1. ## proving derivatives help

recall that a function f is called even if f(-x)=f(x) for all x in its domain and odd if f(-x)=-f(x) for all such x. prove each of the following.

a.) the derivative of an even function is an odd function
b.) the derivative of an odd function is even function

2. ## Re: proving derivatives help

Originally Posted by pnfuller
recall that a function f is called even if f(-x)=f(x) for all x in its domain and odd if f(-x)=-f(x) for all such x. prove each of the following.
a.) the derivative of an even function is an odd function
b.) the derivative of an odd function is even function
You know that the derivative $D_x\left[f(-x)\right]=-f^{\prime}(-x)~.$

3. ## Re: proving derivatives help

and how does that help me?

4. ## Re: proving derivatives help

If you know what "even" and "odd" function mean that pretty much is the answer to your question.

5. ## Re: proving derivatives help

I'll just show the f even case. Like many problems, the way to do it is kinda forced on you.

Look at the problem:

Given: f(-x) = f(x) (f is even) and f is differentiable (implicitly given, otherwise the problem makes no sense).
(Ideally, the domain should be specified, but we'll just it this and assume f is defined whenever needs be in our calculations.)
Prove: f'(-x) = -f'(x) (f' is odd)

Observe: We need to say something about the derivative of f, but know nothing about f's derivative other than it exists. This strongly suggests using the definition of the derivative. Also, we're going to need to compare f'(x) with f'(-x), so we'll have to apply the definition in both those case. These observation pretty much tell you exactly how to proceed.
I'll start with f'(-x), since it's ususally easier to see how to go from the more complex to the simple (i.e. simplify) than to see how to massage the simple into a desired more complex form. Here, I'm thinking negatives are "more complex".
Thus I'll start with the definition of f'(-x), and try to manipulate it into some expression for f'(x) so that I can compare the two. How can I manipulate it? The only thing I know about f is that f(x) = f(-x), so that's going to be the manipulation I'll have to use.

$f'(-x) = \lim_{h \rightarrow 0} \frac{f((-x)+h) - f(-x)}{h}$

Good - I see f inside the definition of f'! That means I can use the f(-x) = f(x) to manipulate it. That's what motivates the following:

$f'(-x) = \lim_{h \rightarrow 0} \frac{f(-x+h) - f(-x)}{h} = \lim_{h \rightarrow 0} \frac{f( -(x-h) ) - f(-x)}{h}= \lim_{h \rightarrow 0} \frac{f( (x-h) ) - f(x)}{h}$.

The last equality was using that f is even. This is great, because that looks almost like the definition of f(x) that we're seeking. However, the definition is usually stated with a numerator of f(x+h) - f(x), and there we have f(x-h) - f(x). I'll fix that up by letting s = -h (hence h = -s), and then rewritting the problem in terms of s. Note that as h goes to 0, s goes to 0, and visa versa.

Thus $f'(-x) = \lim_{h \rightarrow 0} \frac{f( x - h ) - f(x)}{h} = \lim_{s \rightarrow 0} \frac{f( x - (-s) ) - f(x)}{(-s)} = \lim_{s \rightarrow 0} \frac{f( x+ s ) - f(x)}{-s}$

$= -\left(\lim_{s \rightarrow 0} \frac{f( x+ s ) - f(x)}{s}\right) = -f'(x)$.

That shows that f'(-x) = -f'(x), so you're done.

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You should understand the last equality. For derivatives sometimes people use h, sometimes they use $\Delta x$. Either way, it's a "dummy variable" that has meaning only as you're computing the limit. Once you've "taken the limit", that dummy variable disappears.

$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} = \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} = \lim_{s \rightarrow 0} \frac{f(x+s) - f(x)}{s}$

$= \lim_{q \rightarrow 0} \frac{f(x+q) - f(x)}{q} = \lim_{whatever \rightarrow 0} \frac{f(x+whatever) - f(x)}{whatever}$.

6. ## Re: proving derivatives help

Oh my goodness! While the is nothing incorrect about reply #5, why use a cannon to kill a fly?
Or is it just being pedantic?

If we know that $f(x)=f(-x)$ then $f^{\prime}(x)=-f^{\prime}(-x)$ which is the definition of an odd function.

7. ## Re: proving derivatives help

That's relying on the chain rule - but the chain rule might not have been discussed at that point in a course when this problem is given.

Let u(x) = -x. Then f even says that f(x) = f(-x) = f(u(x)), so $f = f \circ u$. Since u'(x) = -1 (the constant function -1), yes, it follows quickly from the chain rule:

$f'(x) = \frac{df}{dx} = \frac{d(f \circ u)}{dx}(x) = \left(\frac{df(u)}{du}\right)\left(\frac{du}{dx}(x )\right) = f'(u)(-1) = -f'(-x)$