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Finding composite derivatives

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data
    a) (f ° g)′(−2) = ?
    b) (g ° f)′(2) = ?

    2. Relevant equations
    f(−2) = −3,
    g(−2) = −4,
    f(2) = 3,
    g(2) = −3,
    f ′(−2) = −1,
    f ′(−4) = −2,
    f ′(2) = 5,
    g ′(−2) = 1,
    g ′(2) = 2,
    g ′(3) = −4.

    3. The attempt at a solution
    I have no idea how to do it every thing ive tried doesnt work, how do you work it out. Could someone explain the processes so i know how to do any derivative of a composite function
     
  2. jcsd
  3. Apr 26, 2010 #2

    gabbagabbahey

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    Just use the chain rule...

    [tex](f\circ g)(x)=f(g(x))\implies\frac{d}{dx}(f\circ g)(x)=\frac{d}{dx}f(g(x))=\frac{df(g)}{dg}\frac{dg}{dx}=f'(g(x))g'(x)=(f'\circ g)(x)g'(x)[/tex]
     
  4. Apr 26, 2010 #3

    lanedance

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    i would take the composition of functions to mean:
    [tex]f \circ g(x) = f(g(x))[/tex]

    using the chain rule, what is
    [tex]\frac{d}{dx}f(g(x))=?[/tex]
     
  5. Apr 26, 2010 #4
    that makes the answer (f o g)'(-2)= f(-2)(g(-2))(g'(-2))
    = -1(-4)x(1)
    which = 4 which is the wrong answer =(
     
  6. Apr 26, 2010 #5

    gabbagabbahey

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    No, [itex](f'\circ g)(-2)=f'(g(-2))=f'(-4)[/itex]
     
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