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Discover Calculus Single-Variable Calculus Topics with Motivating Activities

Peter Keep

Section 2.3 Some Early Derivative Rules

We are going to break this topic into two parts:
  1. We will try to find some common patterns or connections between derivatives and specific functions. For instance, when we use Definition 2.1.2 The Derivative Function to build a derivative, are there patterns in the work of evaluating that limit that will allow us to get through the limit work quickly? Can we group some functions together based on how we might deal with the limit?
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  2. We will try to think about derivatives a bit more generally and show how we can build some basic properties to help us think about differentiating variations of the functions that we recognize.
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Subsection Derivatives of Common Functions

Activity 2.3.1. Derivatives of Power Functions.

We’re going to do a bit of pattern recognition here, which means that we will need to differentiate several different power functions. For our reference, a power function (in general) is a function in the form \(f(x)=a(x^n)\) where \(n\) and \(a\) are real numbers, and \(a\neq 0\text{.}\)
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Let’s begin our focus on the power functions \(x^2\text{,}\) \(x^3\text{,}\) and \(x^4\text{.}\) We’re going to use Definition 2.1.2 The Derivative Function a lot, so feel free to review it before we begin.
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(a)
Find \(\dfrac{d}{dx}\left(x^2\right)\text{.}\) As a brief follow up, compare this to the derivative \(j'(x)\) that you found in Activity 2.1.3 Calculating a Bunch of Slopes. Why are they the same? What does the difference, the \(-4\text{,}\) in the \(j(x)\) function do to the graph of it (compared to the graph of \(x^2\)) and why does this not impact the derivative?
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Hint.
Remember that the graph of \(x^2-4\) has the same shape as the graph of \(x^2\text{,}\) but has been shifted down by 4 units. Why does this vertical shift not change the value of the derivative at any \(x\)-value?
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(b)
Find \(\dfrac{d}{dx}\left(x^3\right)\text{.}\)
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Hint.
Remember that \((x+\Delta x)^3 = (x+\Delta x)(x+\Delta x)(x+\Delta x)\)
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(c)
Find \(\dfrac{d}{dx}\left(x^4\right)\text{.}\)
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Hint.
Remember that \((x+\Delta x)^4 = (x+\Delta x)(x+\Delta x)(x+\Delta x)(x+\Delta x)\)
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(d)
Notice that in these derivative calculations, the main work is in multiplying \((x+\Delta x)^n\text{.}\) Look back at the work done in all three of these derivative calculations and find some unifying steps to describe how you evaluate the limit/calculate the derivative after this tedious multiplication was finished. What steps did you do? Is it always the same thing?
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Another way of stating this is: if I told you that I knew what \((x+\Delta x)^5\) was, could you give me some details on how the derivative limit would be finished?
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(e)
Finish the following derivative calculation:
\begin{align*} \frac{d}{dx}\left(x^5\right) \amp =\lim_{\Delta x\to 0} \left(\frac{(x+\Delta x)^5 - x^5}{\Delta x}\right) \\ \amp = \lim_{\Delta x\to 0} \left(\frac{(x^5+5x^4\Delta x + 10x^3\Delta x^2 + 10x^2\Delta x^3 + 5 x\Delta x^4 + \Delta x^5)-x^5}{\Delta x}\right)\\ \amp = \leadsto ... \end{align*}
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(f)
Make a conjecture about the derivative of a power function in general, \(\frac{d}{dx}(x^n)\text{.}\)
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Something to notice here is that the calculation in this limit is really dependent on knowing what \((x+\Delta x)^n\) is. When \(n\) is an integer with \(n\geq 2\text{,}\) this really just translates to multiplication. If we can figure out how to multiply \((x+\Delta x)^n\) in general, then this limit calculation will be pretty easy to do. We noticed that:
  1. The first term of that multiplication will combine with the subtraction of \(x^n\) in the numerator and subtract to 0.
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  2. The rest of the terms in the multiplication have at least one copy of \(\Delta x\text{,}\) and so we can factor out \(\Delta x\) and “cancel” it with the \(\Delta x\) in the denominator.
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  3. Once this is done, we’ve escaped the portion of the limit that was giving us the \(\frac{0}{0}\) indeterminate form, and so we can evaluate the limit as \(\Delta x\to 0\text{.}\) The result is just whatever terms still have at least one remaining copy of \(\Delta x\) in them “go to” 0, and we’re left with just the terms that do not have any copies of \(\Delta x\) in them.
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There is a really nice theorem that helps to understand how the \((x+\Delta x)^n\) multiplication goes (for different positive integer exponents, \(n\)) called the Binomial Theorem, but we won’t introduce it here, fully. What we’ll do, instead, is summarize the results:
  1. In each term, the exponents on \(x\) count down from \(n\) to \(0\text{.}\)
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  2. Similarly, in each term, the exponents on \(\Delta x\) will count up from \(0\) to \(n\text{.}\)
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  3. The coefficients for each term are generated from a very cool pattern, displayed below.
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Triangle with the binomial coefficients, ending with the row that begins 1, 8, 28, ...
Figure 2.3.1.
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In each row, the coefficients are build by the sum of the two coefficients above it. We have highlighted the 5th row, in order to compare it to the work we did in Activity 2.3.1.
\begin{equation*} (x+\Delta x)^5 = x^5+5x^4\Delta x + 10x^3\Delta x^2 + 10x^2\Delta x^3 + 5 x\Delta x^4 + \Delta x^5 \end{equation*}
We can see the coefficients on each term match the highlighted row!
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Triangle with the binomial coefficients, ending with the row that begins 1, 8, 28, ... In each row, the first term (1) is red. The second number (1, 2, 3, 4, ..., 8) is red. The remaining numbers are blue.
Figure 2.3.2.
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We can now think of each row, and figure out what was happening in the limit work we did.
  1. The first term of that multiplication will combine with the subtraction of \(x^n\) in the numerator and subtract to 0.
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    We can designate this in the triangle by shading in the first term in red.
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  2. The rest of the terms in the multiplication have at least one copy of \(\Delta x\text{,}\) and so we can factor out \(\Delta x\) and “cancel” it with the \(\Delta x\) in the denominator.
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    We can designate the terms in the triangle without any remaining \(\Delta x\)’s in them by shading the second term in green. The rest, that have at least one \(\Delta x\) in it, are shaded in blue.
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  3. Once this has done, we’ve escaped the portion of the limit that was giving us the \(\frac{0}{0}\) indeterminate form, and so we can evaluate the limit as \(\Delta x\to 0\text{.}\) The result is just that whatever terms still have at least one remaining copy of \(\Delta x\) in it “go to” 0, and we’re left with just the terms that do not have any copies of \(\Delta x\) in them.
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    Note that all of the blue terms go to \(0\) in the limit, and we are left with the green terms, but without any \(\Delta x\text{.}\) Since these are the second terms, we have the coefficient and one fewer exponent on \(x\text{.}\)
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We have shown that this is true for \(n=2,3,4,...\text{,}\) but this is also true for any value of \(n\) (including \(n=1\text{,}\) non-integers, and non-positives). We will prove this more formally later (in Section 3.3), and until then we will be free to use this result.
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Example 2.3.4.

Let’s confirm this Power Rule for two examples that we are familiar with.
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(a)
Find the derivative \(\frac{d}{dx}\left(\sqrt{x}\right)\) using the limit definition of the derivative function. Note that \(\sqrt{x}= x^{1/2}\) and \(\frac{1}{2\sqrt{x}} = \frac{1}{2}x^{-1/2}\text{.}\)
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(b)
Find the derivative \(\frac{d}{dx}\left(\frac{1}{x}\right)\) using the limit definition of the derivative function. Note that \(\frac{1}{x}= x^{-1}\) and \(-\frac{1}{x^2} = -x^{-2}\text{.}\)
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In this activity, we also found one other result.
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Activity 2.3.2. Derivatives of Trigonometric Functions.

Let’s try to think through the derivatives of \(y=\sin(\theta)\) and \(y=\cos(\theta)\text{.}\) In this activity, we’ll look at graphs and try to collect some information about the derivative functions. We’ll be practicing our interpretations, so if you need to brush up on Section 2.2 before we start, that’s fine!
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(a)
The following plot includes both the graph of \(y=\sin(x)\text{,}\) and the line tangent to \(y=\sin(x)\text{.}\) Watch as the point where we build the tangent line moves along the graph, between \(x=-2\pi\) and \(x=2\pi\text{.}\)
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Collect as much information about the derivative, \(y'\text{,}\) as you can. What do you know about it? What are some facts about the slopes of the tangent lines in this animation?
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Hint.
What kinds of values do the slopes take? Are there some values that these slopes never will be? Can you find any special points on this graph where you can actually tell what the slope is?
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(b)
We’re going to get more specific here: let’s find the coordinates of points that are on both the graph of \(y=\sin(x)\) and it’s derivative \(y'\text{.}\) Remember, to get the values for \(y'\text{,}\) we’re really looking at the slope of the tangent line at that point.
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(c)
Let’s repeat this process using the \(y=\cos(x)\) function instead.
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The following plot includes both the graph of \(y=\cos(x)\text{,}\) and the line tangent to \(y=\cos(x)\text{.}\) Watch as the point where we build the tangent line moves along the graph, between \(x=-2\pi\) and \(x=2\pi\text{.}\)
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Collect as much information about the derivative, \(y'\text{,}\) as you can. What do you know about it? What are some facts about the slopes of the tangent lines in this animation?
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Hint.
What kinds of values do the slopes take? Are there some values that these slopes never will be? Can you find any special points on this graph where you actually can tell what the slope is?
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(d)
We’re going to get more specific here: let’s find the coordinates of points that are on both the graph of \(y=\cos(x)\) and it’s derivative \(y'\text{.}\) Remember, to get the values for \(y'\text{,}\) we’re really looking at the slope of the tangent line at that point.
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Proof.

In order to show why \(\dfrac{d}{d\theta}\left(\sin(\theta)\right) = \cos(\theta)\) and \(\dfrac{d}{dx}\left(\cos(\theta)\right) = -\sin(\theta)\text{,}\) we will work with the limit definitions of both. Consider both:
\begin{align*} \frac{d}{d\theta}\left(\sin(\theta)\right) \amp = \lim_{\Delta \theta \to 0}\left(\frac{\sin(\theta+\Delta \theta)-\sin(\theta)}{\Delta \theta}\right)\\ \frac{d}{d\theta}\left(\cos(\theta)\right) \amp = \lim_{\Delta \theta \to 0}\left(\frac{\cos(\theta+\Delta \theta)-\cos(\theta)}{\Delta \theta}\right) \end{align*}
Our goal is to rewrite the numerators in both of these limits as something more usable. So far, we have been evaluating these types of limits (First Indeterminate Forms) using some algebraic manipulations. Instead of using algebra, we will use geometry.
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Consider the unit circle below. We have plotted the angle \(\theta\) and are reminded that the point on the circle that corresponds with the terminal side of the angle \(\theta\) has coordinates \((\cos(\theta),\sin(\theta))\text{.}\) We can label the sides of the triangle pictured below.
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A unit circle, with an angle theta placed in the first quadrant. A triangle is formed using the angle, where the hypotenuse is the radius of the circle (1) the height is sin(theta) and the base is cos(theta).
Figure 2.3.7. Unit circle with the angle \(\theta\text{.}\)
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Now we consider a second point on the circle, this one formed by the terminal side of the angle \((\theta+\Delta \theta)\text{.}\) This point has coordinates \((\cos(\theta+\Delta \theta), \sin(\theta+\Delta \theta))\text{.}\) Note below, we want to find expressions for \(\sin(\theta+\Delta\theta)-\sin(\theta)\) and \(\cos(\theta+\Delta\theta) - \cos(\theta)\text{.}\) We can find these geometrically.
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A triangle on a unit circle, with standard lengths 1, sin(theta), and cos(theta). Another point is labeled on the unit circle. There is a triangle formed by the line connecting the two points, and then the vertical and horizontal components of the distance between the two points. The vertical distance is labeled sin(theta+Delta theta)-sin(theta) and the horizontal distance is labeled cos(theta)-cos(theta+Delta theta). The hypotenuse is labeled h, but it is very close to the same as the arclength between the two points, labeled Delta theta.
Figure 2.3.8. Angles \(\theta\) and \(\theta+\Delta\theta\text{.}\)
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Note, then, that the two triangles look to be similar triangles. We can say, then, that:
\begin{align*} \sin(\theta+\Delta\theta)-\sin(\theta)\amp \approx h\cos(\theta) \\ \cos(\theta)-\cos(\theta+\Delta\theta)\amp \approx h\sin(\theta) \end{align*}
In fact, we will find that in the limit as \(\Delta \theta\to 0\text{,}\) the length \(h\) matches the arc length \(\Delta \theta\) perfectly, and thus lays at a right angle to the terminal side of the angle \(\theta+\Delta \theta\text{.}\) Since as \(\Delta \theta\to 0\) we have \(h\to\Delta \theta\text{,}\) we can say:
\begin{align*} (\sin(\theta+\Delta\theta)-\sin(\theta))\amp\to \Delta \theta \cos(\theta)\\ (\cos(\theta)-\cos(\theta+\Delta\theta))\amp\to \Delta \theta \sin(\theta) \end{align*}
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Consider, then, the limits involved in the derivative calculations that we built earlier.
\begin{align*} \frac{d}{d\theta}\left(\sin(\theta)\right) \amp = \lim_{\Delta \theta\to 0}\left(\frac{\sin(\theta+\Delta \theta)-\sin(\theta)}{\Delta \theta}\right)\\ \amp =\lim_{\Delta \theta \to 0} \left(\frac{\Delta \theta \cos(\theta)}{\Delta \theta}\right)\\ \amp = \lim_{\Delta \theta \to 0} (\cos (\theta))\\ \amp = \cos(\theta)\\ \frac{d}{d\theta}\left(\cos(\theta)\right) \amp = \lim_{\Delta \theta \to 0}\left(\frac{\cos(\theta+\Delta \theta)-\cos(\theta)}{\Delta \theta}\right)\\ \amp = \lim_{\Delta \theta \to 0}\left(\frac{-(\cos(\theta)-\cos(\theta+\Delta \theta)) }{\Delta \theta}\right)\\ \amp = \lim_{\Delta \theta \to 0}\left(\frac{-\Delta \sin(\theta)}{\Delta \theta}\right)\\ \amp = \lim_{\Delta \theta \to 0} (-\sin(\theta))\\ \amp = -\sin(\theta) \end{align*}
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So we have shown that \(\dfrac{d}{d\theta}(\sin(\theta)) = \cos(\theta)\) and \(\dfrac{d}{d\theta}(\cos(\theta)) = -\sin(\theta)\) as we claimed.
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Activity 2.3.3. Derivative of the Exponential Function.

We’re going to consider a maybe-unfamiliar function, \(f(x)=e^x\text{.}\) We’ll explore this function similarly to how we thought of the derivatives of sine and cosine in Activity 2.3.2: we’ll look at a tangent line at different points, and think about the slope.
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(a)
The plot below includes both the graph of \(y=e^x\) and the line tangent to \(y=e^x\text{.}\) Watch as the point moves along the curve.
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Collect as much information about the derivative, \(y'\text{,}\) as you can. What do you know about it? What are some facts about the slopes of the tangent lines in this animation?
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Hint.
Are there any \(x\)-values where the slope is negative? Are there any where the slope is equal to 0? What happens to the slopes as \(x\) increases?
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(b)
There is a slightly hidden fact about slopes and tangent lines in this animation. In the following animation, we’ll add (and label) one more point. Let’s look at this again, this time noting the point at which this tangent line crosses the \(x\)-axis. This will make it easier to think about slopes!
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What information does this reveal about the slopes?
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Hint.
It might be especially helpful to think about the slopes and their relationship to the \(y\)-value of the point we are building the tangent line at.
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(c)
Make a conjecture about the slope of the line tangent to the exponential function \(y=e^x\) at any \(x\)-value. What do you believe the formula/equation for \(y'\) is then?
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Subsection Some Properties of Derivatives in General

We can think about each of these properties through the lense of how combining these functions impacts the slopes. For instance, if we wanted to visualize the property about coefficients (that \(\ddx{kf(x)}=k\ddx{f(x)}\)), we can visualize this coefficient as a scaling factor.
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Example 2.3.11. Putting These Together.

Find the following derivatives:
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(a)
\(\displaystyle \ddx{4x^5 - \dfrac{5x}{2}+6\cos(x)-1}\)
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Solution.
\begin{align*} \ddx{4x^5 - \dfrac{5x}{2}+6\cos(x)-1}\amp = \ddx{4x^5} - \ddx{\dfrac{5x}{2}} + \ddx{6\cos(x)} - \ddx{1} \\ \amp =4\ddx{x^5} - \dfrac{5}{2}\ddx{x} + 6\ddx{\cos(x)} - \ddx{1}\\ \amp =4(5x^4) - \dfrac{5}{2}(1) + 6(-\sin(x)) - 0\\ \amp =20x^4 - \dfrac{5}{2} - 6\sin(x) \end{align*}
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Practice Problems Practice Problems

1.

Consider the following table of values of the derivative for \(f\) and \(g\text{:}\)
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Table 2.3.12.
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\(x\) \(-1\) \(0\) \(1\) \(2\)
\(f'(x)\) \(4\) \(\dfrac{1}{3}\) \(-\dfrac{5}{2}\) \(0\)
\(g'(x)\) \(-3\) \(-7\) \(\dfrac{1}{2}\) \(-2\)
For each version of \(h(x)\text{,}\) find the derivative indicated.
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(a)
\(h(x) = 3f(x)\text{;}\) Find \(h'(0)\)
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(b)
\(h(x) = -2g(x)\text{;}\) Find \(h'(-1)\)
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(c)
\(h(x) = \frac{1}{3}f(x)-\frac{2}{5}g(x)\text{;}\) Find \(h'(2)\)
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(d)
\(h(x) = g(x)-\frac{f(x)}{6}\text{;}\) Find \(h'(1)\)
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2.

For each of the following functions, find \(\dydx\text{.}\)
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(d)
\(y = \sqrt{x}+4x^{15} + 400\)
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(f)
\(y = \dfrac{\sqrt{x^3}}{5} - \sqrt[3]{x^2}\)
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(g)
\(y = \dfrac{5}{x^3} + \dfrac{x^3}{5}\)
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(j)
\(y = 4e^x + \dfrac{2}{\sqrt{x^5}}\)
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3.

We have not yet learned how to differentiate products and quotients of functions.
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For each of the following functions including products or quotients, rewrite the function and find the derivative.
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(a)
\(f(x) = \dfrac{x^2+1}{x}\)
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(c)
\(p(x) = \dfrac{x^2-9}{x+3}\)
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(d)
\(r(h) = \dfrac{h-2}{\sqrt{h}-\sqrt{2}}\)
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(e)
\(\Gamma(x) = \sqrt{x}(x^{5/2}-5)\)
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(f)
\(w(t) = \dfrac{te^t-1}{t}\)
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(g)
\(g(w) = w(w+3)^2(2w-5)^2\)
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(h)
\(s(t) = \dfrac{6t^{3/2}+4t^{5/3}}{5t^{1/3}}\)
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4.

For each of the following functions, find \(f'(x)\text{,}\) and then \(\ddx{f'(x)} = f''(x)\text{.}\)
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(a)
\(f(x) = 7x^5 - 2x^4 + x^3 - 7x^2+14x-10\)
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(b)
\(f(x) = 4e^x - \dfrac{3}{x^5}+ x\)
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5.

For each of the following functions, use \(f'(a)\) to find the equation of the line tangent to \(f(x)\) at \(a\) (for the given value of \(a\)).
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\(f(x) = 4x^2-x^3\) for \(a=3\)
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(b)
\(f(x) = e^x-4x\) for \(a=0\)
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(c)
\(f(x) = x^2-x+1\) for \(a=-2\)
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(d)
\(f(x) = 4\sqrt{x}-x\) for \(a=9\)
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6.

Consider the function \(f(x)= 4e^x - 5x\text{.}\)
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(a)
Find the derivative \(f'(x)\text{.}\)
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(b)
Interpret \(f'(x)\text{.}\) What does it tell us about \(f(x)\text{?}\)
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(c)
Find the value of \(x\) where the slope of the line tangent to \(f(x)\) is 6.
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(d)
Find the value for \(x\) where \(f(x)\) has a horizontal tangent line.
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7.

Consider the function \(f(x)= \sqrt[3]{x}+x\text{.}\)
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(a)
Find the derivative \(f'(x)\text{.}\)
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(b)
Find the value for \(x\) where the slope of the line tangent to \(f(x)\) is 2.
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(c)
Find the value for \(x\) where \(f(x)\) has a vertical tangent line.
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8.

Give an example of a function that meets the following requirements.
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(a)
The equation of the line tangent to the curve at \(x=2\) is \(y=-3x+4\text{.}\)
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(b)
The tangent line is horizontal at \(x=5\text{,}\) and the function also passes through the point \((1,-2)\text{.}\)
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(c)
The derivative is positive when \(x\lt-1\) and negative when \(x\gt-1\text{.}\)
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