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Even and Odd functions

  1. Oct 22, 2004 #1


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    If a function is even, prove that the derivative is odd.

    Look at a graph of x^2 we can clearly see why.

    This is how I would approach this...

    If we solve d/dx x^n and n is an even integer, we get the derivative nx^(n-1). Since n is even, n-1 is odd.

    Because n-1 is odd, the derivative nx^(n-1) becomes odd because f'(-x)= - f'(x). Therefore the derivative of an even function becomes an odd function.

    Note: The TA couldn't solve this... :uhh:

    I excluded the proof of d/dx x^n = nx^(n-1) because it is not necessary because I know how to do that.
  2. jcsd
  3. Oct 22, 2004 #2


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    Since [tex]f(x)[/tex] is even we have:
    now, let's take the derivative of both sides. The LHS is easy:
    and we can do the RHS using the chain rule:
    [tex]\frac{d}{dx}f(-x)=- f'(-x)[/tex]
    so, the original equation turns into:
    multiply both sides by [tex]-1[/tex]
    which indicates that the function is odd.
  4. Oct 22, 2004 #3


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    Not all functions are powers of x!

    If f(x)= sin(x), an odd function, its derivative is cos(x), an even function.

    NateTG's response is the way to go.
  5. Oct 22, 2004 #4


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    Thanks NateTG.
    Last edited: Oct 22, 2004
  6. Feb 25, 2011 #5
    Yes but this would be sufficient if the function can be expanded in a power series. sin x = x - x^3/3! + x^5/5! +....

    d sinx/dx = 1 - x^2/2! + x^4/4 +... = cos x.

    Thus, his proof is correct for all analytic functions.
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