How do you show that the derivative of an odd function is even?

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Answer 1 · 2018-05-05T03:09:39

For a given function $f$ $f$ , its derivative is given by

$g(x)=lim_(h->0)(f(x+h)-f(x))/h$

Now we need to show that, if $f(x)$ is an odd function (in other words, $-f(x)=f(-x)$ for all $x$ ) then $g(x)$ is an even function ( $g(-x)=g(x)$ ).

With this in mind, let's see what $g(-x)$ is:

$g(-x)=lim_(h->0)(f(-x+h)-f(-x))/h$

Since $f(-x)=-f(x)$ , the above is equal to

$g(-x)=lim_(h->0)(-f(x-h)+f(x))/h$

Define a new variable $k=-h$ . As $h->0$ , so does $k->0$ . Therefore, the above becomes

$g(-x)=lim_(k->0)(f(x+k)-f(k))/k=g(x)$

Therefore, if $f(x)$ is an odd function, its derivative $g(x)$ will be an even function.

$"Q.E.D."$