Derivative exponential problem

[MacLaddy]

Homework Statement

Compute the derivative of the following function

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

Homework Equations

Product Rule and Quotient rule

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?

 

[fauboca]

Homework Statement

Compute the derivative of the following function

(1-2x)e^(-x)

Homework Equations

Product Rule and Quotient rule

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. ( And that I can't figure out the Latex for e^-x... If someone can show me how to do that I'll clean this post up)
e^-x

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 (-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)= (-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?

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) = ??

 

[SammyS]

... ( And that I can't figure out the Latex for e^-x... If someone can show me how to do that I'll clean this post up)

e^-x

...

Put the whole exponent in braces { } .

[itex ]e^{-x}[/itex ] gives   e^{-x}

 

[MacLaddy]

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

 

[eumyang]

Trying it with the product rule

f(x)= (1-2x)e^-x
f'(x)= -2e^-x + e^-x(1-2x)

Here's the problem. It should be a minus above, not a plus. (Remember, chain rule.)

With the quotient rule

f'(x)= (e^-x(1-2x) - (-2e^-x)) / [e^-x]^2

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}
= ...

 

[MacLaddy]

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.

 

[MacLaddy]

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.

Thank you very much, I appreciate the help.