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

$\displaystyle

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

$

How do I find g'(x)?

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- May 16th 2010, 09:56 AMAlderDragonDerivative of an Integral problem
I need to find the derivative of an integral, with a limit of integration being a function:

$\displaystyle

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

$

How do I find g'(x)? - May 16th 2010, 10:16 AMCaptainBlack
Put:

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

Then:

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

and so by the chain rule:

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

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

CB - May 16th 2010, 05:49 PMAlderDragon
Upon closer inspection of my past work I found another method of doing this type of problem. Substitute $\displaystyle x^4$ in for $\displaystyle t$ and multiply by the derivative of $\displaystyle x^4$, $\displaystyle 4x^3$:

$\displaystyle

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

$

$\displaystyle

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? - May 16th 2010, 07:42 PMCaptainBlack
- May 17th 2010, 03:18 AMHallsofIvy
By the way, here is the general Leibniz formula:

$\displaystyle \frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} F(x,t)dt= $$\displaystyle F(x, \beta(x))\frac{d\beta(x)}{dx}- F(x,\alpha(t))\frac{d\alpha(x)}{dx}+$$\displaystyle \int_{\alpha(x)}^{\beta(x)}$$\displaystyle \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.