TITLE: Differentiate $f(x)=\int_x^{10}e^{-xy^2}dy$ with respect to $x$
QUESTION [7 upvotes]: I am trying to find $f'(x)$ when $0\leq x\leq 10$. I know I could use the formula given on this wikipedia page: http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign but I have been asked to justify all steps of the calculation so this isn't allowed.
I have been given a hint to let $I(a,b,c)=\int_a^bf(x,c)dx$ and then told to show that $f$ satisfies all conditions necessary for FTC1 and the theorem of differentiation of integrals depending on a parameter.
The problem I am having is translating $f(x)$ into something of the same form as $I(a,b,c)$. Can anyone help?
EDIT: I think I've done it now using the method described by @mvggz . Is this the final answer once the $u$ has been substituted back out:
$$ f'(x)=-\frac{1}{x} \int_x^{10} e^{-xy^2} dy + \frac{5}{x} e^{-100x}-\frac{3}{2}e^{-x^3}$$

REPLY [2 votes]: Put $$\Phi(a,b) = \int_a^{10}e^{-by^2}dy$$
Then $f(x) = \Phi(x,x).$ By the chain rule, you have
$$f'(x) = \partial_a  \Phi \cdot{a'(x)} + \partial_{b}\Phi\cdot b'(x)$$
By the fundamental theorem of calculus,
$\partial_a \Phi = e^{-a^2 b}$. By the Leibnitz rule, you have
$$\partial_b \Phi = \int_a^{10} -y^2e^{-by^2}\, dy$$
Now put $a(x) = x$ and $b(x) = x$.