Section 2.3 The product and quotient rules
ΒΆMotivating 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?
Example 2.3.1.
Differentiate
Because \(f\) is a sum of basic functions, we can now quickly say that \(f'(x) = 77x^{10} - 4 \cdot 9^x \ln(9) + \pi \cos(x) + \sqrt{3} \sin(x)\text{.}\)
Preview Activity 2.3.1.
Let and be the functions defined by and
Determine and
Let and observe that Rewrite the formula for by distributing the term. Then, compute using the sum and constant multiple rules.
True or false:
Let and observe that Rewrite the formula for by dividing each term in the numerator by the denominator and simplify to write as a sum of constant multiples of powers of Then, compute using the sum and constant multiple rules.
True or false:
Subsection 2.3.1 The product rule
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. Say that an investor is regularly purchasing stock in a particular company. Let represent the number of shares owned on day where represents the first day on which shares were purchased. Let give the value of one share of the stock on day note that the units on are dollars per share. To compute the total value of the stock on day we take the product- 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 and
- 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.
Product Rule.
If and are differentiable functions, then their product is also a differentiable function, and
Example 2.3.2.
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,
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.
Let Find
Let Find
Determine the slope of the tangent line to the curve at the point where if is given by the rule
Find the tangent line approximation to the function at the point where if is given by the rule
Let the first function be \(3w^{17}\text{.}\)
Let the first function be \((\sin(t) + \cos(t))\text{.}\)
Remember that the slope of the tangent line to \(y = f(x)\) at \((a,f(a))\) is given by \(f'(a)\text{.}\)
Remember that \(L(x) = g(a) + g'(a)(x-a)\text{.}\)
\(m'(w) = 3w^{17} \cdot 4^w \ln(4) + 4^w \cdot 51w^{16}\text{.}\)
\(h'(t) = (\sin(t) + \cos(t)) \cdot 4t^3 + t^4 \cdot (\cos(t) - \sin(t))\text{.}\)
\(f'(1) = e(\cos(1) + \sin(1)) \approx 3.756\text{.}\)
\(L(x) = -\frac{1}{2}(x+1)\text{.}\)
-
By the product rule,
\begin{equation*} m'(w) = 3w^{17} \cdot 4^w \ln(4) + 4^w \cdot 51w^{16}\text{.} \end{equation*} -
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*} 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, \(f'(1) = e^1 \cdot \cos(1) + \sin(1) \cdot e^1 = e(\cos(1) + \sin(1)) \approx 3.756\text{.}\)
-
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*}
Subsection 2.3.2 The 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 be defined by where and are both differentiable functions. It turns out that is differentiable everywhere that We would like a formula for in terms of and multiplying both sides of the formula by we observe thatQuotient Rule.
If and are differentiable functions, then their quotient is also a differentiable function for all where and
Example 2.3.3.
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,
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
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 clearly label the formula you find for It is not necessary to algebraically simplify any of the derivatives you compute.
Let Find
Let Find
Determine the slope of the tangent line to the curve at the point where
-
When a camera flashes, the intensity of light seen by the eye is given by the function
where is measured in candles and is measured in milliseconds. Compute and include appropriate units on each value; and discuss the meaning of each.
When applying the quotient rule, use parentheses around the bottom function, \(z^4 + 1\text{,}\) to ensure that when you compute βthe bottom times the derivative of the topβ that the rule is applied correctly.
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.
Remember one of the key interpretations of the derivative.
Let the top function be \(100t\) and simply use the constant multiple rule to find its derivative.
\(r'(z)=\frac{(z^4+1) 3^z \ln(3) - 3^z(4z^3)}{(z^4 + 1)^2}\text{.}\)
\(v'(t) = \frac{(\cos(t) + t^2)\cos(t) - \sin(t)(\sin(t) + 2t)}{(\cos(t) + t^2)^2}\text{.}\)
\(R'(0) = \frac{2}{9}\text{.}\)
\(I'(0.5) = \frac{50}{e^{0.5}} \approx 30.327\text{,}\) \(I'(2) = \frac{-100}{e^{2}} \approx -13.534\text{,}\) and \(I'(5) = \frac{-400}{e^4} \approx -2.695\text{,}\) each in candles per millisecond.
-
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*} -
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*} -
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 \(x = 0\text{.}\)
-
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 \(t = 2\text{.}\)
Subsection 2.3.3 Combining rules
In order to apply the derivative shortcut rules correctly we must recognize the fundamental structure of a function.Example 2.3.4.
Determine the derivative of the function
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 ruleβ2β to get
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
Example 2.3.5.
Differentiate
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,
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
While some simplification is possible, we are content to leave \(s'(y)\) in its current form.
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.
Let Find
Let Find
Let Find
A moving particle has its position in feet at time in seconds given by the function Find the particle's instantaneous velocity at the moment
Suppose that and are differentiable functions and it is known that and If and calculate and
Observe that \(f\) is fundamentally a product. Which is the first function? The second?
Note that \(p\) has the overall structure of a quotient.
Think about how \(g\) is a sum of three functions. What is the structure of each of the three functions in the sum?
How is the velocity of a moving object related to its position?
Since we know \(p(x) = f(x) \cdot g(x)\text{,}\) it follows \(p'(x) = f(x) g'(x) + g(x) f'(x)\text{.}\)
\(f'(r) = (5r^3 + \sin(r))[4^r \ln(4) + 2\sin(r)] + (4^r - 2\cos(r))[15r^2 + \cos(r)]\text{.}\)
\(p'(t) = \frac{t^6 \cdot 6^t [-\sin(t)] - \cos(t) [t^6 \cdot 6^t \ln(6) + 6^t \cdot 6t^5]}{(t^6 \cdot 6^t)^2}\text{.}\)
\(g'(z) = 3 [z^7 e^z + 7z^6e^z] - 2[z^2 \cos(z) + 2z\sin(z)] + \frac{(z^2+1) 1 - z(2z)}{(z^2 + 1)^2}\text{.}\)
\(s'(1) = \frac{-2\sin(1)-4\cos(1)}{e^1} \approx -1.414\) feet per second.
\(p'(3) = 30\) and \(q'(3) = \frac{13}{8}\text{.}\)
-
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*} -
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*} -
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*} -
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 \(t = 1\text{.}\)
-
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*}
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 is a product of differentiable functions and according to the rule then
-
The quotient rule tells us that if is a quotient of differentiable functions and according to the rule then
-
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 has the form
then is a quotient, in which the numerator is a sum of constant multiples and the denominator is a product. This, the derivative of 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.
Exercises 2.3.5 Exercises
ΒΆ1. Derivative of a basic product.
2. Derivative of a product.
3. Derivative of a quotient of linear functions.
4. Derivative of a rational function.
5. Derivative of a product of trigonometric functions.
6. Derivative of a product of power and trigonmetric functions.
7. Derivative of a sum that involves a product.
8. Product and quotient rules with graphs.
9. Product and quotient rules with given function values.
10.
Let and be differentiable functions for which the following information is known:
Let be the new function defined by the rule Determine and
Find an equation for the tangent line to at the point (where is the function defined in (a)).
Let be the function defined by the rule Is increasing, decreasing, or neither at Why?
Estimate the value of (where is the function defined in (c)) by using the local linearization of at the point
\(h(2) = -15\text{;}\) \(h'(2) = 23/2\text{.}\)
\(L(x) = -15 + 23/2(x-2)\text{.}\)
Increasing.
\(r(2.06) \approx -0.5796\text{.}\)
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, \(h'(2) = g(2) \cdot f'(2) + f(2) \cdot g'(2) = (-3)(-1/2) + (5)(2) = 23/2\text{.}\)
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 \(L(x) = -15 + 23/2(x-2)\text{.}\)
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 \(a = 2\text{.}\)
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 \(r(2.06) \approx L(2.06) = -\frac{3}{5} + \frac{17}{50}(2.06-2) = -0.5796\text{.}\)
11.
Consider the functions and for which you are given the facts that and Do not be concerned with where these derivative formulas come from. We restrict our interest in both functions to the domain
Let Determine
Find an equation for the tangent line to at the point
Let Is increasing or decreasing at the instant Why?
\(w'(t) = t^t\left(\ln t+1\right)\cdot\left(\arccos t\right)+t^t\cdot\frac{-1}{\sqrt{1-t^2}} \text{.}\)
\(L(t) \approx 0.740-0.589(t-0.5)\text{.}\)
Increasing.
-
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*} -
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, \(L(t) \approx 0.740-0.589(t-0.5)\text{.}\)
-
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.
12.
Let functions and be the piecewise linear functions given by their respective graphs in Figure 2.3.6. Use the graphs to answer the following questions.
Let Determine and
Are there values of for which does not exist? If so, which values, and why?
Find an equation for the tangent line to at the point
Let Determine and
Are there values of for which does not exist? If so, which values, and why?
\(r'(-2) = 2\) and \(r'(0) = 0.25\text{.}\)
At \(x = -1\) and \(x = 1\text{.}\)
\(L(x) = 2\text{.}\)
\(z'(0) = \frac{1}{16}\) and \(z'(2) = -1\text{.}\)
At \(x = -1\text{,}\) \(x = 1\text{,}\) \(x = -1.5\text{,}\) and \(x = 1\text{.}\)
-
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*} 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.
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 \((2,r(2))\text{.}\)
-
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*} \(z\) 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 \(x = 1\text{.}\)
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.
Say that the present year is that denotes the number of acres the farmer devotes to growing corn in year represents the average yield in year (measured in bushels per acre), and is the total number of bushels of corn the farmer produces. What is the formula for in terms of and Why?
What is the value of What does it measure?
Write an expression for in terms of and Explain your thinking.
What is the value of What does it measure?
Based on the given information and your work above, estimate the value of
\(C(t) = A(t)Y(t)\) bushels in year \(t\text{.}\)
\(1 190 000\) bushels of corn.
\(C'(t) = A(t)Y'(t) + A'(t)Y(t)\text{.}\)
\(C'(0) = 158 000\) bushels per year.
\(C(1) \approx 1 348 000\)bushels.
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 \(t\text{.}\)
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.
The product rule tells us that \(C'(t) = A(t)Y'(t) + A'(t)Y(t)\text{.}\)
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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.
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.
14.
Let be the gas consumption (in liters/km) of a car going at velocity (in km/hour). In other words, tells you how many liters of gas the car uses to go one kilometer if it is traveling at kilometers per hour. In addition, suppose that and
Let be the distance the same car goes on one liter of gas at velocity What is the relationship between and Hence find and
Let be the gas consumption in liters per hour of a car going at velocity In other words, tells you how many liters of gas the car uses in one hour if it is going at velocity What is the algebraic relationship between and Hence find and
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.
\(g(80) = 20\) kilometers per liter, and \(g'(80) = -0.16\text{.}\) kilometers per liter per kilometer per hour.
\(h(80) = 4\) liters per hour and \(h'(80) = 0.082\) liters per hour per kilometer per hour.
Think carefully about units and how each of the three pairs of values expresses fundamentally the same facts.
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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.''
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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.
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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.