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OCW Scholar - Single Variable Calculus 18 Online
OpenStudy (anonymous):

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)

OpenStudy (anonymous):

Your limit definition is incorrect btw. Here it is: \[\lim_{h \rightarrow 0}\frac{ f(x+h)-f(x) }{ h }\]

OpenStudy (anonymous):

I think in the video when the professor teaching the "Binomial theorem" since 44:00, he explained your question.

OpenStudy (anonymous):

I attached a picture in the above post with the examples which may help you to understand.

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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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