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This is the question: "Show that the derivative of an even function is odd and that the derivative of an odd function is even. (Write the equation that says f is even, and differentiate both sides, using the chain rule.)"

I already read numerous solutions online. This was the official solution but I didn't quite understand it (particularly, I'm not convinced why exactly $dz/dx=-1$; even though $z=-x$).

Thanks in advance =]

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Well, what what you say that the derivative of $-x$ is, if not $-1$? –  Hans Lundmark Sep 11 '12 at 9:44
    
Wow, can't believe I missed that. Thank you =] (I looked at it in a different way altogether.) –  Py42 Sep 11 '12 at 9:47
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2 Answers 2

up vote 3 down vote accepted

Following the official solution, we have $f(-x) = -f(x)$ by assumption. Thus, by considering the function $g(x) = -x = (-1) \cdot x$, we have $f(g(x)) = (-1)\cdot f(x)$. Differentiating on both sides gives $$\frac{d}{dx} f(g(x)) - \frac{d}{dx} (-1)\cdot f(x) = -1 \cdot \frac{d}{dx} f(x).$$

Now, applying the chain rule, we get $$\frac{d}{dx} f(g(x)) = \frac{df}{dx} g(x) \cdot \frac{d}{dx} g(x) = \frac{df}{dx} (-x) \cdot \frac{d}{dx} (-1 \cdot x) = \frac{df}{dx} (-x) \cdot (-1).$$

Equating both sides and simplifying gives $$\frac{d}{dx} f(-x) = \frac{d}{dx} f(x).$$

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Official, shmofficial: I think the following might prove to be easier to grasp for some: suppose $\,f\,$ is odd, then

$$f'(x_0):=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}\frac{-f(x)+f(x_0)}{-x+x_0}=$$

$$=\lim_{x\to x_0}\frac{f(-x)-f(-x_0)}{(-x)-(-x_0)}\stackrel{-x\to y}=\lim_{y\to -x_0}\frac{f(y)-f(-x_0)}{y-(-x_0)}=:f'(-x_0)$$

The above remains, mutatis mutandis, in case $\,f\,$ is even.

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