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Section 2.3 The product and quotient rules

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So far, we can differentiate power functions (xn), exponential functions (ax), and the two fundamental trigonometric functions (sin⁡(x) and cos⁡(x)). With the sum rule and constant multiple rules, we can also compute the derivative of combined functions.

Differentiate

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

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

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What about a product or quotient of two basic functions, such as

p(z)=z3cos⁡(z),

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or

q(t)=sin⁡(t)2t?

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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.

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Preview Activity 2.3.1.

Let f and g be the functions defined by f(t)=2t2 and g(t)=t3+4t.

  1. Determine f′(t) and g′(t).

  2. Let p(t)=2t2(t3+4t) and observe that p(t)=f(t)⋅g(t). Rewrite the formula for p by distributing the 2t2 term. Then, compute p′(t) using the sum and constant multiple rules.

  3. True or false: p′(t)=f′(t)⋅g′(t).

  4. Let q(t)=t3+4t2t2 and observe that q(t)=g(t)f(t). 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. Then, compute q′(t) using the sum and constant multiple rules.

  5. True or false: q′(t)=g′(t)f′(t).

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Subsection 2.3.1 The product rule

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As part (b) of Preview Activity 2.3.1 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 an example involving meaningful functions.

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Say that an investor is regularly purchasing stock in a particular company. Let N(t) represent the number of shares owned on day t, 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; note that the units on S(t) are dollars per share. To compute the total value of the stock on day t, we take the product

V(t)=N(t)shares⋅S(t)dollars per share,

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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?

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To help us understand the relationship among changes in N, S, and V, 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.
  • 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.

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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?

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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 which total value would rise is the product of the current value of the shares and the rate at which the number of shares is changing. That is, the rate at which total value would change is given by

S(100)⋅N′(100)=27.50dollarsshare⋅12sharesday=330dollarsday.

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Note particularly how the units make sense and show the rate at which the total value V is changing, measured in dollars per day.

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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. The total value is rising at a rate of

N(100)⋅S′(100)=520shares⋅0.75dollars per shareday=390dollarsday.

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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

V′(100)=S(100)⋅N′(100)+N(100)⋅S′(100)=330+390=720dollarsday.

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We expect the total value of the investor's holdings to rise by about $720 on the 100th day. 1 

While 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'd expect that \(V(101) = N(101) \cdot S(101) = 532 \cdot 28.25 = 15029\text{.}\) If, as noted above, we expect the total value to rise by $720, then with \(V(100) = N(100) \cdot S(100) = 520 \cdot 27.50 = 14300\text{,}\) then it seems we should find that \(V(101) = V(100) + 720 = 15020\text{.}\) 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.

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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. If P(x)=f(x)⋅g(x), our work above suggests that P′(x)=f(x)g′(x)+g(x)f′(x). 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.

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Product Rule.

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

P′(x)=f(x)g′(x)+g(x)f′(x).

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In light of the earlier example 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 first times the derivative of the second, plus the second times the derivative of the first.” It is often a helpful mental exercise to say this phrasing aloud when executing the product rule.

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 first, \(z^3\text{,}\) times the derivative of the second, \(-\sin(z)\text{,}\) plus the second, \(\cos(z)\text{,}\) times the derivative of the first, \(3z^2\text{.}\) That is,

\begin{equation*} P'(z) = z^3(-\sin(z)) + \cos(z) 3z^2 = -z^3 \sin(z) + 3z^2 \cos(z)\text{.} \end{equation*}
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Activity 2.3.2.

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)=3w174w. Find m′(w).

  2. Let h(t)=(sin⁡(t)+cos⁡(t))t4. Find h′(t).

  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)=exsin⁡(x).

  4. Find the tangent line approximation L(x) to the function y=g(x) at the point where a=−1 if g is given by the rule g(x)=(x2+x)2x.

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. Remember that \(L(x) = g(a) + g'(a)(x-a)\text{.}\)

Answer

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

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

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

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

Solution

  1. By the product rule,

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

  2. By the product rule,

    \begin{equation*} h'(t) = (\sin(t) + \cos(t)) \cdot 4t^3 + t^4 \cdot (\cos(t) - \sin(t))\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 \(f'(x) = e^x \cdot \cos(x) + \sin(x) \cdot e^x\text{.}\) Thus, \(f'(1) = e^1 \cdot \cos(1) + \sin(1) \cdot e^1 = e(\cos(1) + \sin(1)) \approx 3.756\text{.}\)

  4. First, observe that \(g(-1) = ((-1)^2 - 1) \cdot 2^{-1} = 0\text{.}\) Further, by the product rule, \(g'(x) = (x^2 + x) \cdot 2^x \ln(2) + 2^x \cdot (2x + 1)\text{,}\) and therefore \(g'(-1) = ((-1)^2 - 1) \cdot 2^{-1} \ln(2) + 2^{-1} \cdot (2(-1) + 1) = 0 + \frac{1}{2}(-1) = -\frac{1}{2}\text{.}\) Therefore,

    \begin{equation*} L(x) = g(-1) + g'(-1)(x+1) = 0 - \frac{1}{2}(x+1) = -\frac{1}{2}(x+1)\text{.} \end{equation*}
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Subsection 2.3.2 The quotient rule

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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)=f(x)/g(x), where f and g are both differentiable functions. It turns out that Q is differentiable everywhere that g(x)≠0. We would like a formula for Q′ in terms of f, g, f′, and g′. multiplying both sides of the formula Q=f/g by g, we observe that

f(x)=Q(x)⋅g(x).

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Now we can use the product rule to differentiate f.

f′(x)=Q(x)g′(x)+g(x)Q′(x).

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We want to know a formula for Q′, so we solve this equation for Q′(x).

Q′(x)g(x)=f′(x)−Q(x)g′(x)

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and dividing both sides by g(x), we have

Q′(x)=f′(x)−Q(x)g′(x)g(x).

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Finally, we recall that Q(x)=f(x)g(x). Substituting this expression in the preceding equation, we have

Q′(x)=(f′(x)−f(x)g(x)g′(x)g(x)=(f′(x)−f(x)g(x)g′(x)g(x)⋅g(x)g(x)=(g(x)f′(x)−f(x)g′(x)g(x)2.

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This calculation gives us the quotient rule.

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Quotient Rule.

If f and g are differentiable functions, then their quotient Q(x)=f(x)g(x) is also a differentiable function for all x where g(x)≠0 and

Q′(x)=g(x)f′(x)−f(x)g′(x)g(x)2.

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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, then Q′ is given by “the bottom times the derivative of the top, minus the top times the derivative of the bottom, all over the bottom squared.”

If \(Q(t) = \sin(t)/2^t\text{,}\) we call \(\sin(t)\) the top function and \(2^t\) the bottom function. By the quotient rule, \(Q'\) is given by the bottom, \(2^t\text{,}\) times the derivative of the top, \(\cos(t)\text{,}\) 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{2^t \cos(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*}

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In general, we must be careful in doing any such simplification, as we don't want to execute the quotient rule correctly but then make an algebra error.

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Activity 2.3.3.

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), clearly label the formula you find for f′(x). It is not necessary to algebraically simplify any of the derivatives you compute.

  1. Let r(z)=3zz4+1. Find r′(z).

  2. Let v(t)=sin⁡(t)cos⁡(t)+t2. Find v′(t).

  3. Determine the slope of the tangent line to the curve R(x)=x2−2x−8x2−9 at the point where x=0.

  4. When a camera flashes, the intensity I of light seen by the eye is given by the function

    I(t)=100tet,

    where I is measured in candles and t is measured in milliseconds. Compute I′(0.5), I′(2), and I′(5); 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 bottom times the derivative of the top” 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.

Answer

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

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

  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{(z^4+1) 3^z \ln(3) - 3^z(4z^3)}{(z^4 + 1)^2}\text{.} \end{equation*}

  2. By the quotient rule,

    \begin{equation*} v'(t) = \frac{(\cos(t) + t^2)\cos(t) - \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{(x^2 - 9)(2x - 2) - (x^2 - 2x - 8)(2x)}{(x^2 - 9)^2}\text{.} \end{equation*}

    From this, it follows that \(R'(0) = \frac{(-9)(-2)-(-8)(0)}{(-9)^2} = \frac{2}{9}\text{,}\) 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{e^t 100 - 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 \(t = 2\text{.}\)

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Subsection 2.3.3 Combining rules

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

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 rule 2  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] \end{align*}
When 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 each of the two examples presented here.

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( x \cos(x) + \sin(x) \right) + \frac{(\cos(x) + 2) 2x - x^2(-\sin(x))}{(\cos(x) + 2)^2}\\ =\mathstrut \amp x \cos(x) + \sin(x) + \frac{2x\cos(x) + 4x^2 + x^2\sin(x)}{(\cos(x) + 2)^2}\text{.} \end{align*}

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{(y^2 + 1) \cdot \frac{d}{dy}\left[ y \cdot 7^y \right] - 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{(y^2 + 1) [y \cdot 7^y \ln(7) + 7^y \cdot 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.

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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 to do a large number of examples.

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Activity 2.3.4.

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)=(5r3+sin⁡(r))(4r−2cos⁡(r)). Find f′(r).

  2. Let p(t)=cos⁡(t)t6⋅6t. Find p′(t).

  3. Let g(z)=3z7ez−2z2sin⁡(z)+zz2+1. Find g′(z).

  4. A moving particle has its position in feet at time t in seconds given by the function s(t)=3cos⁡(t)−sin⁡(t)et. Find the particle's instantaneous velocity at the moment t=1.

  5. Suppose that f(x) and g(x) are differentiable functions and it is known that f(3)=−2, f′(3)=7, g(3)=4, and g′(3)=−1. If p(x)=f(x)⋅g(x) and q(x)=f(x)g(x), calculate p′(3) and q′(3).

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 \(p'(x) = f(x) g'(x) + g(x) f'(x)\text{.}\)

Answer

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

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

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

  4. \(s'(1) = \frac{-2\sin(1)-4\cos(1)}{e^1} \approx -1.414\) 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 (5r^3 + \sin(r))\frac{d}{dr}[4^r - 2\cos(r)] + (4^r - 2\cos(r))\frac{d}{dr}[5r^3 + \sin(r)]\\ =\mathstrut \amp (5r^3 + \sin(r))[4^r \ln(4) + 2\sin(r)] + (4^r - 2\cos(r))[15r^2 + \cos(r)] \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{t^6 \cdot 6^t \frac{d}{dt}[\cos(t)] - \cos(t) \frac{d}{dt}[t^6 \cdot 6^t]}{(t^6 \cdot 6^t)^2}\\ =\mathstrut \amp \frac{t^6 \cdot 6^t [-\sin(t)] - \cos(t) [t^6 \cdot 6^t \ln(6) + 6^t \cdot 6t^5]}{(t^6 \cdot 6^t)^2} \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 [z^7 e^z + 7z^6e^z] - 2[z^2 \cos(z) + 2z\sin(z)] + \frac{(z^2+1) 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 simplify by removing a common factor of \(e^t\) to get

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

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

    and

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

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

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Subsection 2.3.4 Summary

  • 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), then

    P′(x)=f(x)g′(x)+g(x)f′(x).

  • The quotient rule tells us that if Q is a quotient of differentiable functions f and g according to the rule Q(x)=f(x)g(x), then

    Q′(x)=g(x)f′(x)−f(x)g′(x)g(x)2.

  • 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

    F(x)=2a(x)−5b(x)c(x)⋅d(x),

    then F is a quotient, in which the numerator is a sum of constant multiples and the denominator is a product. This, the derivative of F can be found by 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.

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Exercises 2.3.5 Exercises

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1. Derivative of a basic product.
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2. Derivative of a product.
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3. Derivative of a quotient of linear functions.
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4. Derivative of a rational function.
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5. Derivative of a product of trigonometric functions.
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6. Derivative of a product of power and trigonmetric functions.
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7. Derivative of a sum that involves a product.
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8. Product and quotient rules with graphs.
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9. Product and quotient rules with given function values.
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10.

Let f and g be differentiable functions for which the following information is known: f(2)=5, g(2)=−3, f′(2)=−1/2, g′(2)=2.

  1. Let h be the new function defined by the rule h(x)=g(x)⋅f(x). Determine h(2) and h′(2).

  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)=g(x)f(x). Is r increasing, decreasing, or neither at a=2? Why?

  4. Estimate the value of r(2.06) (where r is the function defined in (c)) by using the local linearization of r at the point (2,r(2)).

Answer

  1. \(h(2) = -15\text{;}\) \(h'(2) = 23/2\text{.}\)

  2. \(L(x) = -15 + 23/2(x-2)\text{.}\)

  3. Increasing.

  4. \(r(2.06) \approx -0.5796\text{.}\)

Solution

  1. Since \(h(x) = g(x) \cdot f(x)\text{,}\) it follows \(h(2) = g(2) \cdot f(2) = (5) \cdot (-3) = -15\text{.}\) Further, by the product rule, we know that \(h'(x) = g(x) \cdot f'(x) + f(x) \cdot g'(x)\text{.}\) Using the given function and derivative values, \(h'(2) = g(2) \cdot f'(2) + f(2) \cdot g'(2) = (-3)(-1/2) + (5)(2) = 23/2\text{.}\)

  2. The tangent line to \(h\) at \(a=2\) is given by \(L(x) = h(2) + h'(2)(x-2)\text{,}\) so using our results from (a), we have \(L(x) = -15 + 23/2(x-2)\text{.}\)

  3. Since \(r(x) = \frac{g(x)}{f(x)}\text{,}\) but the quotient rule that \(r'(x) = \frac{f(x) \cdot g'(x) - g(x) \cdot f'(x)}{(f(x))^2}\text{,}\) and thus \(r'(2) = \frac{f(2) \cdot g'(2) - g(2) \cdot f'(2)}{(f(2))^2}\text{.}\) Employing the known function and derivative values, \(r'(2) = \frac{(5)(2) - (-3)(-1/2)}{(5)^2} = \frac{17/2}{25} = \frac{17}{50}\text{.}\) Therefore, \(r'(2) \gt 0\text{,}\) and \(r\) is increasing at \(a = 2\text{.}\)

  4. The local linearization of \(r\) at \(a=2\) is \(L(x) = r(2) + r'(2)(x-2) = -\frac{3}{5} + \frac{17}{20}(x-2)\text{,}\) and thus \(r(2.06) \approx L(2.06) = -\frac{3}{5} + \frac{17}{50}(2.06-2) = -0.5796\text{.}\)

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11.

Consider the functions r(t)=tt and s(t)=arccos⁡(t), for which you are given the facts that r′(t)=tt(ln⁡(t)+1) and s′(t)=−11−t2. Do not be concerned with where these derivative formulas come from. We restrict our interest in both functions to the domain 0<t<1.

  1. Let w(t)=ttarccos⁡(t). Determine w′(t).

  2. Find an equation for the tangent line to y=w(t) at the point (12,w(12)).

  3. Let v(t)=ttarccos⁡(t). Is v increasing or decreasing at the instant t=12? Why?

Answer

  1. \(w'(t) = t^t\left(\ln t+1\right)\cdot\left(\arccos t\right)+t^t\cdot\frac{-1}{\sqrt{1-t^2}} \text{.}\)

  2. \(L(t) \approx 0.740-0.589(t-0.5)\text{.}\)

  3. Increasing.

Solution

  1. Using the Product Rule and the information given,

    \begin{align*} w'(t) &= \frac{d}{dt}\left(t^t\right)\cdot\left(\arccos t\right)+t^t\cdot\frac{d}{dt}\left(\arccos t\right)\\ &= t^t\left(\ln t+1\right)\cdot\left(\arccos t\right)+t^t\cdot\frac{-1}{\sqrt{1-t^2}} \end{align*}

  2. Using the local linearization formula, we have \(L(t) = w\left(\frac{1}{2}\right) + w'\left(\frac{1}{2}\right) \left(t-\frac{1}{2}\right)\text{.}\) Substituting \(t=\frac{1}{2}\) into the formulas for \(w\) and \(w'\) and using a computational device gives

    \begin{equation*} w(2) = \left(\frac{1}{2}\right)^{\left(\frac{1}{2}\right)}\arccos\left(\frac{1}{2}\right) \approx 0.740 \end{equation*}

    and

    \begin{equation*} w'(2) = \left[\left(\frac{1}{2}\right)^{\left(\frac{1}{2}\right)}\left(\ln t+1\right)\cdot\left(\arccos \left(\frac{1}{2}\right)\right)+\left(\frac{1}{2}\right)^{\left(\frac{1}{2}\right)}\cdot\frac{-1}{\sqrt{1-\left(\frac{1}{2}\right)^2}}\right] \approx -0.589\text{.} \end{equation*}

    Thus, \(L(t) \approx 0.740-0.589(t-0.5)\text{.}\)

  3. Using the Quotient Rule,

    \begin{equation*} v'(t) = frac{(t)^{t}\left(\ln(t)+1\right)\left(\arccos(t)\right)-(t)^{t}\cdot\left(\frac{-1}{\sqrt{1-(t)^2}}\right)}{\arccos^2(t)}\text{.} \end{equation*}

    Using a computational device to evaluate \(v'(2)\text{,}\) it follows \(v'(2) \approx 0.952\text{.}\) Since this number is positive, the instantaneous rate of change of \(v\) at \(t=\frac{1}{2}\) is positive, which means that \(v\) is increasing at that instant.

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12.

Let functions p and q be the piecewise linear functions given by their respective graphs in Figure 2.3.6. Use the graphs to answer the following questions.

  1. Let r(x)=p(x)⋅q(x). Determine r′(−2) and r′(0).

  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)).

  4. Let z(x)=q(x)p(x). Determine z′(0) and z′(2).

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

Figure 2.3.6. The graphs of p (in blue) and q (in green).
Answer

  1. \(r'(-2) = 2\) and \(r'(0) = 0.25\text{.}\)

  2. At \(x = -1\) and \(x = 1\text{.}\)

  3. \(L(x) = 2\text{.}\)

  4. \(z'(0) = \frac{1}{16}\) and \(z'(2) = -1\text{.}\)

  5. At \(x = -1\text{,}\) \(x = 1\text{,}\) \(x = -1.5\text{,}\) and \(x = 1\text{.}\)

Solution

  1. For all values of \(t\) for which , the product rule applies, and we have \(r'(x) = p(x) \cdot q'(x) + q(x) \cdot p'(x)\text{.}\) Since \(p\) and \(q\) are differentiable at both \(x=-2\) and \(x=0\text{,}\) we can determine the values of \(p'\) and \(q'\) at these points by computing the slope of the function (which is piecewise linear). In particular, \(p'(-2) = -2\text{,}\) \(q'(-2) = 3\text{,}\) \(p'(0) = 0\text{,}\) and \(q'(0) = \frac{1}{2}\text{.}\) In addition, we can read the values of \(p\) and \(q\) at these points from the graph: \(p(-2) = 1\text{,}\) \(q(-2) = -1\text{,}\) \(p(0) = \frac{1}{2}\text{,}\) and \(q(0) = 2\text{.}\) Thus,

    \begin{align*} r'(-2) &= p(-2) \cdot q'(-2) + q(-2) \cdot p'(-2)\\ &= (1)(3) + (-1)(1)\\ &= 2 \end{align*}

    and

    \begin{align*} r'(0) &= p(0) \cdot q'(0) + q(0) \cdot p'(0)\\ &= (0.5)(0.5) + (2)(0)\\ &= 0.25\text{.} \end{align*}

  2. Neither \(r'(-1)\) nor \(r'(1)\) exist because both \(p\) and \(q\) fail to be differentiable at \(x = -1\) and \(x = 1\) due to the sharp corners on their graphs.

  3. To find the tangent line to \(r\) at \((2,r(2))\text{,}\) we first observe that \(r(2) = p(2) \cdot q(2) = 2 \cdot 1 = 2\text{.}\) In addition, we know by the product rule that \(r'(2) = p(2) \cdot q'(2) + q(2) \cdot p'(2) = (2)(-1) + (1)(2) = 0\text{.}\) Thus, the tangent line is given by \(L(x) = r(2) + r'(2)(x-2) = 2 + 0(x-1) = 2\text{.}\) In particular, the tangent line is horizontal at \((2,r(2))\text{.}\)

  4. By the quotient rule,

    \begin{equation*} z'(x) = \frac{p(x)\cdot q'(x) - q(x) \cdot p'(x)}{p(x)^2} \end{equation*}

    Thus, using the values of \(p\) and \(q\) and their derivatives at \(x = 0\) and \(x = 2\) that we have determined in earlier parts of this problem,

    \begin{equation*} z'(0) = \frac{p(0)\cdot q'(0) - q(0) \cdot p'(0)}{p(0)^2} = \frac{(0.5)(0.5)-(2)(0)}{2^2} = \frac{1}{16} \end{equation*}

    and

    \begin{equation*} z'(2) = \frac{p(2)\cdot q'(2) - q(2) \cdot p'(2)}{p(2)^2} = \frac{(2)(-1)-(1)(2)}{2^2} = -1\text{.} \end{equation*}

  5. \(z\) fails to be differentiable at \(x = -1\) and \(x = 1\) because both \(p\) and \(q\) have sharp corners at these \(x\)-values. In addition, since \(p(-1.5) = 0\) and \(p(1) = 0\text{,}\) \(z\) is not differentiable at \(x = -1.5\) and \(x = 1\text{.}\)

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13.

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, that A(t) denotes the number of acres the farmer devotes to growing corn in year t, 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)? Why?

  2. What is the value of C(0)? What does it measure?

  3. Write an expression for C′(t) in terms of A(t), A′(t), Y(t), and Y′(t). Explain your thinking.

  4. What is the value of C′(0)? What does it measure?

  5. Based on the given information and your work above, estimate the value of C(1).

Answer

  1. \(C(t) = A(t)Y(t)\) bushels in year \(t\text{.}\)

  2. \(1 190 000\) bushels of corn.

  3. \(C'(t) = A(t)Y'(t) + A'(t)Y(t)\text{.}\)

  4. \(C'(0) = 158 000\) bushels per year.

  5. \(C(1) \approx 1 348 000\)bushels.

Solution

  1. If we multiply \(A(t)\) acres by \(Y(t)\) bushels per acre, we arrive at \(C(t) = A(t)Y(t)\) bushels as the total number of bushels of corn the farmer produces in year \(t\text{.}\)

  2. We are given that \(A(0) = 7000\) acres with a yield of \(Y(0) = 170\) bushels per acre, giving us \(C(0) = A(0)Y(0) = (7000)(170) = 1190000\) bushels. Thus, the farmer produces \(1 190 000\) bushels of corn in the present year.

  3. The product rule tells us that \(C'(t) = A(t)Y'(t) + A'(t)Y(t)\text{.}\)

  4. We know \(A(0) = 7000\) and \(Y(0) = 170\text{.}\) We are also given that the farmer will 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. This tells us that \(A'(t) = 600\) and \(Y'(0) = 8\text{.}\) So

    \begin{equation*} C'(0) = A(0)Y'(0) + A'(0)Y(0) = (7000)(8) + (600)(170) = 158 000\text{.} \end{equation*}

    Note that the units of \(A(t)\) are acres and the units of \(Y'(t)\) are bushels per acre per year. So the units of \(A(0)Y'(0)\) are bushels per year. Similarly, the units of \(A'(t)\) are acres per year and the units of \(Y(t)\) are bushels per acre, so the units of \(A'(0)Y(0)\) are also bushels per acre. So the units of \(C'(0)\) are bushels per acre. Therefore, \(C'(0) = 158 000\) tells us that the number of bushels the farmer can produce will increase by approximately \(158 000\) bushels for each year past the present year.

  5. The linearization of \(C\) at \(t=0\) is \(L(t) = C(0) + C'(0)(t-0) = 119 0000 + 158 000t\text{,}\) \(C(1) \approx L(1) = 119 0000 + 158 000(1) = 1 348 000\)bushels.

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14.

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.

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

  2. Let h(v) be the gas consumption in liters per hour of a car going at velocity v. 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. What is the algebraic relationship between h(v) and f(v)? Hence find h(80) and h′(80).

  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.

Answer

  1. \(g(80) = 20\) kilometers per liter, and \(g'(80) = -0.16\text{.}\) kilometers per liter per kilometer per hour.

  2. \(h(80) = 4\) liters per hour and \(h'(80) = 0.082\) liters per hour per kilometer per hour.

  3. Think carefully about units and how each of the three pairs of values expresses fundamentally the same facts.

Solution

  1. Observe that the units on \(g(v)\) are ``kilometers per liter.'' Since the units on \(f(v)\) are ``liters per kilometer,'' it follows that \(g\) is the reciprocal of \(f\text{:}\) \(g(v) = \frac{1}{f(v)}\text{.}\) By the quotient rule, we then know that

    \begin{equation*} g'(v) = \frac{f(v) \cdot 0 - 1 \cdot f'(v)}{f(v)^2} = -\frac{f'(v)}{f(v)^2}\text{.} \end{equation*}

    Further, using the given values we can say that \(g(80) = \frac{1}{f(80)} = \frac{1}{0.05} = 20\) kilometers per liter, and \(g'(80) = -\frac{f'(80)}{f(80)^2} = -\frac{0.0004}{0.05^2} = -0.16\text{.}\) Note that the units on \(g'(80)\) are ``kilometers per liter per kilometer per hour.''

  2. Since the units on \(h(v)\) are ``liters per hour,'' and the units on \(f(v)\) are ``liters per kilometer,'' we note that if we multiply \(f(v)\) by \(v\) (whose units are ``kilometers per hour''), we get that the units on \(v \cdot f(v)\) match the units of \(h(v)\text{.}\) Thus,

    \begin{equation*} h(v) = v \cdot f(v)\text{.} \end{equation*}

    By the product rule, it follows that

    \begin{equation*} h'(v) = v \cdot \frac{d}{dv}[f(v)] + \frac{d}{dv}[v] \cdot f(v) = v \cdot f'(v) + 1 \cdot f(v)\text{.} \end{equation*}

    Using the known function and derivative values of \(f\text{,}\) we determine that \(h(80) = 80 \cdot 0.05 = 4\) liters per hour and \(h'(80) = 80 \cdot 0.0004 + 0.05 = 0.082\) liters per hour per kilometer per hour.

  3. To explain the meaning of these function and derivative values to someone who doesn't know calculus, we could say something like the following:

    • The given values of \(f(80) = 0.05\) and \(f'(80) = 0.0004\) tell us that when the car is going \(80\) kilometers per hour, it is using \(0.05\) liters of fuel for every kilometer it travels. In addition, for each additional kilometer per hour of velocity above \(80\) kph, we expect that the car will use about an additional \(0.0004\) liters of fuel per kilometer.

    • The values of \(g(80) = 20\) and \(g'(80) = -0.16\) tell us that when the car is going \(80\) kilometers per hour, it is able to travel \(20\) kilometers for each liter of fuel used. In addition, for each additional kilometer per hour of velocity above \(80\) kph, we expect that the car will be able to travel about \(0.16\) fewer kilometers per liter of fuel.

    • Finally, the values of \(h(80) = 4\) and \(h'(80) = 0.82\) tell us that when the car is going \(80\) kilometers per hour, it is using \(4\) liters of fuel for every hour it travels. In addition, for each additional kilometer per hour of velocity above \(80\) kph, we expect that the car will ues about \(0.082\) more liters of fuel per hour.

    Note that all three of these observations are saying fundamentally the same thing, just from different perspectives: the first regards fuel consumption measured in liters per kilometer; the second regards fuel consumption in kilometers per liter; and the last regards fuel consumption in liters per hour.

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