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The derivative of an even function is an odd function. the derivative of an odd function is an even function. Prove these results from the limit definition of the derivative: lim(as x approaches zero) [f(x) - f(a)]/(x -a)

OCW Scholar - Single Variable Calculus
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Well it takes some math to prove that by that definition... However if you look at the rule f(x)=x^n then f'(x)=nx^(n-1) it's very clear. If you have the even function x^4 it's derivative is 4(x^3) that is an odd function. It's derivative is 12(x^2), an even function.
Unfortunately, the problem was to use the definition. but I think I've got it. thanks for trying/
Another way you could prove this easily is by using trigonometric terms and identities, i.e. if you solved this, \[\lim_{x \rightarrow 0} [\sin(x+h)-\sin (x)]/h\] you would get cos x out as the answer. The sine function is an odd function, cosine an even, so it would be proven by that.

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It comes from the definition of even and odd. If a function is even, f(x) = f(-x). (If it's odd, f(x) = -f(-x).) You can use the product rule, since f(-x) = f(-1 . x), so f' = (-1)' . x + x' . (-1) = -x' Since f'(x) = -f'(-x), then the derivative is an odd function. A similar argument can be used for odd functions.
I think ac7qz has the right idea, but it looks like there might be an error in the calculation.
Let f(x) be an odd function (f(x)=-f(-x)) $$f'(x_0)=\lim_{x \rightarrow x_0} \dfrac{f(x)-f(x_0)}{x-x_0}=\lim_{-x \rightarrow -x_0} \dfrac{-f(-x)+f(-x_0)}{-x_0-(-x)}=\lim_{y \rightarrow- x_0} \dfrac{-f(y)+f(-x_0)}{-x_0-y}= $$ $$ =\lim_{y \rightarrow- x_0} \dfrac{f(y)-f(-x_0)}{y-(-x_0)}=f'(-x_0) $$ So we proved that $$ f′(x_0)=f′(−x_0)$$ That means that the derivative of an odd function is even. Let f(x) be an even function (f(x)=f(-x)) $$f'(x_0)=\lim_{x \rightarrow x_0} \dfrac{f(x)-f(x_0)}{x-x_0}=\lim_{-x \rightarrow -x_0} \dfrac{f(-x)-f(-x_0)}{-x_0-(-x)}=\lim_{y \rightarrow- x_0} \dfrac{f(y)-f(-x_0)}{-x_0-y}= $$ $$ =-\lim_{y \rightarrow- x_0} \dfrac{f(y)-f(-x_0)}{y-(-x_0)}=-f'(-x_0) $$ So we proved that $$ f′(x_0)=-f′(−x_0)$$ That means that the derivative of an even function is odd. The only assumption we need is the existense of the derivatives of the regarded function in some area around f(x0) and f(-x0).
Can we not do this with the definition \[f'(-x)= \lim_{h \to 0}\frac{f(-x+h) - f(x)} {h} = \lim_{h \to 0}\frac{f(-(x-h)) - f(x)} {h} = \lim_{h \to 0}\frac{f((x-h)) - f(x)} {h}=\\ \lim_{-h \to 0}\frac{f((x-(-h)) - f(x)} {-h}= -\lim_{h \to 0}\frac{f((x+h)) - f(x)} {h}= \\-f'(x)\]

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