# derivative of e

• Jul 5th 2011, 03:36 AM
alyosha2
derivative of e
Firstly, isn't the rule that subtraction of powers is division not multiplication?

Attachment 21777

Then on finding the integral

Attachment 21778

why has the e to 3 come out as negative?

Thanks.
• Jul 5th 2011, 03:42 AM
TKHunny
Re: derivative of e
$\displaystyle e^{3-x} = \frac{e^{3}}{e^{x}} = e^{3}e^{-x}$

It is all the same. Please ponder it until you see that it is not scary.

$\displaystyle \int a\cdot e^{-x}dx = a\cdot\int e^{-x}dx$

Feel free simply to remove constants from the integral argument. $\displaystyle e^{3}$ is a constant.

By the way, "the derivative of e" is zero. Be more careful with words.
• Jul 5th 2011, 03:51 AM
Deveno
Re: derivative of e
perhaps it will make more sense if you realize that $\displaystyle \frac{d}{dx}(e^{-x}) = -e^{-x}$.

this is a consequence of the chain rule: $\displaystyle e^{-x}$ is really the composition:

$\displaystyle exp\circ g(x)$, where g(x)= -x = (-1)x.

so when we integrate, we need a minus sign to go inside the integral, and another minus sign to go outside (to balance, (-1)(-1) = 1). the minus sign inside the integral disappears, since when we integrate:

$\displaystyle \int \ -e^{-x} dx = e^{-x} + C$.

the minus sign outside the integral stays, however.
• Jul 5th 2011, 04:10 AM
alyosha2
Re: derivative of e
Got it. Great explanations.

Thanks guys.