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Homework Help: Chain rule help

  1. Feb 12, 2006 #1
    f '(8)=5 g '(8)=3 f(4)=8 g(4)=10 g(4)=10 g(8)=2 f(8)=5

    find (g o f)'(4)

    how do I go about setting up these types of problem.
     
  2. jcsd
  3. Feb 12, 2006 #2

    AKG

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    What have you tried; where are you getting stuck?
     
  4. Feb 12, 2006 #3
    have not tried because I don't know how to set up the problem...I don't really care about the answer I just want to know how you go about setting up these kinds of problems because I have a quiz tomorrow.

    I figure the first step is g(f(x))

    so g'(f(x))*f'(x)
    g'(8)*f'(x)
    3*f'(x)
     
    Last edited: Feb 12, 2006
  5. Feb 12, 2006 #4

    AKG

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    (g o f)'(4)
    = g'(f(4))*f'(4) by chain rule
    = g'(8)*f'(4) since f(4) = 8
    = 3*f'(4) since g'(8) = 3

    And that's all you can do, since they don't tell you what f'(4) is. I suspect they do, and you just copied out the question wrong. Also, why have you given "g(4) = 10" twice?

    Anyways, the way to setting up the problem is this:

    Given a problem, "find X", write:

    X
    = A (by some theorem, or given fact, or logical inference)
    = B (again, give justification)
    = C (justification)
    = D (justification)

    until you get some answer D that you think the teacher will like, like an actual numeral. In this case, your X is (f o g)'(4), and your C is something like 3f'(4). You want a numeral for your D, but you can't get it yet from C because they haven't given you enough information (or you copied the question wrong).
     
  6. Feb 12, 2006 #5
    what about this type of problem:

    For a given functionhttps://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk5/696680/kopko.4-prob5image1.png [Broken]consider[/URL] the composite function https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk5/696680/kopko.4-prob5image2.png [Broken] Suppose we know that https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk5/696680/kopko.4-prob5image3.png [Broken]

    Calculate f ' (x)

    How do I go about setting up this type of problem?
     
    Last edited by a moderator: May 2, 2017
  7. Feb 13, 2006 #6

    HallsofIvy

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    You titled this thread "chain rule"! It ought to occur to you to use the chain rule!
    If h(x)= f(2x3) then h'(x)= f '(2x3)(6x2).

    You are given that h'(x)= 7x5.

    You can easily solve f '(2x3)(6x2)= 7x5 for f '(2x3).

    Now let y= 2x3. What is f(y)?
     
    Last edited by a moderator: May 2, 2017
  8. Feb 14, 2006 #7
    the way our TA showed up, the answer ought to be:

    (7x^3/6)*(y/2)^(1/3)...but my answer must be in terms of x...so could someone please tell me if I went wrong somewhere or how to make it all in terms of x (by x I mean I can't have that "y"). Here are my steps:

    f '(2x^3)(6x^2)= 7x^5

    f '(2x^3)= (7x^5)/(6x^2)

    f '(2x^3)= (7x^3)/6
    ~~~~~~~~~~~~
    2x^3=y

    x^3= y/2

    x= (y/2)^(1/3)
    ~~~~~~~~~~

    answer: (7x^3/6)*(y/2)^(1/3)
     
  9. Feb 15, 2006 #8

    VietDao29

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    It's fine up to here.
    Now sub what you get in the expression:
    f '(2x3)= (7x3)/6, we have:
    f'(y) = 7y / 12.
    So what's f'(x)?
    Can you go from here?
     
  10. Feb 15, 2006 #9
    where did the f'(y) come from?
     
  11. Feb 15, 2006 #10

    NateTG

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    In VietDao's post y is a place holder for [itex]2x^3[/itex]

    I don't know if this will make things any clearer for you:

    The problem gives you
    [tex]h(x)=f(z(x))[/tex]
    So, let's say we have some [itex]a[/itex] so that [itex]x=z^{-1}(a)[/itex] (provided that [itex]z^{-1}[/itex] actually exists). Then we can substitute that in
    [tex]h(z^{-1}(a))=f(z(z^{-1}(a))[/tex]
    then simplify
    [tex]h(z^{-1}(a))=f(a)[/tex]
    Now, we can take the derivative of both sides w.r.t. a
    [tex]h'(z^{-1}(a)) \times \left(z^{-1}\right)' (a) = f'(a)[/tex]

    Now, since [itex]h'[/itex] and [itex]z[/itex] are both known, you should be able to work out what the left hand side of the equation is equal to.
     
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