Results 1 to 4 of 4

Thread: Using chain rule..

  1. #1
    Newbie
    Joined
    Jun 2010
    Posts
    23

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    Well, I would start with the initial equation:

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

    Try differentiating both sides of the initial equation. What does that give you?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member mfetch22's Avatar
    Joined
    Feb 2010
    From
    Columbus, Ohio, USA
    Posts
    168
    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:

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

    So we have:

    $\displaystyle f(x)$

    and

    $\displaystyle g(x) = -x$

    Thus:

    $\displaystyle f(g(x))= f(-x)$

    so differentiating with the cain rule gets:

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

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

    And since:

    $\displaystyle -f(x) = f(x)$

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

    $\displaystyle -f'(x) = f'(-x)$

    and

    $\displaystyle -f(x) = f(-x)$

    thus:

    $\displaystyle f'(x) = f'(-x)$

    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2009
    Posts
    130
    f(-x) = -f(x)

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

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

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

    Was that hard?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: Nov 9th 2010, 01:40 AM
  2. Chain Rule Inside of Chain Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 22nd 2009, 08:50 PM
  3. Replies: 5
    Last Post: Oct 19th 2009, 01:04 PM
  4. Replies: 3
    Last Post: May 25th 2009, 06:15 AM
  5. Replies: 2
    Last Post: Dec 13th 2007, 05:14 AM

Search Tags


/mathhelpforum @mathhelpforum