could you please help?

Answer 1

Hint: Use Leibniz Rule

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

which follows from the chain rule. As an idea

$$\frac{\partial}{\partial x}\int_c ^{a(x)} f(t,x) dt = F_a(a(x))\cdot a'(x) + F_x (a(x), x) $$

where $F(a(x),x) = \int_c^{a(x)} f(t,x) dt$.

Answer 2

Ok, hints. Write: $$g(x) = \int_{a(x)}^{b(x)}f(t,x)\,{\rm d}t,$$for convenience. Rewrite as: $$g(x) = \int_0^{a(x)}f(t,x)\,{\rm d}t - \int_{0}^{b(x)}f(t,x)\,{\rm d}t.$$Now we have: $$g'(x) = \frac{\rm d}{{\rm d} x}\int_0^{a(x)}f(t,x)\,{\rm d}t - \frac{\rm d}{{\rm d} x}\int_{0}^{b(x)}f(t,x)\,{\rm d}t.$$ Your problem boils down to computing each derivative. Because of the chain rule, we will get $a'(x)$ and $b'(x)$ appearing. Now use the initial hint given (Leibniz's rule) in each one. Can you go on?

Answer 3

It might be easier to consider the function of 3 variables $F(a,b,x) = \int_a^b f(t,x) dt$ and apply the chain rule to $F(a(x),b(x),x)$.

Of course, you need to compute $\partial_a F$, $\partial_b F$ and $\partial_x F$.