# Small Question on Integration

• Jan 21st 2010, 05:39 PM
NBrunk
Small Question on Integration
Hello, and thanks in advance for the help.

I know that the derivative of e^x is just e^x. However, if I'm given a function such as e^-x, how is the -x reflected in integrating?

Something tells me it can't be as simple as e^-x ---> e^-x.

Thanks again!
• Jan 21st 2010, 05:50 PM
Sudharaka
Dear NBrunk,

Here you must use the chain rule.

$\displaystyle \frac{d(e^{-x})}{dx}=\frac{d(e^{-x})}{d(-x)}\times\frac{d(-x)}{dx}$

$\displaystyle \frac{d(e^{-x})}{dx}=-e^{-x}$

Hope this helps.
• Jan 21st 2010, 05:55 PM
VonNemo19
Quote:

Originally Posted by NBrunk
Hello, and thanks in advance for the help.

I know that the derivative of e^x is just e^x. However, if I'm given a function such as e^-x, how is the -x reflected in integrating?

Something tells me it can't be as simple as e^-x ---> e^-x.

Thanks again!

Well, almost that simple....

Let's make use of the chain rule here, namely that $\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

So, let $\displaystyle y=e^u$ and let $\displaystyle u=-x$.

Then the derivative of $\displaystyle e^u$ would be $\displaystyle \frac{dy}{dx}=e^u\cdot\frac{du}{dx}=e^{-x}\cdot(-1)=-e^{-x}$
• Jan 21st 2010, 05:56 PM
pickslides
$\displaystyle \int e^{-x}~dx = -e^{-x}+C$

You can differentiate $\displaystyle -e^{-x}+C$ to check