# How do you find the derivative of g(x) = int_3^x e^(4t²-t) dt ?

Apr 30, 2015

Answering this question is one of the main jobs of the Fundamental Theorem of Calculus, Part I.
(Its other main job is to help us prove the Fundamental Theorem of Calculus, Part II.)

FTC Part I: If $f$ is continuous on $\left[a , b\right]$ and $g \left(x\right)$ is defined by:

$\textcolor{w h i t e}{\text{sssssssss}}$ $g \left(x\right) = {\int}_{a}^{x} f \left(t\right) \mathrm{dt}$ for $x \in \left[a , b\right]$, then

$\textcolor{w h i t e}{\text{ss}}$ $g ' \left(x\right) = f \left(x\right)$

So, for g(x) = int_3^x e^(4t²-t) dt, we get

g'(x) = e^(4x²-x) .

Yes. That's it. Yes, really. Just change the $t$'s to $x$'s and move on to the next question.