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

I think you could do it like this:

Let f(x) be the odd function we're gonna work with. So, f(-x) = -f(x), By the definition of the derivative:

f'(x) = ##lim_(h->0)(f(x+h) - f(x))/h##.

We want to prove that f'(-x) = f'(x). Now, f'(-x) is:

##f'(-x) = lim_(h->0)(f(-x+h)-f(-x))/h## .

Since f(x) is and odd function, f(-x) = -f(x), therefore, we can rewrite the above expression as:

##f'(-x) = lim_(h->0)((-f(x-h)) + f(x))/h## = ##lim_(h->0)[-(f(x-h)-f(x))/h]##.

But we can rewrite the above expression as:

##f'(-x) = lim_(h->0)(f(x-h)-f(x))/-h##.

To make things clearer, let t = -h. Therefore:

##f'(-x) = lim_(t->0)(f(x+t)-f(x))/t##, which is just f'(x). That is: f'(-x) = f'(x). And there we have it: if f(x) is an odd function, then f'(x) is an even function.

Hope it helps you!

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