# Thread: taking the derivative of an integral with multiple variables

1. ## taking the derivative of an integral with multiple variables

$F(x)=\int_{h(x)}^{g(x)}f(x,t)dt$

How would I evaluate F'(x) without evaluating the integral? More specifically, I have a problem that states:

$f(x)=\int_{0}^{h(x)}(g(x,t))^4dt$

where g and h are both C1, and asks for a formula for f'(x), but I really have no idea how to tackle this.

2. $F(x)=\int_{h(x)}^{g(x)}f(x,t)dt$

So we take derivatives with respect to x of both sides.

$\frac{\partial}{\partial{x}}F(x)=\frac{\partial}{\ partial{x}}\int_{h(x)}^{g(x)}f(x,t)dt$

apply the fundamental theorem of calculus

$F'(x) = h_{x}(x)*f_{x}(h(x)) - g_{x}(x)*f_{x}(h(x))$

3. I assume you mean

$F'(x)=g_{x}(x)*f_{x}(g(x))-h_{x}(x)*f_{x}(h(x))$?

And $h_{x}(x)$ would denote the derivative of h with respect to x?

Either way, I'm having trouble making much sense of this, or I'm just applying it totally incorrectly.

Let's say I have

$F(x)=\int_{0}^{1}(x+t)dt$

Solutions clearly depend on x so F'(x) shouldn't be 0, but since 0 and 1 are constants, their derivatives are 0 and I would get F'(x) = 0 using the fundamental theorem.

From class, I have that if $F(x)=\int_{a}^{b}f(x,t)dt$, then $F'(x)=\int_{a}^{b}\frac{\partial{f(x,t)}}{\partial {x}}dt$, but I'm not sure what to do when the limits of integration are functions of x.

4. When you apply the bounds of 0 and 1, you won't get constants you will get an expression in terms of variable x. So the partial derivative in terms of x won't give you 0 because there are variable terms. You gave an easy example to integrate. Do the integral, apply the bounds, then differentiate with respect to x. You shouldn't get 0.

I think the derivative you wrote is the correct one, as the lower bound is subtracted from the top bound with integrals. When the bounds are functions of x, then you get a F(g(x)) situation and the chain rule is used. Nothing tricky.

I think you get it but think you don't. Which part is stumping you?

5. I have no idea why this is confusing me so much, as it seems like it should be very simple.

I can see that obviously $\frac{\partial}{\partial{x}}\int_{0}^{1}(x+t)dt=1$ if I evaluate the integral first then differentiate, but I'm unsure of how to get it by using the fundamental theorem directly, which I need to be able to do.

Perhaps a better example would be $\frac{\partial}{\partial{x}}\int_{0}^{x}(x+t)dt$. If you evaluate the integral first then differentiate then it's clear the answer is 3x. But when I use the fundamental theorem (incorrectly), I get $(x+x)=2x$.

6. Well I figured it out, with a little help from wolfram alpha.

Apparently $\frac{\partial}{\partial{x}}\int_{g(x)}^{h(x)}f(x, t)dt=\int_{g(x)}^{h(x)}\frac{\partial}{\partial{x} }f(x,t)dt+h'(x)f(x,h(x))-g'(x)f(x,g(x))$, in case anyone was interested.

As to how that's derived, I have no idea.