Derivatives of Even and Odd Functions

# Derivatives of Even and Odd Functions

Recall that a function $f$ is said to be even on its domain if for every $x$ in the domain of $f$ we have that:

(1)\begin{align} \quad f(x) = f(-x) \end{align}

Similarly, a function $f$ is said to be odd on its domain if for every $x$ in the domain of $f$ we have that:

(2)\begin{align} \quad -f(x) = f(-x) \end{align}

We now state and prove two important results which says that the derivative of an even function is an odd function, and the derivative of an odd function is an even function.

Theorem 1: If $f$ is an even function then $f'$ is an odd function. |

**Proof:**Let $f$ be an even function. Then $f(x) = f(-x)$ for all $x$ in the domain of $f$. We differentiable both sides of this equation and apply the chain rule to get:

\begin{align} \quad f'(x) &= f'(-x) \cdot (-1) \\ \quad &= - f'(-x) \end{align}

- Therefore $-f'(x) = f'(-x)$. Hence $f'$ is an odd function. $\blacksquare$

Theorem 2: If $f$ is an odd function then $f'$ is an even function. |

**Proof:**Let $f$ be an odd function. Then $-f(x) = f(-x)$ for all $x$ in the domain of $f$. We differentiable both sides of this equation and apply the chain rule to get:

\begin{align} \quad -f'(x) &= f'(-x) \cdot (-1) \\ \quad &= -f'(-x) \end{align}

- Therefore $f'(x) = f'(-x)$. Hence $f'$ is an even function. $\blacksquare$