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The product and quotient rules

So far, we can differentiate power functions (x^n), exponential functions (a^x), and the two fundamental trigonometric functions (\sin(x) and \cos(x)).

Introduction

So far, we can differentiate power functions (\(x^n\)), exponential functions (\(a^x\)), 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.

Example

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

What about a product or quotient of two basic functions, such as \[\begin{aligned}\end{aligned}\], or \[\begin{aligned}\end{aligned}\]?

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.

Exploration

The product rule

As part (b) of Preview Activity 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 \(N(t)\) 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 \[\begin{aligned}\end{aligned}\]. 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 \(N\), \(S\), and \(V\), let's consider some specific data.

\(N(100) = 520\)\(S(100) = 27.50\)
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.

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 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 \[\begin{aligned}\end{aligned}\].

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 \(f\) and \(g\) are differentiable functions, then their product \(P(x) = f(x) \cdot g(x)\) is also a differentiable function, and \[\begin{aligned}\end{aligned}\].

Condensed — the full section is in Boelkins, Active Calculus.

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 \(Q(x)\) 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 \(g(x) \ne 0\). 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 \[\begin{aligned}\end{aligned}\].

Applying the product rule to differentiate \(f\), it follows \[\begin{aligned}\end{aligned}\].

Since we want to know a formula for \(Q'\), we solve this most recent equation for \(Q'(x)\) and find \[\begin{aligned}\end{aligned}\]. Dividing both sides by \(g(x)\), we have \[\begin{aligned}\end{aligned}\].

Finally, we recall that \(Q(x) = \frac{f(x)}{g(x)}\). Substituting this expression in the preceding equation and simplifying, it follows that \[\begin{aligned}Q'(x) \amp= \frac{f'(x) - \frac{f(x)}{g(x)} g'(x)}{g(x)} \\ \amp= \frac{f'(x) - \frac{f(x)}{g(x)} g'(x)}{g(x)} \cdot \frac{g(x)}{g(x)} \\ \amp= \frac{g(x) f'(x) - f(x) g'(x)}{g(x)^2}\end{aligned}\].

The preceding argument results in the 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 \(g(x) \ne 0\) and \[\begin{aligned}\end{aligned}\].

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.

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.

Condensed — the full section is in Boelkins, Active Calculus.

Combining rules

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

Example

Determine the derivative of the function \[\begin{aligned}\end{aligned}\]. Clearly state which derivative rules you use and how they were applied.

Solution

To differentiate any complicated function, 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 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 both in this example and the one that follows. to get \[\begin{aligned}f'(x) \amp= \frac{d}{dx} \left[ x\sin(x) + \frac{x^2}{\cos(x) + 2} \right] \\ \amp= \frac{d}{dx} \left[ x\sin(x) \right] + \frac{d}{dx}\left[ \frac{x^2}{\cos(x) + 2} \right]\end{aligned}\]

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 \[\begin{aligned}f'(x) \amp= \left( x \cos(x) + \sin(x) \right) + \frac{(\cos(x) + 2) 2x - x^2(-\sin(x))}{(\cos(x) + 2)^2} \\ \amp= x \cos(x) + \sin(x) + \frac{2x\cos(x) + 4x + x^2\sin(x)}{(\cos(x) + 2)^2}\end{aligned}\].

Example

Determine the derivative of the function \[\begin{aligned}\end{aligned}\]. Clearly state which rules you used and how they were applied.

Solution

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, \[\begin{aligned}\end{aligned}\]. Now, there remain two derivatives to determine. 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 \[\begin{aligned}\end{aligned}\]. While some minor simplification is possible, we are content to leave \(s'(y)\) in its current form.

Condensed — the full section is in Boelkins, Active Calculus.

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 \[\begin{aligned}\end{aligned}\].
The quotient rule tells us that if \(Q\) is a quotient of differentiable functions \(f\) and \(g\) according to the rule \(Q(x) = \frac{f(x)}{g(x)}\), then \[\begin{aligned}\end{aligned}\].
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 \[\begin{aligned}\end{aligned}\], 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.

Practice (12)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Each of the functions on the left below could be described as having a specific algebraic structure as noted on the right. Match each function with its corresponding structure. It is possible that not all structures are used; and it is possible that not all functions have a description for its algebraic structure.
If you do not find a match in the previous exercise, explain why.
Describe the algebraic differences between the two functions \(f(x)=x^2 \sin x\) and \(g(x)=x^2- \sin x\).
A small business makes apple pies. They purchase \(N(t)\) apples at a unit cost of \(C(t)\) dollars per apple. The function \(A(t) = N(t) \cdot C(t)\) measures the total amount the business spends on apples on day \(t\), since \((N(t) \text{ apples}) \cdot (C(t) \text{ dollars/apple})\) has units of dollars.
Each of the quantities on the left below has units that explain the quantity's meaning. For each, identify the corresponding units given on the right. It is possible that not every set of units gets used, and also that some functions and expressions have the same units.
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 \(h(x) = g(x) \cdot f(x)\). 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 \(r(x) = \frac{g(x)}{f(x)}\). 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 \((2,r(2))\).
Revelar a resposta
1. Since \(h(x) = g(x) \cdot f(x)\), it follows \(h(2) = g(2) \cdot f(2) = (5) \cdot (-3) = -15\). Further, by the product rule, we know that \(h'(x) = g(x) \cdot f'(x) + f(x) \cdot g'(x)\). Using the given function and derivative values, \(h'(2) = g(2) \cdot f'(2) + f(2) \cdot g'(2) = (-3)(-1/2) + (5)(2) = 23/2\).
2. The tangent line to \(h\) at \(a=2\) is given by \(L(x) = h(2) + h'(2)(x-2)\), so using our results from (a), we have \(L(x) = -15 + 23/2(x-2)\).
3. Since \(r(x) = \frac{g(x)}{f(x)}\), but the quotient rule that \(r'(x) = \frac{f(x) \cdot g'(x) - g(x) \cdot f'(x)}{(f(x))^2}\), and thus \(r'(2) = \frac{f(2) \cdot g'(2) - g(2) \cdot f'(2)}{(f(2))^2}\). Employing the known function and derivative values, \(r'(2) = \frac{(5)(2) - (-3)(-1/2)}{(5)^2} = \frac{17/2}{25} = \frac{17}{50}\). Therefore, \(r'(2) \gt 0\), and \(r\) is increasing at \(a = 2\).
4. The local linearization of \(r\) at \(a=2\) is \(L(x) = r(2) + r'(2)(x-2) = -\frac{3}{5} + \frac{17}{50}(x-2)\), and thus \(r(2.06) \approx L(2.06) = -\frac{3}{5} + \frac{17}{50}(2.06-2) = -0.5796\).
Let functions \(p\) and \(q\) be the piecewise linear functions given by their respective graphs in Figure. Use the graphs to answer the following questions.
1. Let \(r(x) = p(x) \cdot q(x)\). 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 \(z(x) = \frac{q(x)}{p(x)}\). 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?
Revelar a resposta
1. For all values of \(t\) for which , the product rule applies, and we have \(r'(x) = p(x) \cdot q'(x) + q(x) \cdot p'(x)\). Since \(p\) and \(q\) are differentiable at both \(x=-2\) and \(x=0\), 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, \(p'(-2) = -2\), \(q'(-2) = 3\), \(p'(0) = 0\), and \(q'(0) = \frac{1}{2}\). In addition, we can read the values of \(p\) and \(q\) at these points from the graph: \(p(-2) = 1\), \(q(-2) = -1\), \(p(0) = \frac{1}{2}\), and \(q(0) = 2\). Thus, \[\begin{aligned}r'(-2) &= p(-2) \cdot q'(-2) + q(-2) \cdot p'(-2) \\ &= (1)(3) + (-1)(-2) \\ &= 5\end{aligned}\] and \[\begin{aligned}r'(0) &= p(0) \cdot q'(0) + q(0) \cdot p'(0) \\ &= (-0.5)(0) + (2)(0.5) \\ &= 1\end{aligned}\].
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 \((2,r(2))\), we first observe that \(r(2) = p(2) \cdot q(2) = 2 \cdot 1 = 2\). In addition, we know by the product rule that \(r'(2) = p(2) \cdot q'(2) + q(2) \cdot p'(2) = (2)(-1) + (1)(2) = 0\). Thus, the tangent line is given by \(L(x) = r(2) + r'(2)(x-2) = 2 + 0(x-1) = 2\). In particular, the tangent line is horizontal at \((2,r(2))\).
4. By the quotient rule, \[\begin{aligned}\end{aligned}\] 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, \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\].
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 \(p(1) = 0\), \(z\) is not differentiable at \(x = -1.5\) and \(x = 1\).
Consider the functions \(r(t) = t^t\) and \(s(t) = \arccos(t)\), for which you are given the facts that \(r'(t) = t^t(\ln(t) + 1)\) and \(s'(t) = -\frac{1}{\sqrt{1-t^2}}\). Do not be concerned with where these derivative formulas come from. We restrict our interest in both functions to the domain \(0 \lt t \lt 1\).
1. Let \(w(t) = t^t \arccos(t)\). Determine \(w'(t)\).
2. Find an equation for the tangent line to \(y = w(t)\) at the point \((\frac{1}{2}, w(\frac{1}{2}))\).
3. Let \(v(t) = \frac{t^t}{\arccos(t)}\). Is \(v\) increasing or decreasing at the instant \(t = \frac{1}{2}\)? Why?
Revelar a resposta
1. Using the Product Rule and the information given, \[\begin{aligned}w'(t) &= \frac{d}{dt}\left(t^t\right)\cdot\left(\arccos t\right)+t^t\cdot\frac{d}{dt}\left(\arccos t\right) \\ &= t^t\left(\ln t+1\right)\cdot\left(\arccos t\right)+t^t\cdot\frac{-1}{\sqrt{1-t^2}}\end{aligned}\]
2. Using the local linearization formula, we have \(L(t) = w\left(\frac{1}{2}\right) + w'\left(\frac{1}{2}\right) \left(t-\frac{1}{2}\right)\). Substituting \(t=\frac{1}{2}\) into the formulas for \(w\) and \(w'\) and using a computational device gives \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\]. Thus, \(L(t) \approx 0.740-0.589(t-0.5)\).
3. Using the Quotient Rule, \[\begin{aligned}\end{aligned}\]. Using a computational device to evaluate \(v'(2)\), it follows \(v'(2) \approx 0.952\). 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.
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.
Revelar a resposta
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 \(f\): \(g(v) = \frac{1}{f(v)}\). By the quotient rule, we then know that \[\begin{aligned}\end{aligned}\]. Further, using the given values we can say that \(g(80) = \frac{1}{f(80)} = \frac{1}{0.05} = 20\) kilometers per liter, and \(g'(80) = -\frac{f'(80)}{f(80)^2} = -\frac{0.0004}{0.05^2} = -0.16\). 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 \(h(v)\). Thus, \[\begin{aligned}\end{aligned}\]. By the product rule, it follows that \[\begin{aligned}\end{aligned}\]. Using the known function and derivative values of \(f\), 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.
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 \(h(x) = g(x) \cdot f(x)\). 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 \(r(x) = \frac{g(x)}{f(x)}\). 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 \((2,r(2))\).
Revelar a resposta
1. Since \(h(x) = g(x) \cdot f(x)\), it follows \(h(2) = g(2) \cdot f(2) = (5) \cdot (-3) = -15\). Further, by the product rule, we know that \(h'(x) = g(x) \cdot f'(x) + f(x) \cdot g'(x)\). Using the given function and derivative values, \(h'(2) = g(2) \cdot f'(2) + f(2) \cdot g'(2) = (-3)(-1/2) + (5)(2) = 23/2\).
2. The tangent line to \(h\) at \(a=2\) is given by \(L(x) = h(2) + h'(2)(x-2)\), so using our results from (a), we have \(L(x) = -15 + 23/2(x-2)\).
3. Since \(r(x) = \frac{g(x)}{f(x)}\), but the quotient rule that \(r'(x) = \frac{f(x) \cdot g'(x) - g(x) \cdot f'(x)}{(f(x))^2}\), and thus \(r'(2) = \frac{f(2) \cdot g'(2) - g(2) \cdot f'(2)}{(f(2))^2}\). Employing the known function and derivative values, \(r'(2) = \frac{(5)(2) - (-3)(-1/2)}{(5)^2} = \frac{17/2}{25} = \frac{17}{50}\). Therefore, \(r'(2) \gt 0\), and \(r\) is increasing at \(a = 2\).
4. The local linearization of \(r\) at \(a=2\) is \(L(x) = r(2) + r'(2)(x-2) = -\frac{3}{5} + \frac{17}{50}(x-2)\), and thus \(r(2.06) \approx L(2.06) = -\frac{3}{5} + \frac{17}{50}(2.06-2) = -0.5796\).
Let functions \(p\) and \(q\) be the piecewise linear functions given by their respective graphs in Figure. Use the graphs to answer the following questions.
1. Let \(r(x) = p(x) \cdot q(x)\). 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 \(z(x) = \frac{q(x)}{p(x)}\). 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?
Revelar a resposta
1. For all values of \(t\) for which , the product rule applies, and we have \(r'(x) = p(x) \cdot q'(x) + q(x) \cdot p'(x)\). Since \(p\) and \(q\) are differentiable at both \(x=-2\) and \(x=0\), 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, \(p'(-2) = -2\), \(q'(-2) = 3\), \(p'(0) = 0\), and \(q'(0) = \frac{1}{2}\). In addition, we can read the values of \(p\) and \(q\) at these points from the graph: \(p(-2) = 1\), \(q(-2) = -1\), \(p(0) = \frac{1}{2}\), and \(q(0) = 2\). Thus, \[\begin{aligned}r'(-2) &= p(-2) \cdot q'(-2) + q(-2) \cdot p'(-2) \\ &= (1)(3) + (-1)(-2) \\ &= 5\end{aligned}\] and \[\begin{aligned}r'(0) &= p(0) \cdot q'(0) + q(0) \cdot p'(0) \\ &= (-0.5)(0) + (2)(0.5) \\ &= 1\end{aligned}\].
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 \((2,r(2))\), we first observe that \(r(2) = p(2) \cdot q(2) = 2 \cdot 1 = 2\). In addition, we know by the product rule that \(r'(2) = p(2) \cdot q'(2) + q(2) \cdot p'(2) = (2)(-1) + (1)(2) = 0\). Thus, the tangent line is given by \(L(x) = r(2) + r'(2)(x-2) = 2 + 0(x-1) = 2\). In particular, the tangent line is horizontal at \((2,r(2))\).
4. By the quotient rule, \[\begin{aligned}\end{aligned}\] 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, \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\].
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 \(p(1) = 0\), \(z\) is not differentiable at \(x = -1.5\) and \(x = 1\).
Consider the functions \(r(t) = t^t\) and \(s(t) = \arccos(t)\), for which you are given the facts that \(r'(t) = t^t(\ln(t) + 1)\) and \(s'(t) = -\frac{1}{\sqrt{1-t^2}}\). Do not be concerned with where these derivative formulas come from. We restrict our interest in both functions to the domain \(0 \lt t \lt 1\).
1. Let \(w(t) = t^t \arccos(t)\). Determine \(w'(t)\).
2. Find an equation for the tangent line to \(y = w(t)\) at the point \((\frac{1}{2}, w(\frac{1}{2}))\).
3. Let \(v(t) = \frac{t^t}{\arccos(t)}\). Is \(v\) increasing or decreasing at the instant \(t = \frac{1}{2}\)? Why?
Revelar a resposta
1. Using the Product Rule and the information given, \[\begin{aligned}w'(t) &= \frac{d}{dt}\left(t^t\right)\cdot\left(\arccos t\right)+t^t\cdot\frac{d}{dt}\left(\arccos t\right) \\ &= t^t\left(\ln t+1\right)\cdot\left(\arccos t\right)+t^t\cdot\frac{-1}{\sqrt{1-t^2}}\end{aligned}\]
2. Using the local linearization formula, we have \(L(t) = w\left(\frac{1}{2}\right) + w'\left(\frac{1}{2}\right) \left(t-\frac{1}{2}\right)\). Substituting \(t=\frac{1}{2}\) into the formulas for \(w\) and \(w'\) and using a computational device gives \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\]. Thus, \(L(t) \approx 0.740-0.589(t-0.5)\).
3. Using the Quotient Rule, \[\begin{aligned}\end{aligned}\]. Using a computational device to evaluate \(v'(2)\), it follows \(v'(2) \approx 0.952\). 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.
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.
Revelar a resposta
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 \(f\): \(g(v) = \frac{1}{f(v)}\). By the quotient rule, we then know that \[\begin{aligned}\end{aligned}\]. Further, using the given values we can say that \(g(80) = \frac{1}{f(80)} = \frac{1}{0.05} = 20\) kilometers per liter, and \(g'(80) = -\frac{f'(80)}{f(80)^2} = -\frac{0.0004}{0.05^2} = -0.16\). 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 \(h(v)\). Thus, \[\begin{aligned}\end{aligned}\]. By the product rule, it follows that \[\begin{aligned}\end{aligned}\]. Using the known function and derivative values of \(f\), 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.

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\arcsin,\ \sin^{-1}

inverse sine
The angle whose sine is the given value (and likewise arccos, arctan).

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: The product and quotient rules

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?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

Tente o seu próprio

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

Mais em Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests