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:
$$ f'(x)=-\frac{1}{x} \int_x^{10} e^{-xy^2} dy + \frac{5}{x} e^{-100x}-\frac{3}{2}e^{-x^3}$$
 A: Write the integral as
$$\begin{align}F(x) &= \int_0^{10} dy \, e^{-x y^2} - \int_0^{x} dy \, e^{-x y^2} \\ &= 10 \int_0^{1} du \, e^{-100 \,x\, u^2} - x \int_0^1 du \, e^{-x^3 u^2} \end{align}$$
so that
$$F'(x) = -1000 \int_0^{1} du \, u^2 \, e^{-100 \,x\, u^2} - \int_0^1 du \, e^{-x^3 u^2} + 3 x^3 \int_0^{1} du \, u^2 \, e^{-x u^2} $$
These may be expressed in terms of error functions.
A: 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 $b(x) = x$.
A: The integral is explicitly given by the error function, so by change of variables ($z = xy$) we have:
$$f(x) = \frac{\sqrt\pi\ \text{erf}\left(\sqrt{x}y\right)}{2\sqrt x}\left. \right|^{10}_x= \frac{\sqrt\pi\ \text{erf}\left(10\sqrt{x}\right)-\sqrt\pi\text{erf}\left(\sqrt{x^3}\right)}{2\sqrt x}$$
Now you can find the derivative, which will involve $e^{g(x)}$ as well as the error function.
Symbolically, if:
$$f(x)=\int_x^a g(x,y)dy, \quad G(u,v)=\int g(u,v)dv$$
Then:
$$f(x)=G(x,a)-G(x,x)$$
And using the chain rule:
$$f'(x) = \frac{dG}{du}(x,a)-\frac{dG}{du}(x,x)-\frac{dG}{dv}(x,x)$$
A: 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.
A: The trick in this cases is writing the integral as a composition:
$$H(x,z)=-\int_{10}^z e^{-xy^2}dy$$
$$\Delta(x)=(x,x)$$
$$\int_x^{10} e^{-xy^2}dy=I(x)=H(x,x)=H(\Delta(x)).$$
The partial derivatives of $\Delta$ are obvious, the partial derivatives of $H$
 can be calculated using differentiation under the integral sign (x) and the Fundamental theorem of the Calculus (z) .
Now you can calculate $I'(x)$ using the chain rule.
A: 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}$
A: You may start your solution by giving an explanation for Lebnitz's rule for integration then using it in solution, than avoiding it altogether...
According to Lebnitz's rule, $\frac{d}{dx}[\int^{\eta(x)}_{\phi(x)}f(x,t)\space dt] = \frac{d}{dx}\eta(x) f(\eta(x)) - \frac{d}{dx}\phi(x) f(\phi(x))$
[Derivation: Let $\int f(x)dz = F(x)$, or in other words, $\frac{d}{dx}F(x) = f(x)$. 
Now, by definition of definite integral, $\int^{\eta(x)}_{\phi(x)}f(x,t)\space dt=F(\eta(x))-F(\phi(x))$
Differentiating both sides,  $\frac{d}{dx}[\int^{\eta(x)}_{\phi(x)}f(x,t)\space dt] = \frac{d}{dx}\eta(x) f(\eta(x)) - \frac{d}{dx}\phi(x) f(\phi(x))$ (by chain rule of differentiation)  ]
Let $t=y^2 \implies t^{\frac{1}{2}}=y \implies \frac{1}{2t^{\frac{1}{2}}}{dt} = dy$
Therefore, $\frac{d}{dx}[\int^{10}_x e^{-xy^2}\space dy] = \frac{d}{dx}[\int^{100}_{x^2} e^{-xt}\frac{1}{2t^{\frac{1}{2}}}dt]\space \space $(limit changes as $t=y^2$)
$\implies \frac{d}{dx}[\int^{10}_x e^{-xy^2}\space dy] = 0 - \frac{e^{-x^3}}{2x}2x$ (by Lebnitz's rule)
$\implies \frac{d}{dx}[\int^{10}_x e^{-xy^2}\space dy] = - e^{-x^3}$
