# Math Help - Derivative of an Integral problem

1. ## Derivative of an Integral problem

I need to find the derivative of an integral, with a limit of integration being a function:

$

g(x) = \int_1^{x^{4}}(t^3+2)dt
$

How do I find g'(x)?

2. Originally Posted by AlderDragon
I need to find the derivative of an integral, with a limit of integration being a function:

$

g(x) = \int_1^{x^{4}}(t^3+2)dt
$

How do I find g'(x)?
Put:

$h(x)=\int_1^{x}(t^3+2)dt$

Then:

$g(x)=h(x^4)$

and so by the chain rule:

$\frac{d}{dx}g(x)=4x^3 \left. \frac{dh}{dx}\right|_{x^4}$

where the derivative of $h$ is obtained by applying the fundamental theorem of calculus.

CB

3. Upon closer inspection of my past work I found another method of doing this type of problem. Substitute $x^4$ in for $t$ and multiply by the derivative of $x^4$, $4x^3$:

$

g(x) = \int_1^{x^{4}}(t^3+2)dt
$

$
g'(x) = (x^{12} + 2)4x^3
$

Is this the correct answer? And, if so, is this a different answer than you might obtain using a different method?

4. Originally Posted by AlderDragon
Upon closer inspection of my past work I found another method of doing this type of problem. Substitute $x^4$ in for $t$ and multiply by the derivative of $x^4$, $4x^3$:

$

g(x) = \int_1^{x^{4}}(t^3+2)dt
$

$
g'(x) = (x^{12} + 2)4x^3
$

Is this the correct answer? And, if so, is this a different answer than you might obtain using a different method?
That is what you get if you finish what I posted in my last post, so yes it is correct. However why you do what you say you do (and what it means for that matter) needs explaining.

CB

5. By the way, here is the general Leibniz formula:
$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} F(x,t)dt=$ $F(x, \beta(x))\frac{d\beta(x)}{dx}- F(x,\alpha(t))\frac{d\alpha(x)}{dx}+$ $\int_{\alpha(x)}^{\beta(x)}$ $\frac{\partial F(x,t)}{\partial x} dt$.

It can be proved from the fundamental theorem of calculus and using the chain rule for both the upper and lower limits.