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Second Derivative of a Composition of Functions

  1. Jan 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Derive an expression for the composition of 2 functions.


    2. Relevant equations



    3. The attempt at a solution
    I started with supposing that y(x) = f(g(x)). I know that dy/dx = df/dg * dg/dx (via the chain rule). Doing the derivative again, I started with the product rule:
    d2y/dx2 = d/dx (df/dg * dg/dx) = df/dg * d2g/dx2 + dg/dx * d/dx (df/dg).

    I don't know what d/dx (df/dg) means. I think I'm supposed to do the chain rule again on this part, but I'm not totally clear. Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 20, 2012 #2

    SammyS

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    Hello skyline01. Welcome to PF !

    One suggestion is to to look at some specific examples, such as y(x) = sin(x5).

    Then f(x) = sin(x), g(x) = x5 .
     
  4. Jan 20, 2012 #3
    Thank you, SammyS! I tried the sample function you suggested, but I'm still not sure if I am expressing d/dx (df/dg) correctly. Here is how your sample function worked out:

    y(x) = sin(x5)
    So, f(x) = sin(x) and g(x) = x5.
    Therefore,
    dy/dx = cos(x5) 5x4
    and
    d2y/dx2 = cos(x5) 20x3 + 5x4 (-sin(x5)) 5x4.

    Generalizing this to any composition of 2 functions, we have
    d2y/dx2 = df/dg * d2g/dx2 + dg/dx * d2f/dg2.

    Does this look right?
     
  5. Jan 21, 2012 #4

    SammyS

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    Almost correct.

    d2y/dx2 = df/dg * d2g/dx2 + (dg/dx)2 * d2f/dg2.
     
  6. Jan 21, 2012 #5
    Yes, you are correct. Thank you for catching that. So,
    d/dx(df/dg) = dg/dx * d2f/dg2.

    I find it strange that in every calculus book I have ever read (including real analysis books), they always stop at the first derivative when discussing the chain rule.
     
  7. Jan 21, 2012 #6

    SammyS

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    Yes, that is strange.

    The notation can be confusing, even for 1st derivatives, but as you mentioned in your initial post, it gets quite a bit more confusing with higher order derivatives. That's why I suggested differentiating a concrete example.

    Here's a link from Wikipedia: http://en.wikipedia.org/wiki/Chain_rule#Higher_derivatives.

    Have a good one !
    SammyS
     
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