Prove that the derivative of an even differentiable function is odd, and the derivative of an odd differentiable function is even.

Question

Prove that the derivative of an even differentiable function is odd, and the derivative of an odd differentiable function is even.

Answer 1

If $f$ is even: $$\begin{align*} f'(-x) & = \lim_{h \rightarrow 0} \frac{f(-x + h) - f(-x)}{h} = \lim_{h \rightarrow 0} \frac{f(x - h) - f(x)}{h} = \\ & = -\lim_{h \rightarrow 0} \frac{f(x) - f(x - h)}{h} = -f'(x) \end{align*}$$

If $f$ is odd: $$\begin{align*} f'(-x) & = \lim_{h \rightarrow 0} \frac{f(-x + h) - f(-x)}{h} = \lim_{h \rightarrow 0} \frac{-f(x - h) + f(x)}{h} = \\ & = \lim_{h \rightarrow 0} \frac{f(x) - f(x - h)}{h} = f'(x) \end{align*}$$

Answer 2

$f$ is even : $f(-x) = f(x)$.

Example: $f(x) =x^2$.

Differentiate both sides:

$-f'(-x) = f'(x)$, chain rule,

or $f'(-x) = -f'(x).$

2) $f$ is odd: $f(-x) = - f(x)$.

Example: $f(x) = x^3$.

Differentiate both sides:

$-f'(-x) = -f'(x)$, chain rule,

or $f'(-x)= f'(x)$.

Answer 3

We will prove that, the derivative of an odd function is even Suppose f is an odd function Therefore f(-x) = - f(x) , for every x in R

Taking Derivatives of both the sides with respect to x , we get d/dx f(-x) = d/dx [-f(x)]

Using chain Rule , we get, f'(-x).d/dx (-x) = - f'(x)

i.e f'(-x) .(-1) = - f'(x) Therefore - f'(-x) = - f'(x) , cancelling -ve signs from both the sides, we get f'(-x) = f'(x) , for all x in R This proves that, f is an even function.

Similarly we can prove that, the derivative of an even function is odd