derivative of e

**alyosha2** · July 5th 2011, 03:36 AM

Firstly, isn't the rule that subtraction of powers is division not multiplication?

derivative of e-ql_4e2312437c4e3c4537d41da80397d509_l3.png

Then on finding the integral

derivative of e-ql_9b2f7c21e23574eb10ad8610db2b2fa7_l3.png

why has the e to 3 come out as negative?

Thanks.

**TKHunny** · July 5th 2011, 03:42 AM

$e^{3-x} = \frac{e^{3}}{e^{x}} = e^{3}e^{-x}$

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

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

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

By the way, "the derivative of e" is zero. Be more careful with words.

**Deveno** · July 5th 2011, 03:51 AM

perhaps it will make more sense if you realize that $\frac{d}{dx}(e^{-x}) = -e^{-x}$ .

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

$exp\circ g(x)$ , 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:

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

the minus sign outside the integral stays, however.

**alyosha2** · July 5th 2011, 04:10 AM

Got it. Great explanations.

Thanks guys.

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