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Use the Chain Rule to prove the following.
(a) The derivative of an even function is an odd function.
(b) The derivative of an odd function is an even function.

My attempt:

I can easily prove these using the definition of a derivative, but I'm having trouble showing them using the chain rule.

(a) $f(x)$ is even $ \therefore f(-x) = f(x)$.
We need to show that $f'(-x) = -f'(x)$.

Let $u = -x$.
My reasoning for the next step is that if we want to find the derivative of $f$ with respect to $x$ from $f(u)$ we need to use the chain rule and first find the derivative of $f$ with respect to $u$ and multiply that with the derivative of $u$ with respect to $x$.
$f'(x) = f'(u) \cdot u' = - f'(u) = -f'(-x)$.
$f'(x) = -f'(-x)$.
$f'(-x) = -f'(x)$.

The problem with this is that I never used the fact that $f$ is even and I feel like there is a mistake in my approach, most likely where I explained my reasoning. I just can't see it. Can you please point out the mistake and point me to the right direction?

Thanks!

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  • $\begingroup$ $f'(x)=f'(u)u'$ is not correct. Take the example $f(x)=x$ to see why this is not true. $\endgroup$ – Kavi Rama Murthy Jan 24 at 9:34
  • $\begingroup$ See math.stackexchange.com/questions/2472684/… $\endgroup$ – Kavi Rama Murthy Jan 24 at 9:37
  • $\begingroup$ @KaviRamaMurthy I can show this using the definition of the derivative, I need to use the chain rule $\endgroup$ – Ani Khachatryan Jan 24 at 9:40
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    $\begingroup$ This is fine; as mentioned in the answers, you used the fact that $f$ is even in your first equality (${d\over dx} f(x)={d\over dx} f(-x)$). $\endgroup$ – David Mitra Jan 24 at 9:52
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$f$ being even means for every $x\in\Bbb{R}$, $f(x)=f(-x)$. Or, if we define the function $u:\Bbb{R}\to\Bbb{R}$ as $u(x)=-x$, then the condition for being an even function is that $f=f\circ u$. So, actually, you did use the evenness assumption when you started calculating derivatives, even if you didn't explain it very explicitly.

Now, the chain rule tells us that $f'=(f\circ u)' = (f'\circ u)\cdot u'$. Explicitly, this says for every $x\in\Bbb{R}$, \begin{align} f'(x)&= f'(u(x))\cdot u'(x)=f'(-x)\cdot (-1) = -f'(-x). \end{align} But this is precisely what it means for $f'$ to be odd. The other question is similar.

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Since $(-x)^\prime=-1$, we can concisely prove both parts viz.$$f(-x)=\pm f(x)\implies f^\prime(-x)=-1\cdot\pm f^\prime(x)=\mp f^\prime(x).$$

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Actually, you are using the assumption that $f$ is even.

If $f$ is even, then as you say $f(-x) = f(x)$. By differentiating, the left hand side is $-f'(-x)$ by the chain rule, and the right hand side is simply $f'(x)$. Thus $-f'(-x) = f'(x)$, or $f'(-x) = -f'(x)$, meaning that $f'$ is odd. You can try with $f$ odd now.

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You did use that $f$ is even. Let us look at the following more general situation that $$ f = v \circ f \circ u .$$ For an even function we have $u(x) = -x$ and $v(x) = x$, for an odd function $u(x) = v (x) = -x$. Then $$f'(x) = v'(f(u(x)))f'(u(x))u'(x) .$$ For $u(x) = -x$ and $v(x) = (-1)^\epsilon x$ we get $$f'(x) = (-1)^{\epsilon +1}f'(-x).$$

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