The nth Derivative of a Function

Let $f : [a, b] \to \mathbb{R}$ be differentiable. Then $f'(x)$ exists and is itself a function. It is very possible that $f'$ is also differentiable and we can differentiate $f'$ to get $(f')' = f''$.

Definition: Let $f : [a, b] \to \mathbb{R}$ be differentiable. If $f'$ is also differentiable then $f$ is said to be $2$-times Differentiable its derivative is called the Second Derivative of $f$ and is denoted $f'' = (f')'$.

We can define an $n$-times differentiable function and the $n^{\mathrm{th}}$ derivative of $f$, denoted $f^{(n)}$, similarly, provided that the previous derivatives exist.

For example, if $f(x) = x^3$ then $f'(x) = 3x^2$ and $f''(x) = [f'(x)]' = [3x^2]' = 6x$.

Theorem 1: If $f$ is a polynomial of degree $n$ then $f^{(n)}$ is a constant function and $f^{(n+1)}$ is the zero function.

• Proof: Let $f$ be a polynomial of degree $n$. Then $f$ has the form:

(1)

\begin{align} \quad f(x) = a_0 + a_1x + ... + a_nx^n \end{align}

• If we differentiate $f$ we get:

(2)

\begin{align} \quad f'(x) = a_1 + 2a_2x + ... + na_nx^{n-1} \end{align}

• So differentiating a polynomial reduces the degree of all of its terms by $1$. So $f^{(n)}$ will be a constant. Namely:

(3)

\begin{align} \quad f^{(n)}(x) = n! a_n \end{align}

• And furthermore, $f^{(n+1)}(x) = 0$ since the derivative of a constant is equal to $0$. $\blacksquare$

Theorem 2: a) $\sin^{(4n)}(x) = \sin(x)$ for all $n \in \mathbb{N}$. b) $\cos^{(4n)}(x) = \cos(x)$ for all $n \in \mathbb{N}$.