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

I would try this : eliminate the x inside the integral.
Use a change of variable: $u = \sqrt{x}*y$ => $ du = \sqrt{x}*dy $
You get : $ f(x) = \frac{1}{\sqrt{x}}\int_{x^{\frac{3}{2}}}^{10*\sqrt{x}} e^{-u^2} du $
Let : $F(x) = \int_0^x e^{-u^2}du $
=> $ f(x) = \frac{1}{\sqrt{x}}*[F(10*\sqrt{x}) - F(x^{\frac{3}{2}})] $
Now you can derive just like regular functions : 
$ f'(x) = -\frac{1}{2*x^{\frac{3}{2}}}*[F(10*\sqrt{x}) - F(x^{\frac{3}{2}})] + \frac{1}{\sqrt{x}}*[\frac{5}{\sqrt{x}}*e^{-(10\sqrt{x})^2} -\frac{3}{2}\sqrt{x}*e^{-x^3}  ]$
=> $ f'(x) = -\frac{x^{-\frac{3}{2}}}{2}*\int_{x^{\frac{3}{2}}}^{10*\sqrt{x}} e^{-u^2} du +\frac{5}{x}e^{-100x} - \frac{3}{2}*e^{-x^3}$
