Differentiate $f(x)=\int_x^{10}e^{-xy^2}dy$ with respect to $x$ 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:
$$ f'(x)=-\frac{1}{x} \int_x^{10} e^{-xy^2} dy + \frac{5}{x} e^{-100x}-\frac{3}{2}e^{-x^3}$$

There are indeed formula for the differentiation of functions of the form $x\mapsto \int_{a(x)}^{b(x)}f(x,t)\mathrm{dt}$ under some good conditions on the functions $a$, $b$ and $f$. 
However, in our case, the function $f$ has a quite nice form. We can start from the substitution $xy^2=t^2$, hence $t=\sqrt x\cdot y$ which gives $\mathrm dt=\sqrt x\cdot\mathrm dy$. Therefore
$$f(x)=\int_x^{10}e^{-xy^2}\mathrm dy=\frac 1{\sqrt x}\int_{x^{3/2}}^{10 \sqrt x}e^{-t^2}\mathrm dt=\frac 1{\sqrt x}\int_{0}^{10 \sqrt x}e^{-t^2}\mathrm dt-
\frac 1{\sqrt x}\int_{0}^{10 \sqrt x}e^{-t^2}\mathrm dt.$$
The derivative of the rand hand side is easier to compute using the fundamental theorem of analysis and the chain rule.
