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

Proof of the Chain Rule

  1. Feb 15, 2014 #1
    Hello everyone,

    I am reading a proof of the chain rule given in this link: http://kruel.co/math/chainrule.pdf

    Here is the portion I am troubled with:

    "We know use these equations to rewrite f(g(x+h)). In particular, use the first equation to obtain

    f(g(x+h)) = f(g(x) + [g'(x) + v]h),

    and use the second equation applied to the right-hand-side with k = [g'(x) + v]h..."


    How do they arrive at this, k = [g'(x) + v]h. Based above previous equations and definitions, I don't see how it is possible to write k in terms of the derivative of g(x), v, and h.

    Could someone help me?
     
  2. jcsd
  3. Feb 15, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    Start with:
    [tex]
    f(y + k) = f(y) + k(f'(y) + w) \\
    g(x + h) = g(x) + h(g'(x) + v)
    [/tex]

    We want to calculate [itex]f(g(x+h)) - f(g(x))[/itex], so first we need [itex]f(g(x+h))[/itex]. From the second equation,
    [tex]
    f(g(x + h)) = f(g(x) + h(g'(x) + v))
    [/tex]
    and now we apply the first equation with [itex]y = g(x)[/itex] and [itex]k = h(g'(x) + v)[/itex].
     
    Last edited: Feb 15, 2014
  4. Feb 15, 2014 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    What "previous" equation or definition did they have involving k? It looks to me like they are defining k to be [g'(x)+ v]h. Had they already defined it as something else?
     
  5. Feb 16, 2014 #4
    Well, if they are defining k as [g'(x) + v]h, that would seem awfully arbitrary. What is the justification for such a definition?
     
  6. Feb 16, 2014 #5

    pasmith

    User Avatar
    Homework Helper

    Go back to here:
    [tex]
    f(g(x+h))=f(g(x)+h(g′(x)+v))
    [/tex]
    We also have
    [tex]
    f(y + k) = f(y) + k(f'(y) + w))
    [/tex]
    which holds for all [itex]y[/itex] and for all [itex]k[/itex].

    Thus, to express [itex]f(g(x+h))[/itex] in the form [itex]f(g(x)) + (\mbox{something involving $x$ and $h$})[/itex], which is what we must do to attain our ultimate goal of finding [itex]f(g(x+h))- f(g(x))[/itex], we need to choose [itex]y[/itex] and [itex]k[/itex] subject to
    [tex]y + k = g(x) + h(g'(x) + v).[/tex]
    What choice for [itex]y[/itex] and [itex]k[/itex] would you make here?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof of the Chain Rule
  1. Chain rule proof (Replies: 5)

  2. Chain rule proof (Replies: 2)

Loading...