The second derivative

Differentiation: Applications of derivatives

The second derivative

The derivative $f'$ of a function $f$ can be differentiated again. We call this the second derivative of $f$ .

For a function $\blue{f(x)}$ , we denote the second derivative as:

$\purple{f''(x)}=\frac{\dd}{\dd x}\green{f'(x)}=\frac{\dd}{\dd x}\left(\frac{\dd}{\dd x}\blue{f(x)}\right)=\frac{\dd^2}{\dd x^2} \,\blue{f(x)}$

Example

$\begin{array}{rcl}\blue{f(x)}&\blue{=}&\blue{3x^2}\\ \green{f'(x)}&\green{=}&\green{6x}\\\purple{f''(x)}&\purple{=}&\purple{6}\end{array}$

We have seen before that the derivative $\green{f'(x)}$ of a function $\blue{f(x)}$ measures its rate of change. If $\blue{f(x)}$ is increasing in $x$ , then $\green{f'(x)}>0$ and if $\blue{f(x)}$ is decreasing in $x$ then $\green{f'(x)}<0$ .

The same reasoning can be extended to the first and second derivatives, meaning that the second derivative $\purple{f''(x)}$ measures the rate of change of the first derivative $\green{f'(x)}$ . Here we have that if $\green{f'(x)}$ is increasing in $x$ , then $\purple{f''(x)}>0$ and if $\green{f'(x)}$ is decreasing in $x$ then $\purple{f''(x)}<0$ .

The second derivative is useful when one wants to find the extreme values of a function $f(x)$ . We saw earlier that the condition $f'(c)=0$ does not immediately imply that $c$ corresponds to an extreme value. The following theorem will help us determine whether stationary points, which are points with $f'(c)=0$ , correspond to an extreme value or not without sketching the graph or making a sign analysis chart.

Let $\blue{f(x)}$ be a function and $\orange{c}$ a stationary point in the domain of $\blue{f(x)}$ .

If $\purple{f''(}\orange{c}\purple{)}\neq 0$ , then $\blue{f(x)}$ has an extreme value in $\orange{c}$ .

More specifically,

If $\purple{f''(}\orange{c}\purple{)}>0$ , then $\orange{c}$ corresponds to a local minimum,
If $\purple{f''(}\orange{c}\purple{)}<0$ , then $\orange{c}$ corresponds to a local maximum.

Example

$\begin{array}{rcl}\blue{f(x)}&=&\blue{2x^2+x}\\ \green{f'(x)}&=&\green{4x+1}\\ \purple{f''(x)}&=&\purple{4}\\ \green{f'(}\orange{-\frac{1}{4}}\green{)}&=&0\\ \purple{f''(}\orange{-\frac{1}{4}}\purple{)}&=&4\neq 0\end{array}$ $\textstyle\purple{f''(}\orange{-\frac{1}{4}}\purple{)}> 0$ , so $\textstyle\blue{f(}\orange{-\frac{1}{4}}\blue{)}=\frac{3}{8}$
is a local minimum of $\blue{f(x)}$ .

The step-by-step method for finding the extreme values of a function $\blue{f(x)}$ gets much easier using this theorem. We can partially replace step $3$ by just checking if the second derivative at the found zeroes is smaller, larger, or equal to $0$ .

To decide if the boundary points of the domain of $\blue{f(x)}$ are extreme values, we still need to sketch the graph or make a sign analysis chart, as described in step $3$ of the step-by-step method.

The case where $\purple{f''(}\orange{c}\purple{)}= 0$ will be discussed later. In this case, we are dealing with inflection points.

Calculate $f''(x)$ for $f(x)=7\cdot x^4-2\cdot x^2-4$ .

Simplify your answer as much as possible.

$f''(x)=$ $84\cdot x^2-4$

We first calculate the first derivative using the power rule.
$f'(x)=28\cdot x^3-4\cdot x$
Then we calculate the second derivative in the same way.
$f''(x)=84\cdot x^2-4$

New example