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Math Help - Even/odd function differentiability

  1. #1
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    Even/odd function differentiability

    How would you prove that if f is an even function ( f(-x)=f(x) ) and has a derivative at every point, then the derivative f' is an odd function ( f'(-x) = -f'(x) ) ?
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  2. #2
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    Quote Originally Posted by frenchguy87 View Post
    How would you prove that if f is an even function ( f(-x)=f(x) ) and has a derivative at every point, then the derivative f' is an odd function ( f'(-x) = -f'(x) ) ?

    Apply the chain rule to f(-x):\,\,\,f'(-x)=-f'(-x) , and thus ...

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Apply the chain rule to f(-x):\,\,\,f'(-x)=-f'(-x) , and thus ...

    Tonio
    How do you get to f'(-x)=-f'(x) from that ?
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  4. #4
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    Quote Originally Posted by frenchguy87 View Post
    How do you get to f'(-x)=-f'(x) from that ?

    Well, since f(-x)=f(x)\Longrightarrow f'(-x)=f'(x) , right?

    Tonio
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  5. #5
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    I don't know why but I'm not convinced
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  6. #6
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    Quote Originally Posted by frenchguy87 View Post
    I don't know why but I'm not convinced

    Perhaps because you don't understand...?

    If f(-x)=f(x) then  (f(-x))'=f'(x) ; OTOH, applying the chain rule we have that (f(-x))'=-f'(-x) and thus we get f'(x)=-f'(-x)\Longleftrightarrow f'(-x)=-f'(x) ...

    Tonio
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  7. #7
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    Tonio gave the true answer. If you need some intuition, just draw an even function (symmetric around the y-axis) and pick some point x. Then draw tangent lines at x and at -x. Clearly they have opposite slopes.
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