
Section2.3The Product and Quotient Rules

Motivating Questions
• How does the algebraic structure of a function guide us in computing its derivative using shortcut rules?

• How do we compute the derivative of a product of two basic functions in terms of the derivatives of the basic functions?

• How do we compute the derivative of a quotient of two basic functions in terms of the derivatives of the basic functions?

• How do the product and quotient rules combine with the sum and constant multiple rules to expand the library of functions we can differentiate quickly?

So far, we can differentiate power functions $x^n\text{,}$ exponential functions $a^x\text{,}$ and the two fundamental trigonometric functions $\sin(x)$ and $\cos(x)\text{.}$ With the sum rule and constant multiple rules, we can also compute the derivative of combined functions.

Example2.34

Differentiate

\begin{equation*} f(x) = 7x^{11} - 4 \cdot 9^x + \pi \sin(x) - \sqrt{3}\cos(x)\text{.} \end{equation*}
Hint

Use a combination of all the differentiation rules we've learned in this chapter.

$f'(x) = 77x^{10} - 4 \cdot 9^x \ln(9) + \pi \cos(x) + \sqrt{3} \sin(x)\text{.}$

Solution

Because $f$ is a sum of basic functions, we can now quickly say that $f'(x)=77x^{10}-4\cdot9^x\ln(9)+\pi\cos(x)+\sqrt{3}\sin(x)\text{.}$

What about a product or quotient of two basic functions, such as $p(z) = z^3 \cos(z) \text{,}$ or $q(t) = \frac{\sin(t)}{2^t} \text{?}$

While the derivative of a sum is the sum of the derivatives, it turns out that the rules for computing derivatives of products and quotients are more complicated.

Example2.35

Let $f$ and $g$ be the functions defined by $f(t) = 2t^2$ and $g(t) = t^3 + 4t\text{.}$

1. Determine $f'(t)$ and $g'(t)\text{.}$

2. Let $p(t) = 2t^2 (t^3 + 4t)$ and observe that $p(t) = f(t) \cdot g(t)\text{.}$ Rewrite the formula for $p$ by distributing the $2t^2$ term. Then compute $p'(t)$ using the sum and constant multiple rules.

3. True or false: $p'(t) = f'(t) \cdot g'(t)\text{.}$

4. Let $q(t) = \frac{t^3 + 4t}{2t^2}$ and observe that $q(t) = \frac{g(t)}{f(t)}\text{.}$ Rewrite the formula for $q$ by dividing each term in the numerator by the denominator and simplify to write $q$ as a sum of constant multiples of powers of $t\text{.}$ Then compute $q'(t)$ using the sum and constant multiple rules.

5. True or false: $q'(t) = \frac{g'(t)}{f'(t)}\text{.}$

Hint
1. Use a combination of the power rule, the constant multiple rule, and the sum rule.

2. $p(t)$ is a polynomial.

3. Multiply your answers from (a) together; does the result match what you found in (b)?

4. Start by splitting the fraction into two pieces, then simplify.

5. Divide your answers from (a) in the correct order; does this match what you found in (d)?

1. $f'(t)=4t$ and $g'(t)=3t^2+4\text{.}$

2. Since $p(t)=2t^5+8t^3\text{,}$ then $p'(t)=10t^4+24t^2\text{.}$

3. False, $p'(t)\neq f'(t)\cdot g'(t)\text{.}$

4. Since $q(t)=\frac12t+2t^{-1}\text{,}$ then $q'(t)=\frac12-2t^{-2}\text{.}$

5. False, $q'(t)\neq\frac{g'(t)}{f'(t)}\text{.}$

Solution
1. Using the power rule and the constant multiple rule, we see that

\begin{equation*} f'(t)=\frac{d}{dt}[2t^2]=2\frac{d}{dt}[t^2]=2(2t)=4t\text{.} \end{equation*}

Using the same rules together with the sum rule, we see that

\begin{equation*} g'(t)=\frac{d}{dt}[t^3+4t]=\frac{d}{dt}[t^3]+4\frac{d}{dt}[t]=(3t^2)+4(1)=3t^2+4\text{.} \end{equation*}
2. Observe that $p(t)=2t^2(t^3+4t)=2t^5+8t^3\text{.}$ Since $\frac{d}{dt}[t^5]=5t^4$ and $\frac{d}{dt}[t^3]=3t^2\text{,}$ we can use the sum and constant multiple rules to say that the derivative of $p$ with respect to $t$ is

\begin{equation*} p'(t)=\frac{d}{dt}[2t^5+8t^3]=2(5t^4)+8(3t^2)=10t^4+24t^2\text{.} \end{equation*}
3. From our answers in (a), we know

\begin{equation*} f'(t)\cdot g'(t)=(4t)\cdot(3t^2+4)=12t^3+16t\text{.} \end{equation*}

This does not match what we found in (b) as the formula for $p'(t)=10t^4+24t^2\text{.}$ We conclude that even though $p(t)=f(t)\cdot g(t)\text{,}$ it is not the case that $p'(t)=f'(t)\cdot g'(t)\text{.}$ In other words, the derivative of a product of two functions need not be the product of the derivatives.

4. To put $q(t)$ into a form we can differentiate, notice that

\begin{equation*} q(t)=\frac{t^3+4t}{2t^2}=\frac{t^3}{2t^2}+\frac{4t}{2t^2}=\frac12t+2t^{-1}\text{.} \end{equation*}

Now since $\frac{d}{dt}[t]=1$ and $\frac{d}{dt}[t^{-1}]=-t^{-2}\text{,}$ we use the sum and constant multiple rules to find

\begin{equation*} q'(t)=\frac12(1)+2(-t^{-2})=\frac12-2t^{-2}\text{.} \end{equation*}
5. From our answers in (a), we know

\begin{equation*} \frac{g'(t)}{f'(t)}=\frac{3t^2+4}{4t}=\frac34t+t^{-1}\text{.} \end{equation*}

This does not match what we found in (d) as the formula for $q'(t)=\frac12-2t^{-2}\text{.}$ We conclude that even though $q(t)=\frac{g(t)}{f(t)}\text{,}$ it is not the case that $q'(t)=\frac{g'(t)}{f'(t)}\text{.}$ In other words, the derivative of a quotient of two functions need not be the quotient of the derivatives.

SubsectionThe Product Rule

As part (b) of Example2.35 shows, it is not true in general that the derivative of a product of two functions is the product of the derivatives of those functions. To see why this is the case, we consider a situation involving functions with physical context.

Example2.36

Say that an investor is regularly purchasing stock in a particular company. Let $N(t)$ represent the number of shares owned on day $t\text{,}$ where $t = 0$ represents the first day on which shares were purchased. Let $S(t)$ give the value of one share of the stock on day $t\text{;}$ note that the units on $S(t)$ are dollars per share. To compute the total value of the stock on day $t\text{,}$ we take the product

\begin{equation*} V(t) = N(t) \, \text{shares} \cdot S(t) \, \text{dollars per share}\text{,} \end{equation*}

Observe that over time, both the number of shares and the value of a given share will vary. The derivative $N'(t)$ measures the rate at which the number of shares is changing, while $S'(t)$ measures the rate at which the value per share is changing. How do these respective rates of change affect the rate of change of the total value function?

To help us understand the relationship among changes in $N\text{,}$ $S\text{,}$ and $V\text{,}$ let's consider some specific data.

• Suppose that on day 100, the investor owns 520 shares of stock and the stock's current value is $27.50 per share. This tells us that $N(100) = 520$ and $S(100) = 27.50\text{.}$ • On day 100, the investor purchases an additional 12 shares (so the number of shares held is rising at a rate of 12 shares per day). • On that same day the price of the stock is rising at a rate of 0.75 dollars per share per day. In calculus notation, the latter two facts tell us that $N'(100) = 12$ (shares per day) and $S'(100) = 0.75$ (dollars per share per day). At what rate is the value of the investor's total holdings changing on day 100? Observe that the increase in total value comes from two sources: the growing number of shares and the rising value of each share. • If only the number of shares is increasing (and the value of each share is constant), the rate at which the total value would rise is the product of the rate at which the number of shares is changing and the current value of the shares. That is, the rate at which the total value would change when the share value is constant is given by \begin{equation*} N'(100) \cdot S(100) = 12 \, \frac{\text{shares}}{\text{day}} \cdot 27.50 \, \frac{\text{dollars}}{\text{share}} = 330 \, \frac{\text{dollars} }{\text{day} }\text{.} \end{equation*} Note particularly how the units make sense and show the rate at which the total value $V$ is changing, measured in dollars per day. • If instead the number of shares is constant, but the value of each share is rising, the rate at which the total value would rise is the product of the number of shares and the rate of change of share value. In this case, the total value is rising at a rate of \begin{equation*} N(100) \cdot S'(100) = 520 \, \text{shares} \cdot 0.75 \, \frac{\text{dollars per share} }{\text{day} } = 390 \, \frac{\text{dollars} }{\text{day} }\text{.} \end{equation*} • Of course, when both the number of shares and the value of each share are changing, we have to include both of these sources. In that case the rate at which the total value is rising is \begin{equation*} V'(100) = N'(100) \cdot S(100) + N(100) \cdot S'(100) = 330 + 390 = 720 \, \frac{\text{dollars} }{\text{day} }\text{.} \end{equation*} We expect the total value of the investor's holdings to rise by about$720 on the 100th day.5While this example highlights why the product rule is true, there are some subtle issues to recognize. For one, if the stock's value really does rise exactly $0.75 on day 100, and the number of shares really rises by 12 on day 100, then we would expect that $V(101) = N(101) \cdot S(101) = 532 \cdot 28.25 = 15,029\text{.}$ If, as noted above, we expect the total value to rise by$720, then with $V(100) = N(100) \cdot S(100) = 520 \cdot 27.50 = 14,300\text{,}$ then it seems we should find that $V(101) = V(100) + 720 = 15,020\text{.}$ Why do the two results differ by 9? One way to understand why this difference occurs is to recognize that $N'(100) = 12$ represents an instantaneous rate of change, while our (informal) discussion has also thought of this number as the total change in the number of shares over the course of a single day. The formal proof of the product rule reconciles this issue by taking the limit as the change in the input tends to zero.

Next, we expand our perspective from the specific example above to the more general and abstract setting of a product $P$ of two differentiable functions, $f$ and $g\text{.}$ If $P(x) = f(x) \cdot g(x)\text{,}$ our work above suggests that $P'(x) = f'(x) g(x) + f(x) g'(x)\text{.}$ Indeed, a formal proof using the limit definition of the derivative can be given to show that the following rule, called the product rule, holds in general.

Product Rule

If $f$ and $g$ are differentiable functions, then their product $P(x) = f(x) \cdot g(x)$ is also a differentiable function, and

\begin{equation*} P'(x) = f'(x) g(x) + f(x) g'(x)\text{.} \end{equation*}

In light of Example2.36 involving shares of stock, the product rule also makes sense intuitively: the rate of change of $P$ should take into account both how fast $f$ and $g$ are changing, as well as how large $f$ and $g$ are at the point of interest. In words, the product rule says: if $P$ is the product of two functions $f$ (the first function) and $g$ (the second), then the derivative of $P$ is the derivative of the first times the second, plus the first times the derivative of the second. It is often a helpful mental exercise to say this phrasing aloud when executing the product rule.

Example2.37

If $P(z) = z^3 \cdot \cos(z)\text{,}$ we can use the product rule to differentiate $P\text{.}$ The first function is $z^3$ and the second function is $\cos(z)\text{.}$ By the product rule, $P'$ will be given by the derivative of the first, $3z^2\text{,}$ times the second, $\cos(z)\text{,}$ plus the first, $z^3\text{,}$ times the derivative of the second, $-\sin(z)\text{.}$ That is,

\begin{equation*} P'(z) = 3z^2\cos(z)+z^3(-\sin(z)) = 3z^2 \cos(z)-z^3\sin(z)\text{.} \end{equation*}
Example2.38

Use the product rule to answer each of the questions below. Throughout, be sure to carefully label any derivative you find by name. It is not necessary to algebraically simplify any of the derivatives you compute.

1. Let $m(w)=3w^{17} 4^w\text{.}$ Find $m'(w)\text{.}$

2. Let $h(t) = (\sin(t) + \cos(t))t^4\text{.}$ Find $h'(t)\text{.}$

3. Determine the slope of the tangent line to the curve $y = f(x)$ at the point where $a = 1$ if $f$ is given by the rule $f(x) = e^x \sin(x)\text{.}$

4. Find an equation for the tangent line $L$ to the graph of $y = g(x)$ at the point where $a = -1$ if $g$ is given by the rule $g(x) = (x^2 + x) 2^x\text{.}$

Hint
1. Let the first function be $3w^{17}\text{.}$

2. Let the first function be $(\sin(t) + \cos(t))\text{.}$

3. Remember that the slope of the tangent line to $y = f(x)$ at $(a,f(a))$ is given by $f'(a)\text{.}$

4. Recall that the tangent line to $y=g(x)$ at $(a,g(a))$ is given by the equation $y-g(a)=g'(a)(x-a)\text{.}$ To denote this line by a function $L\text{,}$ we can write $L(x)=g'(a)(x-a)+g(a)\text{.}$

1. $m'(w) = 51w^{16} \cdot 4^w + 3w^{17}\cdot 4^w \ln(4)\text{.}$

2. $h'(t) = (\cos(t) - \sin(t)) \cdot t^4 + (\sin(t) + \cos(t))\cdot 4t^3\text{.}$

3. $f'(1) = e(\sin(1) + \cos(1)) \approx 3.756\text{.}$

4. $L(x) = -\frac{1}{2}(x+1)\text{.}$

Solution
1. By the product rule,

\begin{equation*} m'(w) = 51w^{16}\cdot 4^2+3w^{17}\cdot 4^2\ln(4)\text{.} \end{equation*}
2. By the product rule,

\begin{equation*} h'(t) = (\cos(t) - \sin(t)) \cdot t^4 + (\sin(t) + \cos(t))\cdot 4t^3\text{.} \end{equation*}
3. To determine the slope of the tangent line at $a = 1\text{,}$ we want to find $f'(1)\text{.}$ Since $f(x) = e^x \sin(x)\text{,}$ the product rule tells us that $f'(x) = e^x \cdot \sin(x) + e^x \cdot \cos(x)\text{.}$ Thus, $f'(1) = e^1 \cdot \sin(1) + e^1\cdot \cos(1) = e(\sin(1) + \cos(1)) \approx 3.756\text{.}$

4. First, observe that $g(-1) = ((-1)^2 - 1) \cdot 2^{-1} = 0\text{,}$ so we are looking for the tangent line to $g$ at the point $(-1,0)\text{.}$ Furthermore, by the product rule, $g'(x) = (2x + 1) \cdot 2^x + (x^2+x)\cdot 2^x\ln(2)\text{,}$ and hence

\begin{align*} g'(-1) \amp= (2(-1)+ 1) \cdot 2^{-1} + ((-1)^2 - 1)\cdot 2^{-1}\ln(2) \\ \amp= (-1)\frac{1}{2} +0 = -\frac{1}{2}\text{.} \end{align*}

Therefore, $L(x)-0=-\frac{1}{2}(x-(-1))\text{,}$ or

\begin{equation*} L(x) = -\frac{1}{2}(x+1)\text{.} \end{equation*}

SubsectionThe Quotient Rule

Because quotients and products are closely linked, we can use the product rule to understand how to take the derivative of a quotient. Let $Q(x)$ be defined by $Q(x) = \frac{f(x)}{g(x)}\text{,}$ where $f$ and $g$ are both differentiable functions. It turns out that $Q$ is differentiable everywhere that $g(x) \ne 0\text{.}$ We would like a formula for $Q'$ in terms of $f\text{,}$ $g\text{,}$ $f'\text{,}$ and $g'\text{.}$ Multiplying both sides of the equation $Q = \frac{f}{g}$ by $g\text{,}$ we observe that

\begin{equation*} f(x) = Q(x) \cdot g(x)\text{.} \end{equation*}

Now we can use the product rule to differentiate $f\text{:}$

\begin{equation*} f'(x) = Q'(x) g(x) + Q(x)g'(x)\text{.} \end{equation*}

We want to know a formula for $Q'\text{,}$ so we solve this equation for $Q'(x)\text{.}$ Then

\begin{equation*} Q'(x) g(x) = f'(x) - Q(x) g'(x)\text{,} \end{equation*}

and dividing both sides by $g(x)\text{,}$ we have

\begin{equation*} Q'(x) = \frac{f'(x) - Q(x) g'(x)}{g(x)}\text{.} \end{equation*}

Finally, we recall that $Q(x) = \frac{f(x)}{g(x)}\text{.}$ Substituting this expression in the preceding equation, we have

\begin{align*} Q'(x) =\mathstrut \amp \frac{f'(x) - \frac{f(x)}{g(x)} g'(x)}{g(x)}\\ =\mathstrut \amp \frac{f'(x) - \frac{f(x)}{g(x)} g'(x)}{g(x)} \cdot \frac{g(x)}{g(x)}\\ =\mathstrut \amp \frac{f'(x)g(x) - f(x) g'(x)}{[g(x)]^2}\text{.} \end{align*}

This calculation gives us the quotient rule.

Quotient Rule

If $f$ and $g$ are differentiable functions, then their quotient $Q(x) = \frac{f(x)}{g(x)}$ is also a differentiable function for all $x$ where $g(x) \ne 0\text{,}$ and

\begin{equation*} Q'(x) = \frac{f'(x)g(x) - f(x) g'(x)}{[g(x)]^2}\text{.} \end{equation*}

As with the product rule, it can be helpful to think of the quotient rule verbally. If a function $Q$ is the quotient of a top function $f$ and a bottom function $g\text{,}$ then $Q'$ is given by the derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared.6If you have studied differentiation before, you may have seen this rule as $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}\text{,}$ with the mnemonic "low-d-high minus high-d-low all over low squared." If this version is more comfortable for you, you are welcome to use it instead.

Example2.39

If $Q(t) = \frac{\sin(t)}{2^t}\text{,}$ we call $\sin(t)$ the top function and $2^t$ the bottom function. By the quotient rule, $Q'$ is given by the derivative of the top, $\cos(t)\text{,}$ times the bottom, $2^t\text{,}$ minus the top, $\sin(t)\text{,}$ times the derivative of the bottom, $2^t \ln(2)\text{,}$ all over the bottom squared, $(2^t)^2\text{.}$ That is,

\begin{equation*} Q'(t) = \frac{\cos(t)2^t - \sin(t) 2^t \ln(2)}{(2^t)^2}\text{.} \end{equation*}

In this particular example, it is possible to simplify $Q'(t)$ by removing a factor of $2^t$ from both the numerator and denominator, so that

\begin{equation*} Q'(t) = \frac{\cos(t) - \sin(t) \ln(2)}{2^t}\text{.} \end{equation*}

In general, we must be careful in doing any such simplification, as we don't want to follow a correct execution of the quotient rule with an algebraic error.

Example2.40

Use the quotient rule to answer each of the questions below. Throughout, be sure to carefully label any derivative you find by name. That is, if you're given a formula for $f(x)\text{,}$ clearly label the formula you find for $f'(x)\text{.}$ It is not necessary to algebraically simplify any of the derivatives you compute.

1. Let $r(z)=\frac{3^z}{z^4 + 1}\text{.}$ Find $r'(z)\text{.}$

2. Let $v(t) = \frac{\sin(t)}{\cos(t) + t^2}\text{.}$ Find $v'(t)\text{.}$

3. Determine the slope of the tangent line to the curve $y=R(x)$ at the point where $x=0\text{,}$ if $\displaystyle R(x) = \frac{x^2 - 2x - 8}{x^2 - 9}\text{.}$

4. When a camera flashes, the intensity $I$ of light seen by the eye $t$ milliseconds after the initial flash point is given by the function

\begin{equation*} I(t) = \frac{100t}{e^t}\text{,} \end{equation*}

where $I$ is measured in candles. Compute $I'(0.5)\text{,}$ $I'(2)\text{,}$ and $I'(5)\text{;}$ include appropriate units on each value; and discuss the meaning of each.

Hint
1. When applying the quotient rule, use parentheses around the bottom function, $z^4 + 1\text{,}$ to ensure that when you compute the derivative of the top times the bottom that the rule is applied correctly.

2. When applying the quotient rule, use parentheses around the bottom function, $\cos(t) + t^2\text{,}$ and its derivative to ensure that the rule is applied correctly.

3. Remember one of the key interpretations of the derivative.

4. Let the top function be $100t$ and simply use the constant multiple rule to find its derivative.

1. $r'(z)=\frac{3^z \ln(3)(z^4+1) - 3^z(4z^3)}{(z^4 + 1)^2}\text{.}$

2. $v'(t) = \frac{\cos(t)(\cos(t) + t^2) - \sin(t)(\sin(t) + 2t)}{(\cos(t) + t^2)^2}\text{.}$

3. $R'(0) = \frac{2}{9}\text{.}$

4. $I'(0.5) = \frac{50}{e^{0.5}} \approx 30.327\text{,}$ $I'(2) = \frac{-100}{e^{2}} \approx -13.534\text{,}$ and $I'(5) = \frac{-400}{e^4} \approx -2.695\text{,}$ each in candles per millisecond.

Solution
1. By the quotient rule,

\begin{equation*} r'(z)=\frac{3^z \ln(3) (z^4+1) - 3^z(4z^3)}{(z^4 + 1)^2}\text{.} \end{equation*}
2. By the quotient rule,

\begin{equation*} v'(t) = \frac{\cos(t)(\cos(t) + t^2) - \sin(t)(\sin(t) + 2t)}{(\cos(t) + t^2)^2}\text{.} \end{equation*}
3. We first compute $R'(x)\text{.}$ By the quotient rule,

\begin{equation*} R'(x) = \frac{(2x - 2)(x^2 - 9) - (x^2 - 2x - 8)(2x)}{(x^2 - 9)^2}\text{.} \end{equation*}

From this, it follows that $R'(0) = \frac{(-2)(-9)-(-8)(0)}{(-9)^2} = \frac{2}{9}\text{,}$ which is the slope of the tangent line to the curve at the point where $x = 0\text{.}$

4. By the quotient rule and algebraic simplification,

\begin{equation*} I'(t) = \frac{100e^t - 100te^t}{(e^t)^2} = \frac{100-100t}{e^t}\text{.} \end{equation*}

Thus, $I'(0.5) = \frac{50}{e^{0.5}} \approx 30.327\text{,}$ $I'(2) = \frac{-100}{e^{2}} \approx -13.534\text{,}$ and $I'(5) = \frac{-400}{e^4} \approx -2.695\text{,}$ each measured in candles per millisecond. These results show that at $t = 0.5\text{,}$ the intensity of the flash is increasing rapidly, while at $t = 2$ and $t = 5\text{,}$ the intensity is decreasing, with the intensity decreasing more rapidly when $t = 2\text{.}$

SubsectionThe Product and Quotient Rule Using Tables and Graphs

In addition to being used to finding the derivatives of functions given by equations, the product and quotient rule can be useful for finding the derivatives of functions given by tables and graphs. The following examples illustrate this process.

Example2.41

Suppose the functions $f$ and $g$ are described by Table2.42 below, and the function $h$ is given by the formula $h(x)=f(x)g(x)\text{.}$ Then we can calculate $h'(2)=f'(2)g(2)+f(2)g'(2)=4 \times 5 + 2 \times 6=32\text{.}$

Example2.43

Suppose the function $f$ is described by Table2.44 below, the function $g$ is described by the graph in Figure2.45, and the function $h$ is given by the formula $h(x)=\frac{f(x)}{g(x)}\text{.}$ Then for certain values of $x\text{,}$ we can calculate

\begin{equation*} h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}\text{.} \end{equation*}

Suppose we want to find $h'(0)\text{.}$ Table2.44 tells us that $f(0)=4$ and $f'(0)=5\text{.}$ From visual inspection of Figure2.45, we also notice that $g(0)=2$ and $g'(0)=1\text{.}$ Therefore

\begin{equation*} h'(0)=\frac{f'(0)g(0)-f(0)g'(0)}{(g(0))^2}=\frac{5\times 2-4 \times 1}{(2)^2}=\frac{6}{4}=\frac{3}{2}. \end{equation*}
 $x$ -2 -1 0 1 2 $f(x)$ 3 2 4 3 2 $f'(x)$ 1 2 5 2 4

SubsectionCombining Rules

In order to apply the derivative shortcut rules correctly we must recognize the fundamental structure of a function.

Example2.46

Determine the derivative of the function

\begin{equation*} f(x) = x\sin(x) + \frac{x^2}{\cos(x) + 2}\text{.} \end{equation*}

How do we decide which rules to apply? Our first task is to recognize the structure of the function. This function $f$ is a sum of two slightly less complicated functions, so we can apply the sum rule7When taking a derivative that involves the use of multiple derivative rules, it is often helpful to use the notation $\frac{d}{dx} \left[ ~~\right]$ to wait to apply subsequent rules. This is demonstrated in both Example2.46 and Example2.47, and has come up previously in solutions to various examples in this chapter. to get

\begin{align*} f'(x) =\mathstrut \amp \frac{d}{dx} \left[ x\sin(x) + \frac{x^2}{\cos(x) + 2} \right]\\ =\mathstrut \amp \frac{d}{dx} \left[ x\sin(x) \right] + \frac{d}{dx}\left[ \frac{x^2}{\cos(x) + 2} \right]\text{.} \end{align*}

Now, the left-hand term above is a product, so the product rule is needed there, while the right-hand term is a quotient, so the quotient rule is required. Applying these rules respectively, we find that

\begin{align*} f'(x) =\mathstrut \amp \left( \sin(x)+x \cos(x) \right) + \frac{ 2x(\cos(x) + 2) - x^2(-\sin(x))}{(\cos(x) + 2)^2}\\ =\mathstrut \amp \sin(x) + x \cos(x) + \frac{2x\cos(x) + 4x + x^2\sin(x)}{(\cos(x) + 2)^2}\text{.} \end{align*}
Example2.47

Differentiate

\begin{equation*} s(y) = \frac{y \cdot 7^y}{y^2 + 1}\text{.} \end{equation*}

The function $s$ is a quotient of two simpler functions, so the quotient rule will be needed. To begin, we set up the quotient rule and use the notation $\frac{d}{dy}$ to indicate the derivatives of the numerator and denominator. Thus,

\begin{equation*} s'(y) = \frac{\frac{d}{dy}\left[ y \cdot 7^y \right]\cdot(y^2 + 1) - y \cdot 7^y \cdot \frac{d}{dy}\left[y^2 + 1 \right]}{(y^2 + 1)^2}\text{.} \end{equation*}

Now, there remain two derivatives to calculate. The first one, $\frac{d}{dy}\left[ y \cdot 7^y \right]$ calls for use of the product rule, while the second, $\frac{d}{dy}\left[y^2 + 1 \right]$ needs only the sum rule. Applying these rules, we now have

\begin{equation*} s'(y) = \frac{[7^y+y \cdot 7^y \ln(7)](y^2 + 1) - y \cdot 7^y [2y]}{(y^2 + 1)^2}\text{.} \end{equation*}

While some simplification is possible, we are content to leave $s'(y)$ in its current form.

Success in applying derivative rules begins with recognizing the structure of the function, followed by the careful and diligent application of the relevant derivative rules. The best way to become proficient at this process is through continuous practice!

Example2.48

Use relevant derivative rules to answer each of the questions below. Throughout, be sure to use proper notation and carefully label any derivative you find by name.

1. Let $f(r) = (5r^3 + \sin(r))(4^r - 2\cos(r))\text{.}$ Find $f'(r)\text{.}$

2. Let $\displaystyle p(t) = \frac{\cos(t)}{t^6 \cdot 6^t}\text{.}$ Find $p'(t)\text{.}$

3. Let $g(z) = 3z^7 e^z - 2z^2 \sin(z) + \frac{z}{z^2 + 1}\text{.}$ Find $g'(z)\text{.}$

4. A moving particle has its position in feet at time $t$ in seconds given by the function $s(t) = \frac{3\cos(t) - \sin(t)}{e^t}\text{.}$ Find the particle's instantaneous velocity at the moment $t = 1\text{.}$

5. Suppose that $f(x)$ and $g(x)$ are differentiable functions and it is known that $f(3) = -2\text{,}$ $f'(3) = 7\text{,}$ $g(3) = 4\text{,}$ and $g'(3) = -1\text{.}$ If $p(x) = f(x) \cdot g(x)$ and $\displaystyle q(x) = \frac{f(x)}{g(x)}\text{,}$ calculate $p'(3)$ and $q'(3)\text{.}$

Hint
1. Observe that $f$ is fundamentally a product. Which is the first function? The second?

2. Note that $p$ has the overall structure of a quotient.

3. Think about how $g$ is a sum of three functions. What is the structure of each of the three functions in the sum?

4. How is the velocity of a moving object related to its position?

5. Since we know $p(x) = f(x) \cdot g(x)\text{,}$ it follows $p'(x) = f'(x)g(x)+f(x) g'(x)$ for any value of $x\text{.}$

1. $f'(r) = [15r^2 + \cos(r)](4^r - 2\cos(r))+(5r^3 + \sin(r))[4^r \ln(4) + 2\sin(r)]\text{.}$

2. $p'(t) = \frac{[-\sin(t)]t^6 \cdot 6^t - \cos(t) [6t^5\cdot 6^t + t^6 \cdot 6^t \ln(6)] }{(t^6 \cdot 6^t)^2}\text{.}$

3. $g'(z) = 3 [7z^6e^z+z^7 e^z] - 2[2z\sin(z)+z^2 \cos(z)] + \frac{(z^2+1) - z(2z)}{(z^2 + 1)^2}\text{.}$

4. $s'(1) = \frac{-2\sin(1)-4\cos(1)}{e} \approx -1.414$ feet per second.

5. $p'(3) = 30$ and $q'(3) = \frac{13}{8}\text{.}$

Solution
1. Using the product rule, followed by the sum and constant multiple rule, observe that

\begin{align*} f'(r) =\mathstrut \amp \frac{d}{dr}[5r^3 + \sin(r)](4^r - 2\cos(r))+(5r^3 + \sin(r))\frac{d}{dr}[4^r - 2\cos(r)] \\ =\mathstrut \amp [15r^2 + \cos(r)](4^r - 2\cos(r))+(5r^3 + \sin(r))[4^r \ln(4) + 2\sin(r)]\text{.} \end{align*}
2. We use the quotient rule on $p\text{,}$ followed by the product rule to differentiate the denominator, finding that

\begin{align*} p'(t) =\mathstrut \amp \frac{\frac{d}{dt}[\cos(t)]t^6 \cdot 6^t - \cos(t) \frac{d}{dt}[t^6 \cdot 6^t]}{(t^6 \cdot 6^t)^2}\\ =\mathstrut \amp \frac{[-\sin(t)]t^6 \cdot 6^t - \cos(t) [6t^5 \cdot 6^t + t^6\cdot 6^t \ln(6)]}{(t^6 \cdot 6^t)^2}\text{.} \end{align*}
3. Using the sum and constant multiple rules, it follows first that

\begin{equation*} g'(z) = 3 \frac{d}{dz}[z^7 e^z] - 2\frac{d}{dz}[z^2 \sin(z)] + \frac{d}{dz}\left[ \frac{z}{z^2 + 1} \right]\text{.} \end{equation*}

Applying the product rule in the first two terms and the quotient rule in the third, we find that

\begin{equation*} g'(z) = 3 [7z^6e^z+z^7 e^z] - 2[2z\sin(z)+z^2 \cos(z)] + \frac{(z^2+1) - z(2z)}{(z^2 + 1)^2}\text{.} \end{equation*}
4. The particle's instantaneous velocity at the moment $t = 1$ is given by $s'(1)\text{.}$ We use the quotient rule to find $s'(t)\text{,}$ and then simplify by removing a common factor of $e^t$ to get

\begin{align*} s'(t) =\mathstrut \amp \frac{(-3\sin(t) - \cos(t))e^t-(3\cos(t) - \sin(t))e^t}{(e^t)^2}\\ =\mathstrut \amp \frac{-2\sin(t)-4\cos(t)}{e^t}\text{.} \end{align*}

Thus, $s'(1) = \frac{-2\sin(1)-4\cos(1)}{e^1} \approx -1.414\text{,}$ which is the particle's instantaneous velocity in feet per second at the moment $t = 1\text{.}$

5. Since $p(x) = f(x) \cdot g(x)\text{,}$ the product rule tells us

\begin{equation*} p'(x) = f'(x)g(x)+f(x)g'(x)\text{,} \end{equation*}

and since $\displaystyle q(x) = \frac{f(x)}{g(x)}\text{,}$ by the quotient rule we know

\begin{equation*} q'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}\text{.} \end{equation*}

Using the given information ($f(3) = -2\text{,}$ $f'(3) = 7\text{,}$ $g(3) = 4\text{,}$ and $g'(3) = -1$) we now see that

\begin{equation*} p'(3) = f'(3)g(3)+f(3)g'(3) = (7)(4)+(-2)(-1) = 30 \end{equation*}

and

\begin{equation*} q'(3) = \frac{f'(3)g(3)-f(3)g'(3)}{g(3)^2} = \frac{(7)(4) - (-2)(-1)}{4^2} = \frac{13}{8}\text{.} \end{equation*}

As the algebraic complexity of the functions we are able to differentiate continues to increase, it is important to remember that the meaning of the derivative remains the same. Regardless of the structure of the function $f\text{,}$ the value of $f'(a)$ tells us the instantaneous rate of change of $f$ with respect to $x$ at the moment $x = a\text{,}$ as well as the slope of the tangent line to $y = f(x)$ at the point $(a,f(a))\text{.}$

SubsectionSummary

• If a function is a sum, product, or quotient of simpler functions, then we can use the sum, product, or quotient rules to differentiate it in terms of the simpler functions and their derivatives.

• The product rule tells us that if $P$ is a product of differentiable functions $f$ and $g$ according to the rule $P(x) = f(x) g(x)\text{,}$ then

\begin{equation*} P'(x) = f'(x)g(x)+f(x)g'(x)\text{.} \end{equation*}
• The quotient rule tells us that if $Q$ is a quotient of differentiable functions $f$ and $g$ according to the rule $Q(x) = \frac{f(x)}{g(x)}\text{,}$ then whenever $g(x)\neq0\text{,}$

\begin{equation*} Q'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\text{.} \end{equation*}
• Along with the constant multiple and sum rules, the product and quotient rules enable us to compute the derivative of any function that consists of sums, constant multiples, products, and quotients of basic functions. For instance, if $F$ has the form

\begin{equation*} F(x) = \frac{2a(x) - 5b(x)}{c(x) \cdot d(x)}\text{,} \end{equation*}

then $F$ is a quotient, in which the numerator is a sum of constant multiples (of the functions $a$ and $b$), and the denominator is a product (of the functions $c$ and $d$). Thus the derivative of $F$ can be found by first applying the quotient rule, and then using the sum and constant multiple rules to differentiate the numerator, and the product rule to differentiate the denominator.

SubsectionExercises

Let $f$ and $g$ be differentiable functions for which the following information is known: $f(2) = 5\text{,}$ $g(2) = -3\text{,}$ $f'(2) = -1/2\text{,}$ $g'(2) = 2\text{.}$

1. Let $h$ be the new function defined by the rule $h(x) = g(x) \cdot f(x)\text{.}$ Determine $h(2)$ and $h'(2)\text{.}$

2. Find an equation for the tangent line to $y = h(x)$ at the point $(2,h(2))$ (where $h$ is the function defined in (a)).

3. Let $r$ be the function defined by the rule $r(x) = \frac{g(x)}{f(x)}\text{.}$ Is $r$ increasing, decreasing, or neither at $a = 2\text{?}$ Why?

4. Estimate the value of $r(2.06)$ (where $r$ is the function defined in (c)) by using the tangent line to the graph of $y=r(x)$ at the point $(2,r(2))\text{.}$

Consider the functions $r(t) = t^t$ and $s(t) = \arccos(t)\text{,}$ for which you are given the facts that $r'(t) = t^t(\ln(t) + 1)$ and $s'(t) = -\frac{1}{\sqrt{1-t^2}}\text{.}$ Do not be concerned with where these derivative formulas come from. We restrict our interest in both functions to the domain $0 \lt t \lt 1\text{.}$

1. Let $w(t) = t^t \arccos(t)\text{.}$ Determine $w'(t)\text{.}$

2. Find an equation for the tangent line to $y = w(t)$ at the point $(\frac{1}{2}, w(\frac{1}{2}))\text{.}$

3. Let $v(t) = \frac{t^t}{\arccos(t)}\text{.}$ Is $v$ increasing or decreasing at the instant $t = \frac{1}{2}\text{?}$ Why?

Let functions $p$ and $q$ be the piecewise linear functions given by their respective graphs in Figure2.49. Use the graphs to answer the following questions.

1. Let $r(x) = p(x) \cdot q(x)\text{.}$ Determine $r'(-2)$ and $r'(0)\text{.}$

2. Are there values of $x$ for which $r'(x)$ does not exist? If so, which values, and why?

3. Find an equation for the tangent line to $y = r(x)$ at the point $(2,r(2))\text{.}$

4. Let $z(x) = \frac{q(x)}{p(x)}\text{.}$ Determine $z'(0)$ and $z'(2)\text{.}$

5. Are there values of $x$ for which $z'(x)$ does not exist? If so, which values, and why?

A farmer with large land holdings has historically grown a wide variety of crops. With the price of ethanol fuel rising, he decides that it would be prudent to devote more and more of his acreage to producing corn. As he grows more and more corn, he learns efficiencies that increase his yield per acre. In the present year, he used 7000 acres of his land to grow corn, and that land had an average yield of 170 bushels per acre. At the current time, he plans to increase his number of acres devoted to growing corn at a rate of 600 acres/year, and he expects that right now his average yield is increasing at a rate of 8 bushels per acre per year. Use this information to answer the following questions.

1. Say that the present year is $t = 0\text{,}$ that $A(t)$ denotes the number of acres the farmer devotes to growing corn in year $t\text{,}$ $Y(t)$ represents the average yield in year $t$ (measured in bushels per acre), and $C(t)$ is the total number of bushels of corn the farmer produces. What is the formula for $C(t)$ in terms of $A(t)$ and $Y(t)\text{?}$ Why?

2. What is the value of $C(0)\text{?}$ What does it measure?

3. Write an expression for $C'(t)$ in terms of $A(t)\text{,}$ $A'(t)\text{,}$ $Y(t)\text{,}$ and $Y'(t)\text{.}$ Explain your thinking.

4. What is the value of $C'(0)\text{?}$ What does it measure?

5. Based on the given information and your work above, estimate the value of $C(1)\text{.}$

Let $f(v)$ be the gas consumption (in liters/km) of a car going at velocity $v$ (in km/hour). In other words, $f(v)$ tells you how many liters of gas the car uses to go one kilometer if it is traveling at $v$ kilometers per hour. In addition, suppose that $f(80)=0.05$ and $f'(80) = 0.0004\text{.}$

1. Let $g(v)$ be the distance the same car goes on one liter of gas at velocity $v\text{.}$ What is the relationship between $f(v)$ and $g(v)\text{?}$ Hence find $g(80)$ and $g'(80)\text{.}$

2. Let $h(v)$ be the gas consumption in liters per hour of a car going at velocity $v\text{.}$ In other words, $h(v)$ tells you how many liters of gas the car uses in one hour if it is going at velocity $v\text{.}$ What is the algebraic relationship between $h(v)$ and $f(v)\text{?}$ Hence find $h(80)$ and $h'(80)\text{.}$

3. How would you explain the practical meaning of these function and derivative values to a driver who knows no calculus? Include units on each of the function and derivative values you discuss in your response.