Small Question on Integration

**NBrunk** · January 21st 2010, 05:39 PM

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!

**Sudharaka** · January 21st 2010, 05:50 PM

Dear NBrunk,

Here you must use the chain rule.

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

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

Hope this helps.

**VonNemo19** · January 21st 2010, 05:55 PM

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 $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

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

Then the derivative of $e^u$ would be $\frac{dy}{dx}=e^u\cdot\frac{du}{dx}=e^{-x}\cdot(-1)=-e^{-x}$