What is the derivative of $\int_{x^3}^{x^2} e^{y^2} dy$? My only idea is to calculate this integral but I know it is very hard to calculate it so there must be some smarter way. I will appreciate any hint or help.
 A: There's no nice formula for $\int e^{y^2} \; dy$, so you can't solve this directly. This is intentional to get you to use the fundamental theorem of calculus. I'd suggest just calling the antiderivative $F(y)$.

*

*If $F(y)$ is the antiderivative of $e^{y^2}$, what is $\int_{x^3}^{x^2} e^{y^2} \; dy$? (in terms of $F$)

*Take the derivative of your answer above with respect to $x$

*Use the fact that you know what $F'(x)$ is to write your answer in closed form.

A: A consequence of the first fundamental theorem of calculus is that
$$\frac{d}{dx}\int_a^x f(t) \ dt = f(x)$$
To do this problem, you can split the integral into two parts:
$$
\begin{eqnarray}
\int_{x^3}^{x^2}e^{y^2} \ dy &= \int_{x^3}^{0}e^{y^2} \ dy + \int_{0}^{x^2}e^{y^2} \ dy \\
&= -\int_{0}^{x^3}e^{y^2} \ dy + \int_{0}^{x^2}e^{y^2} \ dy \\
&= \int_{0}^{x^2}e^{y^2} \ dy -\int_{0}^{x^3}e^{y^2} \ dy 
\end{eqnarray}
$$
Then you'll need to substitute for the upper limits and use the chain rule when you differentiate.
A: Thanks Ninad Munshi, Tuvasbien, Kman3, TomKern for responding to my question.
So derevative of this function is $2x \cdot e^{x^4}-3x^2 \cdot e^{x^6}$ ?
