Higher Order Differentiation

Higher Order Differentiation

In some applications (soon to come up) and in the physical sciences such as physics, it may be necessary to find what are known as higher order derivatives. Higher order derivatives are simply derivatives of derivatives. A common example occurs in physics where the derivative of displacement $s(t)$ is velocity, that is $s'(t) = v(t)$. Also, the derivative of velocity is acceleration $s''(t) = v'(t) = a(t)$. Alternatively we could say that the second derivative of displacement is acceleration to which a double prime symbol denotes differentiating $s(t)$ twice.

Definition: If $f$ is a function such that $f'(x)$ is differentiable, then the Second Derivative of $f$ denoted $f''(x)$ or $\frac{d^2}{dx^2} f(x)$ is the derivative of the derivative of $f$. If $f''(x)$ exists, then we say that $f$ is Twice Differentiable.

We should note that if it is possible to differentiable again and again (that is, the derivatives are still differentiable), then we can also define the third, fourth, fifth, etc… derivatives in a similar fashion.

Let's now look at some examples of higher order derivatives.

Example 1

Compute the second derivative $\frac{d^2f(x)}{dx^2}$ of the function $f(x) = 6x^2 - 3x^3 + 2x - 3$.

Let's first find $\frac{d}{dx} f(x)$.

\begin{align} f(x) = 6x^2 - 3x^3 + 2x - 3 \\ \frac{d}{dx} f(x) = 12x - 9x^2 - 2 \end{align}

And now let's take the derivative of $\frac{d}{dx} f(x)$:

\begin{align} \frac{d}{dx} f(x) = 12x - 9x^2 - 2 \\ \frac{d}{dx} ( \frac{d f(x)}{dx} ) = 12 - 18x \\ \frac{d^2f(x)}{dx^2} = 12 - 18x \end{align}

Patterns in Higher Order Derivatives

Sometimes a pattern may form when we work with finding derivatives of higher orders of a function. For example, let's look at the higher order derivatives of the function$f(x) = e^x$:

\begin{align} \frac{d}{dx} e^x = e^x \\ \frac{d^2}{dx^2} e^x = e^x \\ \frac{d^3}{dx^3} e^x = e^x \\ . . . \\ \frac{d^n}{dx^n} e^x = e^x \\ \end{align}

Since the derivative of $e^x$ is always $e^x$, then when we continue to take higher order derivatives of the original function $f(x)$, we continue to get $e^x$ as a result.

Recall, a similar pattern occurred with the sine and cosine functions:

\begin{align} \frac{d}{dx} \sin x = \cos x \\ \frac{d^2}{dx^2} \sin x = -\sin x \\ \frac{d^3}{dx^3} \sin x = -\cos x \\ \frac{d^4}{dx^4} \sin x = \sin x \\ \frac{d^5}{dx^5} \sin x = \cos x \end{align}

As you can see, $\frac{d}{dx} \sin x = \frac{d^5}{dx^5} \sin x$. In fact:

\begin{align} \quad \frac{d^{4n + 1}}{dx^{4n + 1}} \sin x = \sin x \quad n \in \mathbb{Z} \quad \mathbf{and} \quad n ≥ 0 \end{align}
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License