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

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

Intuition
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.

Identifying stationary points
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)}#.

Step-by-step 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.

Second derivative is zero 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^3+8\cdot x+3#.

Simplify your answer as much as possible.

#f''(x)=# #42\cdot x#

We first calculate the first derivative using the power rule.
\[f'(x)=21\cdot x^2+8\]
Then we calculate the second derivative in the same way.
\[f''(x)=42\cdot x\]

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