# Derivative exponential problem

1. Feb 25, 2012

1. The problem statement, all variables and given/known data

Compute the derivative of the following function

$(1-2x)e^{-x}$

2. Relevant equations

Product Rule and Quotient rule

3. The attempt at a solution

My problem here is that I come up with two different answers when I use the quotient rule vs. the product rule.

Trying it with the product rule

f(x)= (1-2x)e^-x
f'(x)= -2e^-x + e^-x(1-2x)
f'(x)= e^-x(-2+(1-2x))
$f'(x)= e^{-x}(-1-2x)$ or $\frac{-1-2x}{e^x}$

With the quotient rule

f'(x)= (e^-x(1-2x) - (-2e^-x)) / [e^-x]^2
f'(x)= (e^-x-2e^-x+2e^-x) / [e^-x]^2
f'(x)= (e^-x(-2x+2)) / [e^-x]^2
$f'(x)= \frac{-2x+2}{e^{-x}}$ or $e^x(-2x+2))$

As you can see, these are two different answers. I would think that I should have the same solution either way I do this, so what am I doing wrong?

Last edited: Feb 25, 2012
2. Feb 25, 2012

### fauboca

I don't get either of your answers when I take the derivative.

(1-2x)e^{-x} = e^{-x} -2xe^{-x} = -e^{-x} -2(product rule) = ??

3. Feb 25, 2012

### SammyS

Staff Emeritus
Put the whole exponent in braces { } .

[itex ]e^{-x}[/itex ] gives   $e^{-x}$

4. Feb 25, 2012

Thanks SammyS. I cleaned up some of the above. Hopefully that helps.

5. Feb 25, 2012

### eumyang

Here's the problem. It should be a minus above, not a plus. (Remember, chain rule.)
Whoa. If you want to use the quotient rule, then you need to rewrite f(x) as
$$f(x)=\frac{1-2x}{e^x}$$
... so there shouldn't be any e-x in the first step:
$f'(x) = \frac{e^x(1-2x) - (-2e^x)}{(e^x)^2}$
= ...

6. Feb 25, 2012

Thanks eumyang. I'm actually seeing numerous errors that I've typed above, and I'm working on cleaning it up with Latex.

Please stand by until I can clean up this mess.

7. Feb 25, 2012

Ahh, I didn't need to completely retype that. You are perfectly correct, eumyang, I was making those mistakes on both. I wasn't considering that the derivative of $e^{-x}$ is $-e^{-x}$. And I wasn't playing with the signs right on the quotient rule.