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Applied Calculus

Kevin Gonzales, Eric Hopkins, Catherine Zimmitti, Cheryl Kane, Modified to fit Applied Calculus from Coordinated Calculus by Nathan Wakefield et. al., Based upon Active Calculus by Matthew Boelkins

Section 2.8 Higher-Order Derivatives

This section corresponds to 2.7 Higher Order Derivatives in the workbook.
We are now accustomed to investigating the behavior of a function by examining its derivative.
Because \(f'\) is itself a function, it is perfectly feasible for us to consider the derivative of the derivative, which is the new function \(y = [f'(x)]'\text{.}\) We call this resulting function the second derivative of \(f\text{,}\) and denote the second derivative by \(y = f''(x)\text{.}\) Consequently, we will sometimes call \(f'\) “the first derivative” of \(f\text{,}\) rather than simply “the derivative” of \(f\text{.}\)

Second Derivative.

The second derivative is the derivative of the first derivative. That is,
\begin{equation*} f''(x) = \frac{d}{dx}\left[f'(x)\right]\text{.} \end{equation*}
We read “\(f''(x)\)” as “\(f\)-double prime of \(x\)”, or as “the second derivative of \(f\)”.
The meaning of the derivative function still holds, so when we compute \(f''(x)\text{,}\) this new function measures slopes of tangent lines to the curve \(y = f'(x)\text{,}\) as well as the instantaneous rate of change of \(y = f'(x)\text{.}\) In other words, just as the first derivative measures the rate at which the original function changes, the second derivative measures the rate at which the first derivative changes. The second derivative will help us understand how the rate of change of the original function is itself changing.

Example 2.8.1.

Consider the function \(f(x)=x^5-3x^4+x\text{.}\) The derivative is:
\begin{equation*} f'(x)=5x^4-12x^3+1. \end{equation*}
This creates a new function whose derivative can also be found:
\begin{equation*} f''(x)=\frac{d}{dx}\left[5x^4-12x^3+1\right]=20x^3-36x^2. \end{equation*}

Example 2.8.2.

Consider the function \(g(x)=3(x^3+4x+2)^7\text{.}\) To find the derivative we must use the chain rule. The derivative is:
\begin{equation*} g'(x)=21(x^3+4x+2)^6(3x^2+4). \end{equation*}
To take the second derivative, first we must use the product rule:
\begin{equation*} g''(x)=\frac{d}{dx}\left[21(x^3+4x+2)^6(3x^2+4)\right]=\frac{d}{dx}\left[21(x^3+4x+2)^6\right](3x^2+4)+21(x^3+4x+2)^6\frac{d}{dx}\left[(3x^2+4)\right] \end{equation*}
The derivative of the first term requires using the chain rule again:
\begin{equation*} g''(x)=126(x^3+4x+3)^5(3x^2+4)(3x^2+4)+21(x^3+4x+3)^6(6x). \end{equation*}
Note that the term \((3x^2+4)\) appears twice, once for each derivative, both are the derivative of the inside function.

Example 2.8.3.

Find the second derivative for each each of the following functions.
  1. \(\displaystyle \displaystyle p(x) = 4x^2 + 3\sqrt{x}+\frac{4}{x^2}\)
  2. \(\displaystyle \displaystyle m(x) = \frac{4x}{5x+2}\)
  3. \(\displaystyle \displaystyle h(x) = \sqrt{3x^2+5x}\)
  4. \(\displaystyle \displaystyle c(x) = \frac{3}{(7x^3+5x)^5}\)
Answer.
  1. \(p''(x) = 8-\frac{3}{3x^{3/2}}+\frac{24}{x^4}\text{.}\)
  2. \(m''(x) = \frac{-80}{(5x+2)^3} \text{.}\)
  3. \(\displaystyle h''(x) =\frac{-1}{4}(3x^2+5x)^{-3/2}(6x+5)(6x+5)+\frac{1}{2}(3x^2+5x)^{-1/2}(6)\text{.}\)
  4. \(c''(x)=90(7x^3+5x)^{-7}(21x^2+5)(21x^2+5)+-15(7x^3+5x)^{-6}(42x)\text{.}\)
Solution.
  1. First rewrite this so that the power rule applies to each term:
    \begin{equation*} p(x) = 4x^2 + 3\sqrt{x}+\frac{4}{x^2}=4x^2+3x^{1/2}+4x^{-2}\text{.} \end{equation*}
    Then the first derivative is
    \begin{equation*} p'(x) = 8x + \frac32 x^{-1/2}-8x^{-3}\text{.} \end{equation*}
    Then take the derivative again to get the second derivative:
    \begin{equation*} p''(x) = 8 - \frac34 x^{-3/2}+24x^{-4}=8-\frac{3}{3x^{3/2}}+\frac{24}{x^4}\text{.} \end{equation*}
  2. To get the derivative apply the quotient rule:
    \begin{equation*} m'(x)=\frac{4(5x+2)-(4x)(5)}{(5x+2)^2}=\frac{20x+8-20x}{(5x+2)^2}=\frac{8}{(5x+2)^2}\text{.} \end{equation*}
    Simplify the numerator in order to make finding the second derivative easier. Then we can rewrite as
    \begin{equation*} m'(x)=8(5x+2)^{-2} \end{equation*}
    and use the chain rule to find the second derivative.
    \begin{equation*} m''(x)=-16(5x+2)^{-3}(5)=\frac{-80}{(5x+2)^3}\text{.} \end{equation*}
  3. By the chain rule,
    \begin{equation*} h'(x) = \frac{1}{2}(3x^2+5x)^{-1/2}(6x+5)\text{.} \end{equation*}
    Applying both the product rule and chain rule can find the second derivative
    \begin{equation*} h''(x) =\frac{-1}{4}(3x^2+5x)^{-3/2}(6x+5)(6x+5)+\frac{1}{2}(3x^2+5x)^{-1/2}(6)\text{.} \end{equation*}
  4. First rewrite
    \begin{equation*} c(x)=\frac{3}{(7x^3+5x)^5}=3(7x^3+5x)^{-5}\text{.} \end{equation*}
    Use the chain rule to find the derivative:
    \begin{equation*} c'(x)=-15(7x^3+5x)^{-6}(21x^2+5) \end{equation*}
    Applying both the product rule and chain rule can find the second derivative
    \begin{equation*} c''(x)=90(7x^3+5x)^{-7}(21x^2+5)(21x^2+5)+-15(7x^3+5x)^{-6}(42x)\text{.} \end{equation*}

Subsection 2.8.1 Acceleration

If the function \(s(t)\) gives the position of an object at time \(t\) then \(s'(t)\) gives the change in position, otherwise known as velocity. That is, \(s'(t)=v(t)\text{,}\) where \(v(t)\) is the velocity function. Using the alternate notation, we have \(\displaystyle \frac{ds}{dt}=v(t)\text{.}\)
Following this same idea, \(v'(t)\) gives the change in velocity, more commonly called acceleration. Using derivative notation, \(v'(t)=a(t)\text{.}\) Therefore, \(s''(t)=a(t)\text{.}\) That is, the second derivative of the position function gives acceleration. Using the alternative notation, we write \(\displaystyle \frac{d^2s}{dt^2}=a(t)\text{.}\)
 1 
Notice that in higher order derivatives the exponent occurs in what appear to be different locations in the numerator and denominator. In reality, what is happening is we have \(\displaystyle \frac{d^{n}}{dt^{n}}\) acting as an operator that takes the \(n\)th order derivative of the function.

Example 2.8.4.

Suppose that an object is launched from a height of 12 feet with an initial velocity of 15 feet per second. We can describe the position of the object with the following function:
\begin{equation*} s(t)=-16^2+15t+12. \end{equation*}
  1. Find an equation for the velocity of the object; make sure to include units.
  2. Find an equation for the acceleration of the object; make sure to include units.
Answer.
  1. \(v(t)=-32t+15 \) feet per second.
  2. \(a(t)=-32\) feet per second per second (or \(ft/s^2\)).
Solution.
  1. The velocity is the derivative of position:
    \begin{equation*} v(t)=s'(t)=-32t+15. \end{equation*}
    The units of velocity are feet per second.
  2. The acceleration is the derivative of velocity or the second derivative of position:
    \begin{equation*} a(t)=v'(t)=s''(t)=-32. \end{equation*}
    The units are feet per second per second, often written as \(ft/s^2\text{.}\) Here the acceleration due to gravity is constant, typically in physics acceleration due to gravity is taken as \(-32ft/s^2\text{.}\) This is negative since gravity pulls an object down. If we are working in the metric system then gravity is given as \(-9.8m/s^2\text{.}\)

Subsection 2.8.2 Higher Order Derivatives

We have seen that when we take the derivative of a function a new function \(f'(x)\) is created. We took the derivative of this new function to get the second derivative. We can repeat this process as many times as we wish.
The third derivative of a function \(f(x)\) is the derivative of the second derivative, written as \(f'''(x)\) or more commonly as \(\displaystyle \frac{d^3f}{dx^3}\text{.}\)
The fourth derivative of a function \(f(x)\) is the derivative of the third derivative, written as \(f^{(4)}(x)\) or more commonly as \(\displaystyle \frac{d^4f}{dx^4}\text{.}\)
We can continue to take derivatives in this way.

\(n^{th}\) Derivative.

The \(n^{th}\) derivative is
\begin{equation*} f^{(n)}(x) =\frac{d^nf}{dx^n}= \frac{d^n}{dx^n}\left[f(x)\right]\text{.} \end{equation*}
This is the derivative of \(f(x)\) taken \(n\) times. Alternatively we refer to this as the \(n^{th}\) order derivative.

Example 2.8.5.

Consider the function \(f(x)=x^5-3x^4+x\text{.}\) Find the \(5^{th}\) derivative.
To find the \(5^{th}\) derivative we will take the derivative \(5\) times.
The first derivative is:
\begin{equation*} f'(x)=\frac{df}{dx}=5x^4-12x^3+1. \end{equation*}
The second derivative is:
\begin{equation*} f''(x)=\frac{d^2f}{dx^2}=20x^3-36x^2. \end{equation*}
The third derivative is:
\begin{equation*} f'''(x)=\frac{d^3f}{dx^3}=60x^2-72x. \end{equation*}
The fourth derivative is:
\begin{equation*} f^{(4)}(x)=\frac{d^4f}{dx^4}=120x-72. \end{equation*}
The fifth derivative is:
\begin{equation*} f^{(5)}(x)=\frac{d^5f}{dx^5}=120. \end{equation*}
Two notes from Example 2.8.5:
1) In Example 2.8.5 any further derivatives would end as 0 since the derivative of a constant is 0. This is true of any polynomial, eventually if you take enough derivatives they will end as 0.
2) For a polynomial as you take more derivatives the function becomes simpler.

Example 2.8.6.

Consider the function \(\displaystyle g(x)=\frac{1}{x}\text{,}\) find the \(4^{th}\) derivative of \(g(x)\text{.}\)
Start with the first derivative of \(g(x)\text{:}\)
\begin{equation*} g'(x)=-x^{-2}=\frac{-1}{x^2}. \end{equation*}
The second derivative is:
\begin{equation*} g''(x)=2x^{-3}=\frac{2}{x^3} \end{equation*}
The third derivative is:
\begin{equation*} g'''(x)=-6x^{-4}=\frac{-6}{x^4}. \end{equation*}
The fourth derivative is:
\begin{equation*} g^{(4)}(x)=24x^{-5}=\frac{24}{x^5}. \end{equation*}
Note that unlike the polynomial in Example 2.8.5, the function in Example 2.8.6 has as many derivatives as we like. They never go to 0!

Example 2.8.7.

Consider the function \(h(t)=\sqrt{t^2+4}\text{,}\) find the \(3rd\) derivative of \(h(t)\text{.}\)
Start with the first derivative of \(h(t)\text{;}\) in this case will have to use the chain rule:
\begin{equation*} \frac{dh}{dt}=\frac{1}{2}(t^2+4)^{-1/2}(2t) \end{equation*}
To find the second derivative, we need to use both product rule and chain rule:
\begin{equation*} \frac{d^2h}{dt^2}=\frac{-1}{4}(t^2+4)^{-3/2}(2t)(2t)+\frac{1}{2}(t^2+4)^{-1/2}(2)=(t^2+4)^{-3/2}(-t^2)+(t^2+4)^{-1/2}. \end{equation*}
To take the third derivative, we will have to use the product rule on the first term and the chain twice:
\begin{equation*} \frac{d^3h}{dt^3}=\frac{-3}{2}(t^2+4)^{-5/2}(2t)(-t^2)+(t^2+4)^{-3/2}(-2t)+\frac{-1}{2}(t^2+4)^{-3/2}(2t) \end{equation*}
Note that when you take higher order derivatives involving the chain rule you must be very careful to keep track of every term.

Subsection 2.8.3 Summary

  • By taking the derivative of the derivative of a function \(f\text{,}\) we arrive at the second derivative, \(f''\text{.}\) The second derivative measures the instantaneous rate of change of the first derivative.
  • If \(f(x)\) is polynomial of degree \(n\text{,}\) then the \((n+1)^{st},\) derivative is zero.
  • We can continue taking the derivative of the derivative n times to the get the \(n^{th}\) derivative \(f^{(n)}(x)\text{.}\)

Exercises 2.8.4 Exercises

1. Comparing \(f, f', f''\) values.

Consider the function \(f(x)\) graphed below.
For this function, are the following nonzero quantities positive or negative?
\(f(4)\) is
  • positive
  • negative
\(f'(4)\) is
  • positive
  • negative
\(f''(4)\) is
  • positive
  • negative
(Because this is a multiple choice problem, it will not show which parts of the problem are correct or incorrect when you submit it.)

2. Signs of \(f, f', f''\) values.

At exactly two of the labeled points in the figure below, which shows a function \(f\text{,}\) the derivative \(f'\) is zero; the second derivative \(f''\) is not zero at any of the labeled points. Select the correct signs for each of \(f\text{,}\) \(f'\) and \(f''\) at each marked point.
Point A B C D E
\(f\)
  • positive
  • zero
  • negative
  • positive
  • zero
  • negative
  • positive
  • zero
  • negative
  • positive
  • zero
  • negative
  • positive
  • zero
  • negative
\(f'\)
  • positive
  • zero
  • negative
  • positive
  • zero
  • negative
  • positive
  • zero
  • negative
  • positive
  • zero
  • negative
  • positive
  • zero
  • negative
\(f''\)
  • positive
  • zero
  • negative
  • positive
  • zero
  • negative
  • positive
  • zero
  • negative
  • positive
  • zero
  • negative
  • positive
  • zero
  • negative

3. Acceleration from velocity.

Suppose that an accelerating car goes from 0 mph to 68.2 mph in five seconds. Its velocity is given in the following table, converted from miles per hour to feet per second, so that all time measurements are in seconds. (Note: 1 mph is 22/15 ft/sec.) Find the average acceleration of the car over each of the first two seconds.
\(t\) (s) 0 1 2 3 4 5
\(v(t)\) (ft/s) 0.00 34.09 59.09 77.27 90.91 100.00
average acceleration over the first second =
average acceleration over the second second =

4. Rates of change of stock values.

Let \(P(t)\) represent the price of a share of stock of a corporation at time \(t\text{.}\) What does each of the following statements tell us about the signs of the first and second derivatives of \(P(t)\text{?}\)
(a) The price of the stock is rising faster and faster.
The first derivative of \(P(t)\) is
  • positive
  • zero
  • negative
The second derivative of \(P(t)\) is
  • positive
  • zero
  • negative
(b) The price of the stock is just past where it bottomed out.
The first derivative of \(P(t)\) is
  • positive
  • zero
  • negative
The second derivative of \(P(t)\) is
  • positive
  • zero
  • negative

5. Interpreting a graph of \(f'\).

The graph of \(f'\) (not \(f\)) is given below.
(Note that this is a graph of \(f'\text{,}\) not a graph of \(f\text{.}\))
At which of the marked values of \(x\) is
A. \(f(x)\) greatest? \(x =\)
B. \(f(x)\) least? \(x =\)
C. \(f'(x)\) greatest? \(x =\)
D. \(f'(x)\) least? \(x =\)
E. \(f''(x)\) greatest? \(x =\)
F. \(f''(x)\) least? \(x =\)

6. Higher order derivatives.

Calculate the 4th derivative of \(y\text{.}\)
\begin{equation*} y = 3x^{8}-x^{6}+x^{4} \end{equation*}
\(y'=\)
\(y'' =\)
\(y''' =\)
\(y^{(4)} =\)

7. Chain rule with higher order derivatives.

Suppose
\begin{equation*} f(t) = \sqrt{9t^{2}+11} . \end{equation*}
\(f'(t)=\)
\(f''(t)=\)
Hint.
First rewrite \(f(x)=\sqrt{11+9 t^2 }\) as \(f(x)=(11+9 t^2 )^{1/2}\) then use the chain rule to find the first derivative.
To find the second derivative use the product rule and the chain rule.

8. Quotient rule with higher order derivatives.

Let \(f(x) = \dfrac{9 x}{4-5 x} .\) Then
\(f'(x)\) =
and
\(f''( x )\) = .
Hint.
Use the quotient rule to find the first derivative. Simplify your numerator.
To find the second derivative use the chain rule or quotient rule.

9. Computing velocity and acceleration.

The function \(s(t)\) describes the position of a particle moving along a coordinate line, where \(s\) is in feet and \(t\) is in seconds.
\begin{equation*} s(t) = t^{4}-98t^{2}+2401, \qquad t \ge 0 \end{equation*}
If appropriate, enter answers in radical form. Use inf to represent \(\small{\infty}\text{.}\)
(a) Find the velocity and acceleration functions.
\(\qquad\qquad\quad\;\,v(t)\)=
\(\qquad\qquad\quad\;\,a(t)\)=
(b) Find the position, velocity, speed, and acceleration at \(t = 3\text{.}\)
Position (ft):
Velocity (ft/sec):
Acceleration (ft/sec\(\small{^2}\)):
Hint.
Recall that velocity is the derivative of position.
Acceleration is the second derivative of position or the derivative of velocity.

10. Interpretting a graph of \(f\) based on the first and second derivatives.

Suppose that \(y = f(x)\) is a differentiable function for which the following information is known: \(f(2) = -3\text{,}\) \(f'(2) = 1.5\text{,}\) \(f''(2) = -0.25\text{.}\)
  1. Is \(f\) increasing or decreasing at \(x = 2\text{?}\) Is \(f\) concave up or concave down at \(x = 2\text{?}\)
  2. Do you expect \(f(2.1)\) to be greater than \(-3\text{,}\) equal to \(-3\text{,}\) or less than \(-3\text{?}\) Why?
  3. Do you expect \(f'(2.1)\) to be greater than \(1.5\text{,}\) equal to \(1.5\text{,}\) or less than \(1.5\text{?}\) Why?
  4. Sketch a graph of \(y = f(x)\) near \((2,f(2))\) and include a graph of the tangent line.

11. Interpreting a graph of \(f'\).

For a certain function \(y = g(x)\text{,}\) its derivative is given by the function pictured in Figure 2.8.8.
Figure 2.8.8. The graph of \(y = g'(x)\text{.}\)
  1. What is the approximate slope of the tangent line to \(y = g(x)\) at the point \((2,g(2))\text{?}\)
  2. How many real number solutions can there be to the equation \(g(x) = 0\text{?}\) Justify your conclusion fully and carefully by explaining what you know about how the graph of \(g\) must behave based on the given graph of \(g'\text{.}\)
  3. On the interval \(-3 \lt x \lt 3\text{,}\) how many times does the concavity of \(g\) change? Why?
  4. Use the provided graph to estimate the value of \(g''(2)\text{.}\)

12. Using data to interpret derivatives.

A bungee jumper’s height \(h\) (in feet ) at time \(t\) (in seconds) is given in part by the table:
\(t\) \(0.0\) \(0.5\) \(1.0\) \(1.5\) \(2.0\) \(2.5\) \(3.0\) \(3.5\) \(4.0\) \(4.5\) \(5.0\)
\(h(t)\) \(200\) \(184.2\) \(159.9\) \(131.9\) \(104.7\) \(81.8\) \(65.5\) \(56.8\) \(55.5\) \(60.4\) \(69.8\)
\(t\) \(5.5\) \(6.0\) \(6.5\) \(7.0\) \(7.5\) \(8.0\) \(8.5\) \(9.0\) \(9.5\) \(10.0\)
\(h(t)\) \(81.6\) \(93.7\) \(104.4\) \(112.6\) \(117.7\) \(119.4\) \(118.2\) \(114.8\) \(110.0\) \(104.7\)
  1. Use the given data to estimate \(h'(4.5)\text{,}\) \(h'(5)\text{,}\) and \(h'(5.5)\text{.}\) At which of these times is the bungee jumper rising most rapidly?
  2. Use the given data and your work in (a) to estimate \(h''(5)\text{.}\)
  3. What physical property of the bungee jumper does the value of \(h''(5)\) measure? What are its units?
  4. Based on the data, on what approximate time intervals is the function \(y = h(t)\) concave down? What is happening to the velocity of the bungee jumper on these time intervals?

13. Sketching functions.

For each prompt that follows, sketch a possible graph of a function on the interval \(-3 \lt x \lt 3\) that satisfies the stated properties.
  1. \(y = f(x)\) such that \(f\) is increasing on \(-3 \lt x \lt 3\text{,}\) concave up on \(-3 \lt x \lt 0\text{,}\) and concave down on \(0 \lt x \lt 3\text{.}\)
  2. \(y = g(x)\) such that \(g\) is increasing on \(-3 \lt x \lt 3\text{,}\) concave down on \(-3 \lt x \lt 0\text{,}\) and concave up on \(0 \lt x \lt 3\text{.}\)
  3. \(y = h(x)\) such that \(h\) is decreasing on \(-3 \lt x \lt 3\text{,}\) concave up on \(-3 \lt x \lt -1\text{,}\) neither concave up nor concave down on \(-1 \lt x \lt 1\text{,}\) and concave down on \(1 \lt x \lt 3\text{.}\)
  4. \(y = p(x)\) such that \(p\) is decreasing and concave down on \(-3 \lt x \lt 0\) and is increasing and concave down on \(0 \lt x \lt 3\text{.}\)
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