For a given function #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."#