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Math Help - even/odd function's derivative is odd/even

  1. #1
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    even/odd function's derivative is odd/even

    using differentiation, i want to show

    f is even on (-r,r) if and only if f' is odd on (-r,r) and also odd-even relation...

    in proving f=>f', differentiating the eqation seems sufficient, but

    i dont know how to do the converse, proving f'=>f using differentiation...
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  2. #2
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    Hi

    For all x in (-r,r)
    -f'(-x) = f'(x)
    (f(-x))' = (f(x))'

    Integrating gives
    f(-x) = f(x) + K

    To find K we take x=0
    f(0) = f(0) + K
    K = 0

    And finally
    f(-x) = f(x)
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  3. #3
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    is there any way to prove without using integration?

    i think mean value theorum might work...
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  4. #4
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    You could use the contrapositive of your original statement:

    You're trying to show that: If f' is odd on (-r,r), then f is even on (-r,r).

    The contrapositive would be: If f is not even on (-r,r), then f' is not odd on (-r,r).

    Both of these statements are logically equivalent.
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  5. #5
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    Use the chain rule: for any function f(x), \frac{df(-x)}{dx}= \frac{df(u)}{du}\frac{du}{dx} with u= -x. That is, \frac{df(-x)}{dx}= f'(-x)(-1)= -f'(-x).

    If f is an even function, then f(-x)= f(x) so \frac{df(-x)}{dx}= \frac{df(x)}{dx}= f'(x) and that previous equation becomes f'(x)= -f'(-x) which is the same as -f'(x)= f'(-x) showing that f'(x) is an odd function.

    If f is an odd function, then f(-x)= -f(x) so \frac{df(-x)}{dx}= -\frac{df(x)}{dx}= -f'(x) and that previous equation becomes -f'(x)= -f'(-x) which is the same as f'(x)= f'(-x) showing that f'(x) is an even function.
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