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Section 2.1 Elementary derivative rules

In Chapter 1, we developed the concept of the derivative of a function. We now know that the derivative \(f'\) of a function \(f\) measures the instantaneous rate of change of \(f\) with respect to \(x\text{.}\) The derivative also tells us the slope of the tangent line to \(y=f(x)\) at any given value of \(x\text{.}\) So far, we have focused on interpreting the derivative graphically or, in the context of a physical setting, as a meaningful rate of change. To calculate the value of the derivative at a specific point, we have relied on the limit definition of the derivative,

\begin{equation*} f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\text{.} \end{equation*}

In this chapter, we investigate how the limit definition of the derivative leads to interesting patterns and rules that enable us to find a formula for \(f'(x)\) quickly, without using the limit definition directly. For example, we would like to apply shortcuts to differentiate a function such as \(g(x) = 4x^7 - \sin(x) + 3e^x\)

Preview Activity 2.1.1.

Functions of the form \(f(x) = x^n\text{,}\) where \(n = 1, 2, 3, \ldots\text{,}\) are often called power functions. The first two questions below revisit work we did earlier in Chapter 1, and the following questions extend those ideas to higher powers of \(x\text{.}\)

  1. Use the limit definition of the derivative to find \(f'(x)\) for \(f(x) = x^2\text{.}\)

  2. Use the limit definition of the derivative to find \(f'(x)\) for \(f(x) = x^3\text{.}\)

  3. Use the limit definition of the derivative to find \(f'(x)\) for \(f(x) = x^4\text{.}\) (Hint: \((a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\text{.}\) Apply this rule to \((x+h)^4\) within the limit definition.)

  4. Based on your work in (a), (b), and (c), what do you conjecture is the derivative of \(f(x) = x^5\text{?}\) Of \(f(x) = x^{13}\text{?}\)

  5. Conjecture a formula for the derivative of \(f(x) = x^n\) that holds for any positive integer \(n\text{.}\) That is, given \(f(x) = x^n\) where \(n\) is a positive integer, what do you think is the formula for \(f'(x)\text{?}\)

Subsection 2.1.1 Some Key Notation

In addition to our usual \(f'\) notation, there are other ways to denote the derivative of a function, as well as the instruction to take the derivative. If we are thinking about the relationship between \(y\) and \(x\text{,}\) we sometimes denote the derivative of \(y\) with respect to \(x\) by the symbol

\begin{equation*} \frac{dy}{dx} \end{equation*}

which we read “dee-y dee-x.” For example, if \(y = x^2\text{,}\) we'll write that the derivative is \(\frac{dy}{dx} = 2x\text{.}\) This notation comes from the fact that the derivative is related to the slope of a line, and slope is measured by \(\frac{\Delta y}{\Delta x}\text{.}\) Note that while we read \(\frac{\Delta y}{\Delta x}\) as “change in \(y\) over change in \(x\text{,}\)” we view \(\frac{dy}{dx}\) as a single symbol, not a quotient of two quantities.

We use a variant of this notation as the instruction to take the derivative. In particular,

\begin{equation*} \frac{d}{dx}\left[ \Box \right] \end{equation*}

means “take the derivative of the quantity in \(\Box\) with respect to \(x\text{.}\)” For example, we may write \(\frac{d}{dx}[x^2] = 2x\text{.}\)

It is important to note that the independent variable can be different from \(x\text{.}\) If we have \(f(z) = z^2\text{,}\) we then write \(f'(z) = 2z\text{.}\) Similarly, if \(y = t^2\text{,}\) we say \(\frac{dy}{dt} = 2t\text{.}\) And it is also true that \(\frac{d}{dq}[q^2] = 2q\text{.}\) This notation may also be used for second derivatives: \(f''(z) = \frac{d}{dz}\left[\frac{df}{dz}\right] = \frac{d^2 f}{dz^2}\text{.}\)

In what follows, we'll build a repertoire of functions for which we can quickly compute the derivative.

Subsection 2.1.2 Constant, Power, and Exponential Functions

So far, we know the derivative formula for two important classes of functions: constant functions and power functions. If \(f(x) = c\) is a constant function, its graph is a horizontal line with slope zero at every point. Thus, \(\frac{d}{dx}[c] = 0\text{.}\) We summarize this with the following rule.

Constant Functions.

For any real number \(c\text{,}\) if \(f(x) = c\text{,}\) then \(f'(x) = 0\text{.}\)

If \(f(x) = 7\text{,}\) then \(f'(x) = 0\text{.}\) Similarly, \(\frac{d}{dx} [\sqrt{3}] = 0\text{.}\)

In your work in Preview Activity 2.1.1, you conjectured that for any positive integer \(n\text{,}\) if \(f(x) = x^n\text{,}\) then \(f'(x) = nx^{n-1}\text{.}\) This rule can be formally proved for any positive integer \(n\text{,}\) and also for any nonzero real number (positive or negative).

Power Functions.

For any nonzero real number \(n\text{,}\) if \(f(x) = x^n\text{,}\) then \(f'(x) = nx^{n-1}\text{.}\)

Using the rule for power functions, we can compute the following derivatives. If \(g(z) = z^{-3}\text{,}\) then \(g'(z) = -3z^{-4}\text{.}\) Similarly, if \(h(t) = t^{7/5}\text{,}\) then \(\frac{dh}{dt} = \frac{7}{5}t^{2/5}\text{,}\) and \(\frac{d}{dq} [q^{\pi}] = \pi q^{\pi - 1}\text{.}\)

It will be instructive to have a derivative formula for one more type of basic function. For now, we simply state this rule without explanation or justification; we will explore why this rule is true in one of the exercises. And we will encounter graphical reasoning for why the rule is plausible in Preview Activity 2.2.1.

Exponential Functions.

For any positive real number \(a\text{,}\) if \(f(x) = a^x\text{,}\) then \(f'(x) = a^x \ln(a)\text{.}\)

If \(f(x) = 2^x\text{,}\) then \(f'(x) = 2^x \ln(2)\text{.}\) Similarly, for \(p(t) = 10^t\text{,}\) \(p'(t) = 10^t \ln(10)\text{.}\) It is especially important to note that when \(a = e\text{,}\) where \(e\) is the base of the natural logarithm function, we have that

\begin{equation*} \frac{d}{dx} [e^x] = e^x \ln(e) = e^x \end{equation*}

since \(\ln(e) = 1\text{.}\) This is an extremely important property of the function \(e^x\text{:}\) its derivative function is itself!

Note carefully the distinction between power functions and exponential functions: in power functions, the variable is in the base, as in \(x^2\text{,}\) while in exponential functions, the variable is in the power, as in \(2^x\text{.}\) As we can see from the rules, this makes a big difference in the form of the derivative.

Activity 2.1.2.

Use the three rules above to determine the derivative of each of the following functions. For each, state your answer using full and proper notation, labeling the derivative with its name. For example, if you are given a function \(h(z)\text{,}\) you should write “\(h'(z) =\)” or “\(\frac{dh}{dz} =\)” as part of your response.

  1. \(f(t) = \pi\)

  2. \(g(z) = 7^z\)

  3. \(h(w) = w^{3/4}\)

  4. \(p(x) = 3^{1/2}\)

  5. \(r(t) = (\sqrt{2})^t\)

  6. \(s(q) = q^{-1}\)

  7. \(m(t) = \frac{1}{t^3}\)

Hint

  1. Is \(\pi\) a variable or a constant?

  2. Is \(g\) a power or exponential function?

  3. Is \(h\) a power or exponential function?

  4. Is \(3^{1/2}\) a constant or a variable?

  5. \(\sqrt{2}\) is a constant

  6. Remember the notation here means “take the derivative with respect to \(q\) of \(q^{-1}\text{.}\)”

  7. Rewrite the fraction using a negative exponent.

Answer

  1. \(f'(t) = 0\text{.}\)

  2. \(g'(z) = 7^z \ln(7)\text{.}\)

  3. \(h'(w) = \frac{3}{4} w^{-1/4}\text{.}\)

  4. \(\frac{dp}{dx} = 0\text{.}\)

  5. \(r'(t) = (\sqrt{2})^t \ln (\sqrt{2})\text{.}\)

  6. \(\frac{d}{dq}[q^{-1}] = -q^{-2}\text{.}\)

  7. \(\frac{dm}{dt} = -3t^{-4} = -\frac{3}{t^4}\text{.}\)

Solution

  1. \(f(t) = \pi\) is constant, so \(f'(t) = 0\text{.}\)

  2. \(g(z) = 7^z\) is an exponential function, so \(g'(z) = 7^z \ln(7)\text{.}\)

  3. \(h(w) = w^{3/4}\) is a power function, thus \(h'(w) = \frac{3}{4} w^{-1/4}\text{.}\)

  4. \(p(x) = 3^{1/2}\) is constant, and therefore \(\frac{dp}{dx} = 0\text{.}\)

  5. \(r(t) = (\sqrt{2})^t\) is exponential (since \(\sqrt{2}\) is a constant), and so we have \(r'(t) = (\sqrt{2})^t \ln (\sqrt{2})\text{.}\)

  6. \(\frac{d}{dq}[q^{-1}] = -q^{-2}\text{,}\) by the rule for power functions.

  7. \(m(t) = \frac{1}{t^3} = t^{-3}\text{,}\) so \(\frac{dm}{dt} = -3t^{-4} = -\frac{3}{t^4}\text{.}\)

Subsection 2.1.3 Constant Multiples and Sums of Functions

Next we will learn how to compute the derivative of a function constructed as an algebraic combination of basic functions. For instance, we'd like to be able to take the derivative of a polynomial function such as

\begin{equation*} p(t) = 3t^5 - 7t^4 + t^2 - 9\text{,} \end{equation*}

which is a sum of constant multiples of powers of \(t\text{.}\) To that end, we develop two new rules: the Constant Multiple Rule and the Sum Rule.

How is the derivative of \(y = kf(x)\) related to the derivative of \(y = f(x)\text{?}\) Recall that when we multiply a function by a constant \(k\text{,}\) we vertically stretch the graph by a factor of \(|k|\) (and reflect the graph across \(y = 0\) if \(k \lt 0\)). This vertical stretch affects the slope of the graph, so the slope of the function \(y = kf(x)\) is \(k\) times as steep as the slope of \(y = f(x)\text{.}\) Thus, when we multiply a function by a factor of \(k\text{,}\) we change the value of its derivative by a factor of \(k\) as well. 1 ,

The Constant Multiple Rule can be formally proved as a consequence of properties of limits, using the limit definition of the derivative.
The Constant Multiple Rule.

For any real number \(k\text{,}\) if \(f(x)\) is a differentiable function with derivative \(f'(x)\text{,}\) then \(\frac{d}{dx}[k f(x)] = k f'(x)\text{.}\)

In words, this rule says that “the derivative of a constant times a function is the constant times the derivative of the function.”

If \(g(t) = 3 \cdot 5^t\text{,}\) we have \(g'(t) = 3 \cdot 5^t \ln(5)\text{.}\) Similarly, \(\frac{d}{dz} [5z^{-2}] = 5 (-2z^{-3})\text{.}\)

Next we examine a sum of two functions. If we have \(y = f(x)\) and \(y = g(x)\text{,}\) we can compute a new function \(y = (f+g)(x)\) by adding the outputs of the two functions: \((f+g)(x) = f(x) + g(x)\text{.}\) Not only is the value of the new function the sum of the values of the two known functions, but the slope of the new function is the sum of the slopes of the known functions. Therefore 2 , we arrive at the following Sum Rule for derivatives:

Like the Constant Multiple Rule, the Sum Rule can be formally proved as a consequence of properties of limits, using the limit definition of the derivative.
The Sum Rule.

If \(f(x)\) and \(g(x)\) are differentiable functions with derivatives \(f'(x)\) and \(g'(x)\) respectively, then \(\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)\text{.}\)

In words, the Sum Rule tells us that “the derivative of a sum is the sum of the derivatives.” It also tells us that a sum of two differentiable functions is also differentiable. Furthermore, because we can view the difference function \(y = (f-g)(x) = f(x) - g(x)\) as \(y = f(x) + (-1 \cdot g(x))\text{,}\) the Sum Rule and Constant Multiple Rules together tell us that \(\frac{d}{dx}[f(x) + (-1 \cdot g(x))] = f'(x) - g'(x)\text{,}\) or that “the derivative of a difference is the difference of the derivatives.” We can now compute derivatives of sums and differences of elementary functions.

Using the sum rule, \(\frac{d}{dw} (2^w + w^2) = 2^w \ln(2) + 2w\text{.}\) Using both the sum and constant multiple rules, if \(h(q) = 3q^6 - 4q^{-3}\text{,}\) then \(h'(q) = 3 (6q^5) - 4(-3q^{-4}) = 18q^5 + 12q^{-4}\text{.}\)

Activity 2.1.3.

Use only the rules for constant, power, and exponential functions, together with the Constant Multiple and Sum Rules, to compute the derivative of each function below with respect to the given independent variable. Note well that we do not yet know any rules for how to differentiate the product or quotient of functions. This means that you may have to do some algebra first on the functions below before you can actually use existing rules to compute the desired derivative formula. In each case, label the derivative you calculate with its name using proper notation such as \(f'(x)\text{,}\) \(h'(z)\text{,}\) \(dr/dt\text{,}\) etc.

  1. \(f(x) = x^{5/3} - x^4 + 2^x\)

  2. \(g(x) = 14e^x + 3x^5 - x\)

  3. \(h(z) = \sqrt{z} + \frac{1}{z^4} + 5^z\)

  4. \(r(t) = \sqrt{53} \, t^7 - \pi e^t + e^4\)

  5. \(s(y) = (y^2 + 1)(y^2 - 1)\)

  6. \(q(x) = \frac{x^3 - x + 2}{x}\)

  7. \(p(a) = 3a^4 - 2a^3 + 7a^2 - a + 12\)

Hint

  1. Use the sum rule.

  2. Use the sum rule together with the constant multiple rule.

  3. How can you rewrite \(\sqrt{z}\) using exponents?

  4. Is \(e^4\) a constant or variable?

  5. Expand the product before attempting to find the derivative.

  6. Rewrite the single fraction as a sum of three fractions, and simplify.

  7. Note that “\(a\)” is the independent variable.

Answer

  1. \(f'(x) = \frac{5}{3}x^{2/3} - 4 x^3 + 2^x \ln(2)\text{.}\)

  2. \(g'(x) = 14e^x + 3 \cdot 5x^4 - 1\text{.}\)

  3. \(h'(z) = \frac{1}{2}z^{-1/2} - 4z^{-5} + 5^z \ln(5)\text{.}\)

  4. \(\frac{dr}{dt} = \sqrt{53} \cdot 7 t^6 - \pi e^t\text{.}\)

  5. \(\frac{ds}{dy} = 4y^3\text{.}\)

  6. \(q'(x) = 2x - 2x^{-2}\text{.}\)

  7. \(p'(a) = 12a^3 - 6 a^2 + 14a - 1\text{.}\)

Solution

  1. \(f(x) = x^{5/3} - x^4 + 2^x\text{,}\) so by the sum rule, \(f'(x) = \frac{5}{3}x^{2/3} - 4 x^3 + 2^x \ln(2)\text{.}\)

  2. \(g(x) = 14e^x + 3x^5 - x\text{,}\) so by the sum and constant multiple rules, \(g'(x) = 14e^x + 3 \cdot 5x^4 - 1\text{.}\)

  3. \(h(z) = \sqrt{z} + \frac{1}{z^4} + 5^z = z^{1/2} + z^{-4} + 5^z\text{,}\) thus \(h'(z) = \frac{1}{2}z^{-1/2} - 4z^{-5} + 5^z \ln(5)\text{.}\)

  4. Since \(r(t) = \sqrt{53} \, t^7 - \pi e^t + e^4\) and \(\sqrt{53}\text{,}\) \(\pi\text{,}\) and \(e^4\) are all constants, it follows from the sum and constant multiple rules, as well as the derivative of a constant rule, that \(\frac{dr}{dt} = \sqrt{53} \cdot 7 t^6 - \pi e^t\text{.}\) (Note particularly that \(\frac{d}{dt}[e^4] = 0\) since \(e^4\) is constant.

  5. \(s(y) = (y^2 + 1)(y^2 - 1)= y^4 - 1\text{,}\) thus \(\frac{ds}{dy} = 4y^3\text{.}\)

  6. \(q(x) = \frac{x^3 - x + 2}{x} = \frac{x^3}{x} - \frac{x}{x} + \frac{2}{x} = x^2 - 1 + 2x^{-1}\text{.}\) Now it follows that \(q'(x) = 2x - 2x^{-2}\text{.}\)

  7. \(p(a) = 3a^4 - 2a^3 + 7a^2 - a + 12\text{,}\) so \(p'(a) = 12a^3 - 6 a^2 + 14a - 1\text{.}\)

In the same way that we have shortcut rules to help us find derivatives, we introduce some language that is simpler and shorter. Often, rather than say “take the derivative of \(f\text{,}\)” we'll instead say simply “differentiate \(f\text{.}\)” Similarly, if the derivative exists at a point, we say “\(f\) is differentiable at that point,” or that \(f\) can be differentiated.

As we work with the algebraic structure of functions, it is important to develop a big picture view of what we are doing. Here, we make several general observations based on the rules we have so far.

  • The derivative of any polynomial function will be another polynomial function, and that the degree of the derivative is one less than the degree of the original function. For instance, if \(p(t) = 7t^5 - 4t^3 + 8t\text{,}\) \(p\) is a degree 5 polynomial, and its derivative, \(p'(t) = 35t^4 - 12t^2 + 8\text{,}\) is a degree 4 polynomial.
  • The derivative of any exponential function is another exponential function: for example, if \(g(z) = 7 \cdot 2^z\text{,}\) then \(g'(z) = 7 \cdot 2^z \ln(2)\text{,}\) which is also exponential.
  • We should not lose sight of the fact that all of the meaning of the derivative that we developed in Chapter 1 still holds. The derivative measures the instantaneous rate of change of the original function, as well as the slope of the tangent line at any selected point on the curve.
Activity 2.1.4.

Each of the following questions asks you to use derivatives to answer key questions about functions. Be sure to think carefully about each question and to use proper notation in your responses.

  1. Find the slope of the tangent line to \(h(z) = \sqrt{z} + \frac{1}{z}\) at the point where \(z = 4\text{.}\)

  2. A population of cells is growing in such a way that its total number in millions is given by the function \(P(t) = 2(1.37)^t + 32\text{,}\) where \(t\) is measured in days.

    1. Determine the instantaneous rate at which the population is growing on day 4, and include units on your answer.

    2. Is the population growing at an increasing rate or growing at a decreasing rate on day 4? Explain.

  3. Find an equation for the tangent line to the curve \(p(a) = 3a^4 - 2a^3 + 7a^2 - a + 12\) at the point where \(a=-1\text{.}\)

  4. What is the difference between being asked to find the slope of the tangent line (asked in (a)) and the equation of the tangent line (asked in (c))?

Hint

  1. How would \(h'(z)\) help you answer the question?

  2. Think about finding both \(P'(t)\) and \(P''(t)\text{.}\)

  3. What two important pieces of information do you need to know to determine the equation of a line?

  4. What information do you find in both (a) and (c)?

Answer

  1. \(h'(4) = \frac{3}{16}\text{.}\)

  2. (i.)\(P'(4) = 2(1.37)^4 \ln(1.37) \approx 2.218\) million cells per day; (ii.) the population is growing at an increasing rate.

  3. \(y - 25 = -33(a+1)\text{.}\)

  4. The slope is a number, while the equation is, well, an equation.

Solution

  1. Note that since \(h(z) = z^{1/2} + z^{-1}\text{,}\) we have \(h'(z) = \frac{1}{2}z^{-1/2} - z^{-2}\text{.}\) Thus, \(h'(4) = \frac{1}{2(4)^{1/2}} - \frac{1}{4^2} = \frac{1}{4} - \frac{1}{16} = \frac{3}{16}\text{.}\) Thus, the slope of the tangent line to \(h(z)\) at the point where \(z = 4\) is \(\frac{3}{16}\text{.}\)

  2. For (i.), note that \(P'(t) = 2(1.37)^t \ln(1.37)\text{,}\) and therefore \(P'(4) = 2(1.37)^4 \ln(1.37) \approx 2.218\) million cells per day. For (ii.), we can compute \(P''(t)\) and find that \(P''(t) = 2(1.37)^t \ln(1.37) \ln(1.37)\) and therefore find that \(P''(4) \approx 0.69825\text{.}\) Since \(P''(4)\) is positive, this tells us that the instantaneous rate of change \(P'(t)\) is increasing at \(t = 4\text{,}\) and thus the population is growing at an increasing rate.

  3. Given \(p(a) = 3a^4 - 2a^3 + 7a^2 - a + 12\text{,}\) first observe that \(p(-1) = 3 + 2 + 7 + 1 + 12 = 25\text{,}\) so the tangent line will pass through \((-1,25)\text{.}\) Further, since \(p'(a) = 12a^3 - 6a^2 + 14a - 1\text{,}\) we have \(p'(-1) = -12 - 6 - 14 - 1 = -33\text{,}\) which is the slope of the tangent line. The equation of the tangent line is therefore \(y - 25 = -33(a+1)\text{.}\)

  4. The biggest difference is that (a) asks for the slope of the tangent line, while (c) asks for the equation of the tangent line. The latter requires the slope of the tangent line, but the slope and equation are two different entities.

Subsection 2.1.4 Summary

  • Given a differentiable function \(y = f(x)\text{,}\) we can express the derivative of \(f\) in several different notations: \(f'(x)\text{,}\) \(\frac{df}{dx}\text{,}\) \(\frac{dy}{dx}\text{,}\) and \(\frac{d}{dx}[f(x)]\text{.}\)

  • The limit definition of the derivative leads to patterns among certain families of functions that enable us to compute derivative formulas without resorting directly to the limit definition. For example, if \(f\) is a power function of the form \(f(x) = x^n\text{,}\) then \(f'(x) = nx^{n-1}\) for any real number \(n\) other than 0. This is called the Rule for Power Functions.

  • We have stated a rule for derivatives of exponential functions in the same spirit as the rule for power functions: for any positive real number \(a\text{,}\) if \(f(x) = a^x\text{,}\) then \(f'(x) = a^x \ln(a)\text{.}\)

  • If we are given a constant multiple of a function whose derivative we know, or a sum of functions whose derivatives we know, the Constant Multiple and Sum Rules make it straightforward to compute the derivative of the overall function. More formally, if \(f(x)\) and \(g(x)\) are differentiable with derivatives \(f'(x)\) and \(g'(x)\) and \(a\) and \(b\) are constants, then

    \begin{equation*} \frac{d}{dx} \left[af(x) + bg(x)\right] = af'(x) + bg'(x)\text{.} \end{equation*}

Exercises 2.1.5 Exercises

NOTE: For the WeBWorK problems, you should work the problems using only the derivative rules we have covered. You may need to perform some algebra in order to use only the results we have discussed to this point.

1.

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

  1. Let \(h\) be the new function defined by the rule \(h(x) = 3f(x) - 4g(x)\text{.}\) Determine \(h(2)\) and \(h'(2)\text{.}\)

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

  3. Let \(p\) be the function defined by the rule \(p(x) = -2f(x) + \frac{1}{2}g(x)\text{.}\) Is \(p\) increasing, decreasing, or neither at \(a = 2\text{?}\) Why?

  4. Estimate the value of \(p(2.03)\) by using the local linearization of \(p\) at the point \((2,p(2))\text{.}\)

Answer

  1. \(h(2) = 27\text{;}\) \(h'(2) = -19/2\text{.}\)

  2. \(L(x) = 27 - \frac{19}{2}(x-2)\text{.}\)

  3. \(p\) is increasing at \(x=2\text{.}\)

  4. \(p(2.03) \approx -11.56\text{.}\)

Solution

  1. First, since \(h(x) = 3f(x) - 4g(x)\text{,}\) we know \(h(2) = 3f(2) - 4g(2) = 3(5) - 4(-3) = 27\text{.}\) Next, by the sum and constant multiple rules, \(h'(x) = 3f'(x) - 4g'(x)\text{.}\) Hence, using the given derivative values, \(h'(2) = 3f'(2) - 4g'(2) = 3(-1/2) - 4(2) = -19/2\text{.}\)

  2. The tangent line at the point \((2,h(2))\) has equation \(L(x) = h(2) + h'(2)(x-2)\text{.}\) Using the information determined in (a), we see \(L(x) = 27 - \frac{19}{2}(x-2)\text{.}\)

  3. Since \(p(x) = -2f(x) + \frac{1}{2}g(x)\text{,}\) it follows that \(p'(x) = -2f'(x) + \frac{1}{2}g'(x)\) by the sum and constant multiple rules. Using the given function and derivative values for \(f\) and \(g\) at \(x=2\text{,}\) we see that \(p(2) = -2f(2) + \frac{1}{2}g(2) = -2(5) + \frac{1}{2}(-3) = -\frac{23}{2}\) and \(p'(2) = -2(-\frac{1}{2}) + \frac{1}{2}(2)) = 2\text{.}\) Since \(p'(2) = 2 \gt 0\text{,}\) we conclude that \(p\) is increasing at \(x=2\text{.}\)

  4. First, the linearization of \(p\) at \(a = 2\) is \(L(x) = p(2) + p'(2)(x-2) = -\frac{23}{2} + 2(x-2)\text{.}\) Using the linearization to estimate \(p(2.03)\text{,}\) we find that

    \begin{equation*} p(2.03) \approx L(2.03) = -\frac{23}{2} + 2(2.03-2) = -11.5 + 2(0.03) = -11.56\text{.} \end{equation*}
2.

Let functions \(p\) and \(q\) be the piecewise linear functions given by their respective graphs in Figure 2.1.6. Use the graphs to answer the following questions.

Figure 2.1.6. The graphs of \(p\) (in blue) and \(q\) (in green).

  1. At what values of \(x\) is \(p\) not differentiable? At what values of \(x\) is \(q\) not differentiable? Why?

  2. Let \(r(x) = p(x) + 2q(x)\text{.}\) At what values of \(x\) is \(r\) not differentiable? Why?

  3. Determine \(r'(-2)\) and \(r'(0)\text{.}\)

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

Answer

  1. \(p\) is not differentiable at \(x=-1\) and \(x=1\text{;}\) \(q\) is not differentiable at \(x=-1\) and \(x=1\text{.}\)

  2. \(r\) is not differentiable at \(x=-1\) and \(x=1\text{.}\)

  3. \(r'(-2) = 4\text{;}\) \(r'(0) = 1\text{.}\)

  4. \(y = 4\text{.}\)

Solution

  1. The function \(p\) is differentiable at points where its graph is locally linear or smooth. Since there are sharp corners in the graph of \(p\) at \(x=-1\) and \(x=1\text{,}\) we see that \(p\) is not differentiable at those points. Similarly, the sharp corners in \(q\) at \(x=-1\) and \(x=1\) mean that \(q\) is not differentiable at these points, either.

  2. In order for \(r\) to be differentiable at a point, both \(p\) and \(q\) must be differentiable at the point. So \(r\) is not differentiable at \(x=-1\) and \(x=1\text{.}\)

  3. Using the sum and scalar multiple rules, we have \(r'(x) = p'(x) + 2q'(x)\text{.}\) The slope of \(p\) at \(x=-2\) is \(-2\) and the slope of \(q\) at \(x=-2\) is \(3\text{,}\) so \(p'(-2) = -2\) and \(q'(-2) = 3\text{.}\) It follows that \('(-2) = p'(-2) + 2q'(-2) = (-2) + 2(3) = 4\text{.}\) Similarly, the slope of \(p\) at \(x=0\) is \(1\) and the slope of \(q\) at \(x=0\) is \(0\text{,}\) so \(p'(0) = 1\) and \(q'(0) = 0\text{.}\) It follows that \(r'(0) = p'(0) + 2q'(0) = (1) + 2(0) = 1\text{.}\)

  4. Note that \(r(2) = p(2)+2q(2) = 2+2(1) = 4\) and \(r'(2) = p'(2)+2q'(2) = 2+2(-1) = 0\text{.}\) Thus, the line tangent to the graph of \(r\) at \(x=2\) is \(y = r(2) + r'(2)(x-2) = 4\text{.}\)

3.

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

  1. Let \(w(t) = 3t^t - 2\arccos(t)\text{.}\) Determine \(w'(t)\text{.}\)

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

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

Answer

  1. \(w'(t) = 3t^t(\ln(t) + 1) + 2\frac{1}{\sqrt{1-t^2}}\text{.}\)

  2. \(L(t) = (\frac{3}{\sqrt{2}} - \frac{2\pi}{3}) + (\frac{3}{\sqrt{2}}(\ln(\frac{1}{2}) + 1) + \frac{4}{\sqrt{3}})(t-2)\text{.}\)

  3. \(v\) is increasing at \(t = \frac{1}{2}\text{.}\)

Solution

  1. We note that \(w\) has the structure \(w(t) = 3r(t) - 2s(t)\text{,}\) so by the sum and constant multiple rules, \(w'(t) = 3r'(t) - 2s'(t)\text{.}\) Applying the given information about \(r'\) and \(s'\text{,}\) \(w'(t) = 3t^t(\ln(t) + 1) + 2\frac{1}{\sqrt{1-t^2}}\text{.}\)

  2. We see that \(w(\frac{1}{2}) = 3(\frac{1}{2})^{\frac{1}{2}} - 2\arccos(\frac{1}{2}) = 3 \cdot \frac{1}{\sqrt{2}} - 2 \cdot \frac{\pi}{3} = \frac{3}{\sqrt{2}} - \frac{2\pi}{3}\text{.}\) Similarly,

    \begin{align*} w'\left(\frac{1}{2}\right) &= 3\left(\frac{1}{2}\right)^{\frac{1}{2}}\left(\ln\left(\frac{1}{2}\right) + 1\right) + 2\frac{1}{\sqrt{1-(\frac{1}{2})^2}}\\ &= \frac{3}{\sqrt{2}}\left(\ln\left(\frac{1}{2}\right) + 1\right) + \frac{4}{\sqrt{3}}\text{.} \end{align*}

    Using these values, we find that the tangent line to \(w\) at the point \((\frac{1}{2},w(\frac{1}{2})\) is

    \begin{equation*} L(t) = w\left(\frac{1}{2}\right) + w'\left(\frac{1}{2}\right)\left(t-\frac{1}{2}\right) = \left(\frac{3}{\sqrt{2}} - \frac{2\pi}{3}\right) + \left(\frac{3}{\sqrt{2}}\left(\ln\left(\frac{1}{2}\right) + 1\right) + \frac{4}{\sqrt{3}}\right)(t-2)\text{.} \end{equation*}

  3. To determine whether \(v\) is increasing or decreasing at \(t = \frac{1}{2}\text{,}\) we compute \(v'(\frac{1}{2})\text{.}\) First, \(v'(t) = t^t(\ln(t) + 1) + \frac{1}{\sqrt{1-t^2}}\text{.}\) Hence,

    \begin{equation*} v'\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^{\frac{1}{2}}\left(\ln\left(\frac{1}{2}\right) + 1\right) + \frac{1}{\sqrt{1-(\frac{1}{2})^2}}\text{.} \end{equation*}

    Since \((\ln(\frac{1}{2})+1)\) is positive, it's clear that \(v'(\frac{1}{2}) \gt 0\text{,}\) and thus \(v\) is increasing at \(t = \frac{1}{2}\text{.}\)

4.

Let \(f(x) = a^x\text{.}\) The goal of this problem is to explore how the value of \(a\) affects the derivative of \(f(x)\text{,}\) without assuming we know the rule for \(\frac{d}{dx}[a^x]\) that we have stated and used in earlier work in this section.

  1. Use the limit definition of the derivative to show that

    \begin{equation*} f'(x) = \lim_{h \to 0} \frac{a^x \cdot a^h - a^x}{h}\text{.} \end{equation*}

  2. Explain why it is also true that

    \begin{equation*} f'(x) = a^x \cdot \lim_{h \to 0} \frac{a^h - 1}{h}\text{.} \end{equation*}

  3. Use computing technology and small values of \(h\) to estimate the value of

    \begin{equation*} L = \lim_{h \to 0} \frac{a^h - 1}{h} \end{equation*}

    when \(a = 2\text{.}\) Do likewise when \(a = 3\text{.}\)

  4. Note that it would be ideal if the value of the limit \(L\) was \(1\text{,}\) for then \(f\) would be a particularly special function: its derivative would be simply \(a^x\text{,}\) which would mean that its derivative is itself. By experimenting with different values of \(a\) between \(2\) and \(3\text{,}\) try to find a value for \(a\) for which

    \begin{equation*} L = \lim_{h \to 0} \frac{a^h - 1}{h} = 1\text{.} \end{equation*}

  5. Compute \(\ln(2)\) and \(\ln(3)\text{.}\) What does your work in (b) and (c) suggest is true about \(\frac{d}{dx}[2^x]\) and \(\frac{d}{dx}[3^x]\text{?}\)

  6. How do your investigations in (d) lead to a particularly important fact about the function \(f(x) = e^x\text{?}\)

Answer
  1. \begin{align*} f'(x) &= \lim_{h \to 0} \frac{a^{x+h}-a^x}{h}\\ &= \lim_{h \to 0} \frac{a^{x} \cdot a^{h} - a^x}{h}\\ &= \lim_{h \to 0} \frac{a^{x}(a^{h}-1)}{h}\text{,} \end{align*}

  2. Since \(a^x\) does not depend at all on \(h\text{,}\) we may treat \(a^x\) as constant in the noted limit and thus write the value \(a^x\) in front of the limit being taken.

  3. When \(a = 2\text{,}\) \(L \approx 0.6931\text{;}\) when \(a = 3\text{,}\) \(L \approx 1.0986\text{.}\)

  4. \(a \approx 2.71828\) (for which \(L \approx 1.000\))

  5. \(\frac{d}{dx}[2^x] = 2^x \cdot \ln(2)\) and \(\frac{d}{dx}[3^x] = 3^x \cdot \ln(3)\)

  6. \begin{equation*} \frac{d}{dx}[e^x] = e^x\text{.} \end{equation*}
Solution

  1. By definition,

    \begin{equation*} f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\text{.} \end{equation*}

    Using the fact that \(f(x)=a^x\) along with \(a^{x+h} = a^x \cdot a^h\text{,}\) we have

    \begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ &= \lim_{h \to 0} \frac{a^{x+h}-a^x}{h}\\ &= \lim_{h \to 0} \frac{a^{x} \cdot a^{h} - a^x}{h}\\ &= \lim_{h \to 0} \frac{a^{x}(a^{h}-1)}{h}\text{,} \end{align*}

    where the last equality follows by removing the common factor of \(a^x\) from each term in the numerator.

  2. Since \(a^x\) does not depend at all on \(h\text{,}\) we may treat \(a^x\) as constant in the noted limit. Thus,

    \begin{equation*} f'(x) = \lim_{h \to 0} \frac{a^{x}(a^{h}-1)}{h} = a^{x} \cdot \lim_{h \to 0} \frac{a^{h}-1}{h}\text{.} \end{equation*}

    Note that this tells us that the derivative of \(a^x\) is some constant multiple of \(a^x\) (and that the constant's values must be \(\lim_{h \to 0} \frac{a^{h}-1}{h}\text{.}\))

  3. Using \(h = \pm 0.0001\text{,}\) it appears that when \(a = 2\text{,}\) \(L \approx 0.6931\text{,}\) while when \(a = 3\text{,}\) \(L \approx 1.0986\text{.}\)

  4. By experimenting, we see first that \(a \approx 2.7\) (for which \(L \approx 0.993\text{,}\) so \(a\) needs to be slightly bigger than \(2.7\)). Continuing to try values of \(a\) that result in \(L\) being closer and closer to \(1\text{,}\) we find \(a \approx 2.71828\) (for which \(L \approx 1.000\))

  5. We note that \(\ln(2) \approx 0.6931\) and \(\ln(3) \approx 1.0986\text{.}\) Since these values correspond to the values of

    \begin{equation*} \lim_{h \to 0} \frac{a^{h}-1}{h} \end{equation*}

    when \(a = 2\) and \(a = 3\text{,}\) we conjecture that

    \begin{equation*} \frac{d}{dx}[2^x] = 2^x \cdot \ln(2) \end{equation*}

    and

    \begin{equation*} \frac{d}{dx}[3^x] = 3^x \cdot \ln(3) \end{equation*}

  6. Our work suggests that \(\frac{d}{dx}[a^x] = a^x \cdot \ln(a)\text{,}\) and when \(a = e \approx 2.71828\text{,}\) it follows that

    \begin{equation*} \frac{d}{dx}[e^x] = e^x \cdot \ln(e) = e^x \end{equation*}

    Thus, \(e^x\) is a very special function: its derivative is itself!

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