Finding $f'(x)$ when $f(x)=\int^1_0 e^{xy+y^2}dy$ 
If $f(x) = \int^1_0 e^{xy+y^2}dy$, find $f'(0)$.

I understand that this is function defined by an integral, and $e^{y^{2}}$ does not integrate into an elementary function. So, I will need to take $f'(x)$ which yields:
$$\int^1_0 ye^{xy+y^2}dy$$
I am trying to integrate this, but I am failing. I take it I should use integration by parts, but I can't because I still have $e^{y^2}$ term. Any help?
 A: You are given
$$f(x) =\int_0^1 \exp(xy+y^2)dy $$
Differentiating gives
$$f'(x) =\int_0^1 y\cdot \exp(xy+y^2)dy $$
Since we need $f'(0)$ we might as well plug in $x=0$. This gives
$$f'(0) =\int_0^1 y\cdot \exp (y^2)\cdot dy $$
But this integral is quite striaghtforward, 
$$f'(0) =\int_0^1 y\cdot \exp (y^2)\cdot dy =\left. \frac 1 2 e^{y^2} \right|_0^1 = \frac 1 2 (e-1) $$

By the OP's request:
$$f'(x) =\int_0^1 y\cdot \exp(xy+y^2)dy $$
We complete the square
$$f'(x) = \exp \left( { - \frac{{{x^2}}}{4}} \right)\int_0^1 {y\exp \left[ {{{\left( {y + \frac{x}{2}} \right)}^2}} \right]dy} $$
We change variables
$$\eqalign{
  & y + \frac{x}{2} = u  \cr 
  & dy = du \cr} $$
$$\eqalign{
  & f'(x) = \exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} {\left( {u - \frac{x}{2}} \right)} \exp \left( {{u^2}} \right)du  \cr 
  & f'(x) = \exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} u \exp \left( {{u^2}} \right)du - \frac{x}{2}\exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} {\exp \left( {{u^2}} \right)du}  \cr} $$
Let's focus on the first integral:
$${I_1} = \int_{\frac{x}{2}}^{1 + \frac{x}{2}} u \exp \left( {{u^2}} \right)du = \left. {\frac{1}{2}\exp {u^2}} \right|_{x/2}^{1 + x/2} = \frac{1}{2}\exp \frac{{{x^2}}}{4}\left[ {\exp \left( {x + 1} \right) - 1} \right]$$
So we have
$$f'(x) = \frac{1}{2}\left[ {\exp \left( {x + 1} \right) - 1} \right] - \frac{x}{2}\exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} {\exp \left( {{u^2}} \right)du} $$
The other integral evaluates in terms of the error function
$$\operatorname{erf} (x) = \frac{2}{{\sqrt \pi  }}\int_0^x {{e^{ - {t^2}}}} dt.$$
with a change or variables $u=-v$, 
$$\int_{ - \left( {1 + \frac{x}{2}} \right)}^{ - \frac{x}{2}} {\exp \left( { - {v^2}} \right)dv}  = \frac{{\sqrt \pi  }}{2}\left\{ {\operatorname{erf} \left( { - \frac{x}{2}} \right) - \operatorname{erf} \left( { - 1 - \frac{x}{2}} \right)} \right\}$$
So the function is
$$f'(x) = \frac{1}{2}\left[ {\exp \left( {x + 1} \right) - 1} \right] - \frac{x}{2}\exp \left( { - \frac{{{x^2}}}{4}} \right)\frac{{\sqrt \pi  }}{2}\left\{ {\operatorname{erf} \left( { - \frac{x}{2}} \right) - \operatorname{erf} \left( { - 1 - \frac{x}{2}} \right)} \right\}$$
For large values of $x$, you can neglect the last result (since the value will be smaller and smaller), and you can estimate $f'$ with
$$f'(x) \approx \frac{1}{2}\left[ {\exp \left( {x + 1} \right) - 1} \right]$$
Cheers.
A: You're asked to calculate $f^{\prime}(0)$. You don't have to calculate $f^{\prime}(x)$ first.
\begin{align}
f^{\prime}(0) = \int_0^1 y e^{y^2}dy
\end{align}
Let $u = y^2$, $du = 2ydy$:
\begin{align}
f^{\prime}(0) &= \frac{1}{2} \int_0^1 e^u du = \left. \frac{1}{2}e^u \right|_0^1 \\
&= \frac{1}{2}(e - 1)
\end{align}
As for the general case of $f^\prime(x)$, the integral cannot be solved using elementary functions only. One will have to use the error function.
A: You have got f'(x), so let x be 0, and integrate by parts. Then you get the answer (e-1)/2.
