Derivatives of Even Functions

A recent tweet from @AnalysisFact noted that the derivative of an even function is an odd function.  There are many ways to explore and understand this fact, but here’s a simple algebraic approach that uses a neat little trick in representing even and odd functions.

Claim:  The derivative of a [differentiable] even function is an odd function.

Proof:  Suppose f(x) is an even, differentiable function.  Consider the function

a(x) = \frac {f(x) + f(-x)}{2}

First we show that a(x) = f(x).  Since f(x) is even, we know

f(x) = f(-x).

Thus, a(x) = \frac{f(x)+f(-x)}{2} = \frac{2f(x)}{2} = f(x).

Now let’s differentiate f(x).  We have

f'(x) = a'(x) = \frac{(f(x)+f(-x))'}{2} = \frac{f'(x) - f'(-x)}{2}

where the last step follows by the chain rule.

And since

f'(-x) = a'(-x) = \frac{f'(-x) -f'(x)}{2} = -\frac{(-f'(-x)+f'(x))}{2} = -a'(x) = -f'(x)

we see that the derivative of f(x) is an odd function, as desired.

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https://learning.blogs.nytimes.com/2012/09/12/test-yourself-sept-12-2012/

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