5.7 Higher-Order Derivatives

Now that we mastered using different rules to find derivatives, we have one more thing to learn: higher-order derivatives. As you may have noticed, the derivative of a function is a function. Therefore, we can differentiate a derivative. A derivative of the derivative of a function is called the second derivative because we differentiate it twice. It has some useful features.

The second derivative is often denoted as $f^{\prime \prime }(x),y^{\prime \prime }$, or $\frac{d^{2}y}{d{x}^2}$ . The second derivative is used to find the acceleration at a certain value of $x$ and the curvature of the graph of a function, which we will discuss in the next chapter. Putting the application aside for now, let’s practice finding the second derivative!

Example 1

What is the second derivative of the following equation?

$f(x)=x^5-3x^2-4x+1$

First, we find the first derivative, which is

$f^{\prime }(x)=5x^4-6x-4$

Therefore, the second derivative is

$f^{\prime \prime }(x)=20x^3-6$

Example 2

Let’s practice one more:

$f(x)=e^{x^5-3x^2-4x-1}$

Recall that the derivative of $e^{g(x)}$ is

$g^{\prime }(x)e^{g(x)}$

where

$g(x)=x^5-3x^2-4x-1$

Since

$g^{\prime }(x)=5x^4-6x-4$

the first derivative of $f(x)$ is

$f^{\prime }(x)=(5x^4-6x-4)e^{x^5-3x^2-4x-1}$

For the second derivative, we have to use the product rule. Let

$\begin{array}{rl}& g(x)=5x^4-6x-4\\ & f(x)=e^{x^5-3x^2-4x-1}\end{array}$

since

$f^{\prime }(x)=g(x)f(x)$

The derivatives of $g(x)$ and $f(x)$ are

$\begin{array}{rl}& g^{\prime }(x)=20x^3-6\\ & f^{\prime }(x)=(5x^4-6x-4)e^{x^5-3x^2-4x-1}\end{array}$

Therefore, by applying the product rule, we get

$\begin{array}{rl}f"(x)& =g^{\prime }(x)f(x)+g(x)f^{\prime }(x)\\ & =(20x^3-6)e^{x^5-3x^2-4x-1}+(5x^4-6x-4)(5x^4-6x-4)e^{x^5-3x^2-4x-1}\\ & =[20x^3-6+(5x^4-6x-4{)}^2]e^{x^5-3x^2-4x-1}\end{array}$

In the above function, we just factored by the exponent term. You can further clean up the function within the brackets if you wish to.