Hi, how do you solve this (btw, sorry, I don't know how to make snazzy math grafics :/):
Find G'(x) for G(x) = the integral exp(t^2)dt in the interval 1/x< a< x.
Say that $\displaystyle \int e^{t^2}\,dt=F(t)$. (It doesn't matter what $\displaystyle F(t)$ is.)
So $\displaystyle \int_{1/x}^x e^{t^2}\,dt=F(x)-F(1/x)$ by the Fundamental Theorem of Calculus.
To take the derivative of this expression, use the chain rule to get
$\displaystyle F'(x)\cdot1-F'(1/x)\cdot-\frac{1}{x^2}$
But because of how we defined $\displaystyle F$, we know that $\displaystyle F'(x)=e^{x^2}$.
So the above is $\displaystyle e^{x^2}+\frac{e^{1/x^2}}{x^2}$.