# Find the derivative of $\int_x^{x^2} e^{-t^2}dt$

Hey guys this was given to me as an exercise question and its really confusing. I'm not really sure where to start with this one, and I am assuming that the derivative isn't just $e^{-t^2} dt$. Anyways, any help is appreciated, thank you!.

Find the derivative of $$\int \limits_x^{x^2} e^{-t^2}dt$$

• You want to get rid of that minus sign, right? That aside, do you know Liebniz's rule? en.wikipedia.org/wiki/Leibniz_integral_rule Commented Jan 25, 2014 at 19:18
• well ya I know the rule, but I have never really applied it before. We just learned integration a couple days ago so I'm not really the best at it @EricAuld Commented Jan 25, 2014 at 19:20
• Is the integral written correct? Is it really $- dt$? Commented Jan 25, 2014 at 19:21
• This is quite an advanced example if you just learned integration two days ago. You might want to work up to this one. Commented Jan 25, 2014 at 19:23
• yeah sorry its not, i just edited it @Amzoti Commented Jan 25, 2014 at 19:24

I assume you mean find the derivative of $F(x)$, where

$$F(x) = \int_x^{x^2} e^{-t^2} dt.$$

Let $G(x) = \int_0^x e^{-t^2} dt$. By the fundamental theorem of calculus,

\begin{align*} F'(x) &= \frac{d}{dx}(G(x^2) - G(x)) \\ &= G'(x^2)(2x) - G'(x) \\ &= e^{-x^4}(2x) - e^{-x^2}. \end{align*}

• The first exponential should have $x$ in the 4th power. Commented Jan 25, 2014 at 19:33
• The chain rule. If I asked you to differentiate $\sin(x^2)$, you should say the derivative is $\cos(x^2)$ times the derivative of the inside, i.e. $\frac{d}{dx}\sin(x^2) = \cos(x^2)*(2x).$ Now instead of $\sin(x)$ and $\sin(x^2)$ consider $G(x)$ and $G(x^2)$.