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)?
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
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?
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.