1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivatives, chain rule 2

  1. Nov 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the derivative of y=xe-kx

    2. Relevant equations

    Chain rule

    3. The attempt at a solution

    y = xeu

    [itex]\frac{dy}{du}[/itex] = xeu+eu

    u = -kx

    [itex]\frac{du}{dx}[/itex] = -k

    [itex]\frac{dy}{dx}[/itex] = (xe-kx+e-kx)(-k)

    = e-kx(x+1)(-k)

    = e-kx(-kx-k)

    The answer is e-kx(-kx+1). What did I do wrong?
  2. jcsd
  3. Nov 9, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There is a subtle error here. The problem is that if ##u=-kx## then ##x=-\frac u k## and you have to include ##\frac{dx}{du} = -\frac 1 k## in your second term. But it's best to avoid this error completely by noting that you have a product and should use the product rule before the chain rule. Try$$
    \frac d {dx} xe^u = e^u\frac d {dx} x + x \frac d {dx}e^u$$ and use your method on that last term.
  4. Nov 9, 2013 #3
    I'm sorry, but I don't get what you mean.

    [itex]\frac{d}{dx}[/itex] xeu = eu[itex]\frac{d}{dx}[/itex]x+x[itex]\frac{d}{dx}[/itex]eu

    is still equal to eu + xeu.
  5. Nov 9, 2013 #4
    Okay, I get it. Thanks!
  6. Nov 9, 2013 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No it isn't. By your own steps in your original post you showed$$
    \frac d {dx} e^u = -ke^{-kx}$$and if you put that in you get the correct answer.
  7. Nov 9, 2013 #6
    I get it now. Thanks!
  8. Nov 9, 2013 #7
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Derivatives, chain rule 2