Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: First derivative: chain rule: easy for you guys

  1. Mar 13, 2008 #1
    [SOLVED] first derivative: chain rule: easy for you guys

    Y=E^(-mx)
    f= E^x g= -mx
    f'= E^x g'= o





    E^(-mx) * 0(E^(-mx))
    i think, not sure though



    Y'= 0
    which is wrong
    someone help
     
  2. jcsd
  3. Mar 13, 2008 #2
    [tex]f(x)=e^{g(x)}[/tex]

    [tex]f'(x)=e^{g(x)}g'(x)[/tex]

    [tex]y=e^{-mx}[/tex]

    [tex]y'=-me^{-mx}[/tex]
     
    Last edited: Mar 13, 2008
  4. Mar 13, 2008 #3
    yes!!!!!!!!!!!!!!!!!
     
  5. Mar 13, 2008 #4
    the derivative of e is itself ... times the derivative of it's exponent.
     
  6. Mar 13, 2008 #5
    how did you get g'(x)
    i thought the (m) was constant thus making the the derivative equal to 0
     
  7. Mar 13, 2008 #6
    Do this one for me ...

    [tex]y=ax[/tex]

    What is it's derivative?
     
  8. Mar 13, 2008 #7
    gosh!
    well i would use the product rule or maybe not.
    f= x g=ax
    f'= 1 g'=product rule
    high school calc.
     
  9. Mar 13, 2008 #8
    Don't worry about the product rule when you're differentiating with a constant.

    The derivative of a constant times a function is simply ...

    [tex]\frac{d}{dx}=cf'(x)[/tex]

    [tex]y=ax[/tex]

    [tex]y'=a[/tex] by constant rule

    [tex]y'=a+x\cdot0=a[/tex] by product rule

    Do you see why we don't need the product rule?
     
  10. Mar 13, 2008 #9
    [tex]ysingle-quote=a+x\cdot0=a[/tex]
    times 0 would be zero,right?
     
  11. Mar 13, 2008 #10
    never mind, sorry.
     
  12. Mar 13, 2008 #11
    lets just go back to the original equation
    [tex]fsingle-quote(x)=e^{g(x)}gsingle-quote(x)[/tex]
    what is g'(X) equall to
     
  13. Mar 13, 2008 #12
    read my previous posts!!!
     
  14. Mar 13, 2008 #13
    g'= -m
    Solved.
    thanks for the help
     
  15. Mar 13, 2008 #14
    :) anytime
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook