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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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Your limit definition is incorrect btw. Here it is: \[\lim_{h \rightarrow 0}\frac{ f(x+h)-f(x) }{ h }\]
I think in the video when the professor teaching the "Binomial theorem" since 44:00, he explained your question.
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all question mark is actually delta, I think the system didn't recognize my answer, the 47:06 and below number means the time in the video. Hope this can help you.
The seconde half of the problem is trivial following from the first. So I'll elaborate on the first here. Proof Suppose f(x) is an even function, i.e. \[\forall x, f(x)=f(-x)\] The derivative of f(x) at x \[g(x)=\lim_{\Delta x \rightarrow 0}\frac{ f(x+\Delta x) - f(x)}{ \Delta x } \] The derivative of f(x) at -x \[g(-x)=\lim_{\Delta x \rightarrow 0}\frac{ f((-x)+\Delta x) - f(-x)}{ \Delta x }\] since \[\forall x, f(x)=f(-x)\],we have \[ f((-x)+\Delta x)= f(x- \Delta x)\],thus \[g(-x)=\lim_{\Delta x \rightarrow 0}\frac{ f(x-\Delta x) - f(x)}{ \Delta x }=-\lim_{ -\Delta x \rightarrow 0}\frac{ f(x+(-\Delta x) ) - f(x)}{- \Delta x } \] Thus \[g(-x)= -g(x) \],i.e.g(x) is an odd function.

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