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!

Chain rule for product of functions

  1. Dec 17, 2015 #1
    Here is a simple question :

    let f(g(x)) = h(x)*g(x).

    I want to calculate df/dx.

    If I use the product rule, I get g(x)h'(x) + h(x)g'x).

    Now if I use the composition/chain rule, I get

    df/dx = df/dg * dg/dx = h(x) * g'(x) which is different.

    I guess my df/dg = h is wrong, but I can't see what it should be.

    Any help is welcome:-)
     
  2. jcsd
  3. Dec 17, 2015 #2

    pwsnafu

    User Avatar
    Science Advisor

    I don't want to double up on ##x##, so let's use ##y##.
    What is ##f(y)## in the question? What is it's derivative?
     
  4. Dec 17, 2015 #3
    that's where I'm confused...
    my first thought that in f(g(x)) f(y) is the function that multiplies y by h(x). therefore if I derivate with respect to y, I get h(x). Which apparently is wrong...

    no your question replaces x by y, which in fact doesn't really change anything so I'd say
    f(y) obviously equals to h(y)g(y) so its derivative is h'(y)g(y)+g'(y)h(y)...

    now I'm again confused
     
  5. Dec 17, 2015 #4

    pwsnafu

    User Avatar
    Science Advisor

    This is wrong.
     
  6. Dec 17, 2015 #5
    humm... so you've seen where my problem is... what's f ?
     
  7. Dec 17, 2015 #6
    are there 2 f ? the one if f(x) = h(x)*g(x) and the one if f(g(x)) ?
     
  8. Dec 17, 2015 #7

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, you've got yourself into a terrible tangle here! If you define another function:

    ##k(x) = f(g(x)) = h(x)g(x)##

    You may be able to see the problem. ##f(x) \ne k(x)##
     
  9. Dec 17, 2015 #8
    OK I see that now, but I'm still in trouble to calculate df/dg. I'd like to say f(u) = h(x)*u but then again I end up with h(x) for df/dg
     
  10. Dec 17, 2015 #9

    pwsnafu

    User Avatar
    Science Advisor

    Instead of trying ##k(x) = (f \circ g)(x) = h(x) \cdot g(x)## start with something simpler ##k(x) = (f \circ g)(x) = (l \circ g) (x) \cdot g(x)##
     
  11. Dec 17, 2015 #10
    hum...
    then here too, I can't find what l is although I see l(g(x)) should be h(x)
     
  12. Dec 17, 2015 #11
    Could you guys explain me what is f here please.
    I don't know why but I really have trouble seing where is the composition here
     
  13. Dec 17, 2015 #12

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's easier if you set ##y = g(x)##, so that ##x = g^{-1}(y)##:

    ##f(y) = f(g(x)) = h(x)g(x) = h(g^{-1}(y))y##
     
  14. Dec 17, 2015 #13
    hum ok I can now do

    f'(y) = h(g^{-1}(y)) + yg^{-1}'(y)h'(g^{-1}(y)) = h(x) + yh'(g^{-1}(y))/g'(g^{-1}(y))
    then g'df/dy = g'h+hg'

    ok... thanks !
     
  15. Dec 17, 2015 #14

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You could have got that directly from the initial equation:

    ##f(g(x)) = h(x)g(x) \ \Rightarrow \ f'(g(x))g'(x) = h'(x)g(x) + h(x)g'(x)##
     
  16. Dec 17, 2015 #15
    I'm sorry I must have a very slow mind today but I don't see exactly why the implication and the second equality.
     
  17. Dec 17, 2015 #16

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Simply differentiate both sides wrt ##x##.
     
  18. Dec 17, 2015 #17
    ah ok :-D
    I was trying to see how one could find h'g+hg' from f'(g)*g' "only"
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Chain rule for product of functions
  1. Chain Rule (Replies: 65)

  2. What is the chain rule (Replies: 0)

Loading...