calculating the taylor term of an integral an exercise ask me to calculate the Taylor term at $x = 0$ and degree four. I know how to take a derivative of an integral, but I'm having doubts about this one. 
The function: 
$$\int_0^x e^{-t^2}  dt $$   
The derivative of this integral is $g'(x) = e^{-x^2}$. I know that I have to differentiate this further, but I don't get the correct result. 
Thanks in advance. 
 A: Let
\begin{equation*}
g(x)=\int_{0}^{x}e^{-t^{2}}dt.
\end{equation*}
The Taylor series of $g(x)$ at $x=0$, also called the Maclaurin series, is
\begin{equation*}
g(0)+\frac{g^{\prime }(0)}{1!}x+\frac{g^{\prime \prime }(0)}{2!}x^{2}+\frac{
g^{\prime \prime \prime }(0)}{3!}x^{3}+\frac{g^{(4)}(0)}{4!}x^{4}+\ldots 
\end{equation*}
Hence the  term of degree 4 is $\dfrac{g^{(4)}(0)}{4!}$. 

I know that I have to differentiate this further, but I don't get the correct result. 

The differentiation can be carried out as follows. Your first evaluation is correct: $g^{\prime }(x)=e^{-x^{2}}$. Then we have $g^{\prime \prime }(x)=-2xe^{x^{2}}=-2xg^{\prime }(x)$. Differentiating it further yields $g^{(4)}(0)=0$, because
\begin{eqnarray*}
g^{\prime \prime \prime }(x) &=&-2g^{\prime }(x)-2xg^{\prime \prime
}(x)=-2g^{\prime }(x)+4x^{2}g^{\prime }(x) \\
&& \\
g^{(4)}(x) &=&-2g^{\prime \prime }(x)+8xg^{\prime }(x)+4x^{2}g^{\prime
\prime }(x) \\&=&12xg^{\prime }(x)+4x^{2}g^{\prime \prime }(x)=x\left( 12g^{\prime
}(x)+4xg^{\prime \prime }(x)\right) .
\end{eqnarray*}
Thus $$\dfrac{g^{(4)}(0)}{4!}=0.$$
