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Math Help - Using chain rule..

  1. #1
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    Using chain rule..

    I am asked to:

    Use the chain rule to show that the derivative of an odd function is an even function.

    Basically assuming that f(x) is odd:

    f(-x) = -f(x) and f(x) = -f(x)

    Show that:

    f'(x) = f'(-x)

    All must be done in terms of f. How would I go about in doing this?
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  2. #2
    A Plied Mathematician
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    Well, I would start with the initial equation:

    f(-x)=-f(x). Incidentally, the equation f(x)=-f(x), if true for all x, implies that f(x)=0 for all x. Not sure that's where you want to go.

    Try differentiating both sides of the initial equation. What does that give you?
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  3. #3
    Member mfetch22's Avatar
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    Quote Originally Posted by GameTheory View Post
    I am asked to:

    Use the chain rule to show that the derivative of an odd function is an even function.

    Basically assuming that f(x) is odd:

    f(-x) = -f(x) and f(x) = -f(x)

    Show that:

    f'(x) = f'(-x)

    All must be done in terms of f. How would I go about in doing this?
    Then chain rule says:

    \frac{d}{dx}[f(g(x))] = (f'(g(x)))(g'(x))

    So we have:

    f(x)

    and

    g(x) = -x

    Thus:

    f(g(x))= f(-x)

    so differentiating with the cain rule gets:

    \frac{d}{dx}[f(g(x))] = f'(-x) = [f'(g(x)][g'(x)]

    = [f'(-x)]\frac{d}{dx}[-x] = -f'(-x)

    And since:

    -f(x) = f(x)

    we can move over the negitive in the original equation you gave to get:

    -f'(x) = f'(-x)

    and

    -f(x) = f(-x)

    thus:

    f'(x) = f'(-x)

    I feel like there is an error in this proof, but it seems logical, atleast I think I'm going in the right direction. Good luck.
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  4. #4
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    f(-x) = -f(x)

    f'(-x)*(-x)'=-f'(x)

    f'(-x)*(-1)=-f'(x)

    f'(-x)=f'(x)

    Was that hard?
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