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## Section1.5The Derivative Function

###### Motivating Questions
• How does the limit definition of the derivative of a function $f$ lead to an entirely new (but related) function $f'\text{?}$

• What is the difference between writing $f'(a)$ and $f'(x)\text{?}$

• How is the graph of the derivative function $y=f'(x)$ related to the graph of $y=f(x)\text{?}$

• What are some examples of functions $f$ for which $f'$ is undefined at one or more points?

###### Supplemental Videos

The main topics of this section are also presented in the following videos:

We now know that the instantaneous rate of change of a function $f(x)$ at $x = a\text{,}$ or equivalently the slope of the tangent line to the graph of $y = f(x)$ at $x = a\text{,}$ is given by the value $f'(a)\text{.}$ In all of our examples so far, we have identified a particular value of $a$ as our point of interest: $a = 1\text{,}$ $a = 3\text{,}$ etc. But it is not hard to imagine that we will often be interested in the derivative value for more than just one $a$-value, and possibly for many of them. In this section, we explore how we can move from computing the derivative at a single point to computing a formula for $f'(a)$ at any point $a\text{.}$ Indeed, the process of taking the derivative generates a new function, denoted by $f'(x)\text{,}$ derived from the original function $f(x)\text{.}$

###### Example1.62

Consider the function $f(x) = 4x - x^2\text{.}$

1. Use the limit definition of the derivative to compute the derivative values: $f'(0)\text{,}$ $f'(1)\text{,}$ $f'(2)\text{,}$ and $f'(3)\text{.}$

2. Observe that the work to find $f'(a)$ is the same, regardless of the value of $a\text{.}$ Based on your work in (a), what do you conjecture is the value of $f'(4)\text{?}$ How about $f'(5)\text{?}$ (Note: you should not use the limit definition of the derivative to find either value.)

3. Conjecture a formula for $f'(a)$ that depends only on the value $a\text{.}$ That is, in the same way that we have a formula for $f(x)$ (recall $f(x) = 4x - x^2$), see if you can use your work above to guess a formula for $f'(a)$ in terms of $a\text{.}$

Hint
1. $f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$
2. Is there a pattern with the four values you found in (a)?
3. Based on (a) and (b), what type of familiar function might the derivative be? What do formulas for this type of function look like?
1. $f'(0)=4,\, f'(1)=2,\, f'(2)=0,\, f'(3)=-2\text{.}$
2. $f'(4)=-4,\, f'(5)=-6\text{.}$
3. $f'(a)=-2a+4\text{.}$
Solution
1. Since $f(0)=0$ and $f(0+h)=f(h)=4h-h^2\text{,}$ we can say that

\begin{align*} f'(0)=\mathstrut \amp \lim_{h\to0}\frac{f(0+h)-f(0)}{h}\\ =\mathstrut \amp \lim_{h\to0}\frac{(4h-h^2)-0}{h}\\ =\mathstrut \amp \lim_{h\to0}(4-h)\\ =\mathstrut \amp 4\text{.} \end{align*}

Since $f(1)=4(1)-(1)^2=3$ and

\begin{align*} f(1+h)=\mathstrut \amp 4(1+h)-(1+h)^2\\ =\mathstrut \amp 4+4h-(1+2h+h^2)\\ =\mathstrut \amp 3+2h-h^2\text{,} \end{align*}

we can say that

\begin{align*} f'(1)=\mathstrut \amp \lim_{h\to0}\frac{f(1+h)-f(1)}{h}\\ =\mathstrut \amp \lim_{h\to0}\frac{(3+2h-h^2)-(3)}{h}\\ =\mathstrut \amp \lim_{h\to0}\frac{2h-h^2}{h}\\ =\mathstrut \amp \lim_{h\to0}(2-h)\\ =\mathstrut \amp 2\text{.} \end{align*}

Similar work shows that $f'(2)=0$ and $f'(3)=-2\text{.}$

2. In part (a), each time we increased the input value by $1\text{,}$ the derivative value decreased by $2\text{.}$ This leads us to believe that $f'(4)=-4$ and $f'(5)=-6\text{,}$ assuming the linear pattern continues.
3. The derivative values seem to fit a linear pattern, with a slope of $-2$ and a $y$-intercept of $4\text{.}$ A reasonable formula for $f'(a)\text{,}$ then, is $f'(a)=-2a+4\text{.}$

### SubsectionHow the Derivative is Itself a Function

In your work in Example1.62 with $f(x) = 4x - x^2\text{,}$ you may have found several patterns. One comes from observing that $f'(0) = 4\text{,}$ $f'(1) = 2\text{,}$ $f'(2) = 0\text{,}$ and $f'(3) = -2\text{.}$ That sequence of values leads us naturally to conjecture that $f'(4) = -4$ and $f'(5) = -6\text{.}$ We also observe that the particular value of $a$ has very little effect on the process of computing the value of the derivative through the limit definition. To see this more clearly, we compute $f'(a)$ where $a$ is a variable that is, $a$ represents a number to be named later. Following the now standard process of using the limit definition of the derivative,

\begin{align*} f'(a) =\mathstrut \amp \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \\ = \lim_{h \to 0} \frac{4(a + h) - (a + h)^2 - (4a-a^2)}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{4a + 4h - a^2 - 2ha - h^2 - 4a+a^2}{h} \\ = \lim_{h \to 0} \frac{4h - 2ha - h^2}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{h(4 - 2a - h)}{h} \\ = \lim_{h \to 0} (4 - 2a - h)\text{.} \end{align*}

Here we observe that neither $4$ nor $2a$ depend on the value of $h\text{,}$ so as $h$ tends to $0\text{,}$ the quantity $(4 - 2a - h)$ approaches $(4 - 2a)\text{.}$ Thus $f'(a) = 4 - 2a\text{.}$

This result is consistent with the specific values we found above: e.g., $f'(3) = 4 - 2(3) = -2\text{.}$ And indeed, our work confirms that the value of $a$ has almost no bearing on the process of computing the derivative. We note further that the letter being used is immaterial: whether we call it $a\text{,}$ $x\text{,}$ or anything else, the derivative at a given value is simply given by 4 minus 2 times the value. We choose to use $x$ for consistency with the original function given by $y = f(x)\text{,}$ as well as for the purpose of graphing the derivative function. For the function $f(x) = 4x - x^2\text{,}$ it follows that $f'(x) = 4 - 2x\text{.}$

Because the value of the derivative function is linked to the graph of the original function, it makes sense to look at both of these functions plotted on the same domain.

The left half of Figure1.63 above shows a plot of $y=f(x)\text{,}$ where $f(x) = 4x - x^2\text{,}$ together with a selection of tangent lines at the points we considered in Example1.62. The right half of the figure shows a plot of $y=f'(x)\text{,}$ where $f'(x) = 4 - 2x\text{,}$ with emphasis on the $y$-coordinates of the derivative graph at the same selection of points. Notice the connection between colors in the left and right graphs: the green tangent line on the original graph is tied to the green point on the right graph in the following way: the slope of the tangent line at a point on the left-hand graph is the same as the $y$-value at the corresponding point on the right-hand graph. That is, at each respective value of $x\text{,}$ the slope of the tangent line to the original function is the same as the output of the derivative function. Do note, however, that the units on the vertical axes are different: in the left graph, the vertical units are simply the output units of $f\text{.}$ On the righthand graph of $y = f'(x)\text{,}$ the units on the vertical axis are units of $f$ per unit of $x\text{.}$

An excellent way to explore how the graph of $f(x)$ generates the graph of $f'(x)$ is through a java applet. See, for instance, the applets at http://gvsu.edu/s/5C or http://gvsu.edu/s/5D, via the sites of Austin and Renault.4David Austin, http://gvsu.edu/s/5r; Marc Renault, http://gvsu.edu/s/5p.

When we first defined the derivative in Section1.4, we wrote the definition in terms of a value $a$ to find $f'(a)\text{.}$ As we have seen above, the letter $a$ is merely a placeholder and it often makes more sense to use $x$ instead. For the record, here we restate the definition of the derivative.

###### Derivative Function

Let $f$ be a function and $x$ a value in the function's domain. We define a new function called $f'$ to be the derivative of $f\text{,}$ where $f'$ is given by the formula

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

provided this limit exists.

We now have two different ways of thinking about the derivative function:

1. Given a graph of $y = f(x)\text{,}$ how does this graph lead to the graph of the derivative function $y = f'(x)\text{?}$ and
2. Given a formula for $y = f(x)\text{,}$ how does the limit definition of derivative generate a formula for $y = f'(x)\text{?}$

Both of these issues are explored in the following examples.

###### Example1.64

Below you will find the graphs of eight different functions, each on a grid with scale $1\times1\text{.}$ For each given graph, sketch an approximate graph of its derivative function on the axes immediately below. Assume the horizontal scale of the grid for each derivative graph is identical to that for the original function. If necessary, adjust the vertical scale on the axes for the graphs of each derivative. Label all axes (vertical and horizontal) with the scales you use.

When you are finished with all eight graphs, write several sentences that describe your overall process for sketching the graph of the derivative function given the graph of the original function. What are the values of the derivative function that you tend to identify first? What do you do thereafter? How do key traits of the graph of the derivative function exemplify properties of the graph of the original function?

Hint

Points where the slope of the tangent line is equal to zero are particularly important. Try finding these points first and plotting those zero values on the axes where you'll graph $y = f'(x)\text{.}$

Solution

The graphs of each given function with its derivative are shown below. In graphing each derivative, it was easiest to first identify the points of horizontal tangency on the original graph in order to plot the zeros of the derivative. From there, we identified intervals where the original function was increasing (or decreasing), and plotted positive (or negative) values for the derivative on those same intervals. We ensured that the derivative values had larger magnitude at those points where the original graph had the steepest slope. In the case where the original graph had a corner or a cusp, we placed a jump discontinuity in the derivative graph. Asymptotic behavior of the original function also was reflected in the graph of the corresponding derivative function.

For a dynamic investigation that allows you to experiment with graphing $f'$ when given the graph of $f\text{,}$ see http://gvsu.edu/s/8y.5Marc Renault, Calculus Applets Using Geogebra.

Now, recall the opening example of this section: we began with the function $f(x) = 4x - x^2$ and used the limit definition of the derivative to show that $f'(a) = 4 - 2a\text{,}$ or equivalently that $f'(x) = 4 - 2x\text{.}$ We subsequently graphed the functions $f$ and $f'$ in Figure1.63. Following Example1.64, we now understand that we could have constructed a fairly accurate graph of $f'(x)$ without knowing a formula for either $f$ or $f'\text{.}$ Even so, it is useful to know a formula for the derivative function whenever it is possible to find one.

In the next example, we further explore the more algebraic approach to finding a derivative: given a formula for $y = f(x)\text{,}$ the limit definition of the derivative will be used to develop a formula for $f'(x)\text{.}$

###### Example1.65

For each of the listed functions, determine a formula for the derivative function. For the first two, determine the formula for the derivative by thinking about the nature of the given function and its slope at various points; do not use the limit definition. For the latter four, use the limit definition. Pay careful attention to the function names and independent variables. It is important to be comfortable with using letters other than $f$ and $x\text{.}$ For example, given a function $p(z)\text{,}$ we call its derivative $p'(z)\text{.}$

1. $f(x) = 1$

2. $g(t) = t$

3. $p(z) = z^2$

4. $q(s) = s^3$

5. $F(t) = \frac{1}{t}$

6. $G(y) = \sqrt{y}$

Hint
1. What is the slope of the function at every point?

2. What is the slope of the function at every point?

3. $p(z+h) = (z+h)^2\text{.}$

4. $q(s+h) = (s+h)^3\text{.}$

5. $F(t+h) = \frac{1}{t+h}\text{.}$

6. $G(y+h) = \sqrt{y+h}\text{;}$ you may find it useful to recall that

\begin{equation*} \left(\sqrt{y+h}-\sqrt{y}\right)\left(\sqrt{y+h}+\sqrt{y}\right)=(y+h)-y\text{.} \end{equation*}
1. $f'(x) = 0\text{.}$

2. $g'(t) = 1\text{.}$

3. $p'(z) = 2z\text{.}$

4. $q'(s) = 3s^2\text{.}$

5. $F'(t) = \frac{-1}{t^2}\text{.}$

6. $G'(y) = \frac{1}{2\sqrt{y}}\text{.}$

Solution
1. The instantaneous rate of change of a linear function at a given point is the same regardless of the point chosen. This is because linear functions have a constant slope, so any tangent line to a linear function coincides with the function itself (i.e., they are the same line). In this case, $f'(x) = 0$ because the slope of the tangent line to the horizontal line given by $f(x) = 1$ is zero at every value of $x\text{.}$

2. $g'(t) = 1$ because the slope of the tangent line to the line given by $g(t) = t$ is $1$ at every value of $t\text{.}$

3. \begin{align*} p'(z) =\mathstrut \amp \lim_{h \to 0} \frac{p(z+h) - p(z)}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{(z+h)^2 - z^2}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{(z^2 + 2zh + h^2) - z^2}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{2zh + h^2}{h}\\ =\mathstrut \amp \lim_{h \to 0} (2z + h)\\ =\mathstrut \amp 2z \end{align*}
4. \begin{align*} q'(s) =\mathstrut \amp \lim_{h \to 0} \frac{q(s+h) - q(s)}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{(s+h)^3 - s^3}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{(s^3 + 3s^2h + 3sh^2 + h^3) - s^3}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{3s^2h + 3sh^2 + h^2}{h}\\ =\mathstrut \amp \lim_{h \to 0} (3s^2 + 3sh + h)\\ =\mathstrut \amp 3s^2 \end{align*}
5. \begin{align*} F'(t) =\mathstrut \amp \lim_{h \to 0} \frac{F(t+h) - F(s)}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{\left(\frac{1}{t+h} - \frac{1}{t}\right)}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{\left(\frac{t - (t+h)}{t(t+h)}\right)}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{-h}{ht(t+h)}\\ =\mathstrut \amp \lim_{h \to 0} \frac{-1}{t(t+h)}\\ =\mathstrut \amp \frac{-1}{t^2} \end{align*}
6. \begin{align*} G'(y) =\mathstrut \amp \lim_{h \to 0} \frac{G(y+h) - G(y)}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{\sqrt{y+h}-\sqrt{y}}{h}\\ =\mathstrut \amp \lim_{h \to 0} \left(\frac{\sqrt{y+h}-\sqrt{y}}{h}\right) \cdot \left(\frac{\sqrt{y+h} + \sqrt{y}}{\sqrt{y+h} + \sqrt{y}}\right)\\ =\mathstrut \amp \lim_{h \to 0} \frac{(y+h)-y}{h(\sqrt{y+h} + \sqrt{y})}\\ =\mathstrut \amp \lim_{h \to 0} \frac{h}{h(\sqrt{y+h} + \sqrt{y})}\\ =\mathstrut \amp \lim_{h \to 0} \frac{1}{\sqrt{y+h} + \sqrt{y}}\\ =\mathstrut \amp \frac{1}{2\sqrt{y}} \end{align*}

### SubsectionSummary

• The limit definition of the derivative, $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\text{,}$ produces a value for each $x$ at which the derivative is defined, and this leads to a new function $y = f'(x)\text{.}$ It is especially important to note that taking the derivative is a process that starts with a given function $f$ and produces a new, related function $f'\text{.}$

• There is essentially no difference between writing $f'(a)$ (as we did regularly in Section1.4) and writing $f'(x)\text{.}$ In either case, the variable is just a placeholder that is used to define the rule for the derivative function.

• Given the graph of a function $y = f(x)\text{,}$ we can sketch an approximate graph of its derivative $y = f'(x)$ by observing that $y$-coordinates on the derivative's graph correspond to slopes on the original function's graph.

• In Example1.64, we encountered some functions that had sharp corners on their graphs, such as the shifted absolute value function. At such points, the derivative fails to exist, and we say that $f$ is not differentiable there. For now, it suffices to understand this as a consequence of the jump that must occur in the derivative function at a sharp corner on the graph of the original function.

### SubsectionExercises

Let $f$ be a function with the following properties: $f$ is differentiable at every value of $x$ (that is, $f$ has a derivative at every point), $f(-2) = 1\text{,}$ and $f'(-2) = -2\text{,}$ $f'(-1) = -1\text{,}$ $f'(0) = 0\text{,}$ $f'(1) = 1\text{,}$ and $f'(2) = 2\text{.}$

1. On the axes provided at left in Figure1.66, sketch a possible graph of $y = f(x)\text{.}$ Explain why your graph meets the stated criteria.

2. Conjecture a formula for the function $y = f(x)\text{.}$ Use the limit definition of the derivative to determine the corresponding formula for $y = f'(x)\text{.}$ Discuss both graphical and algebraic evidence for whether or not your conjecture is correct.

Consider the function $g(x) = x^2 - x + 3\text{.}$

1. Use the limit definition of the derivative to determine a formula for $g'(x)\text{.}$

2. Use a graphing utility to plot both $y = g(x)$ and your result for $y = g'(x)\text{;}$ does your formula for $g'(x)$ generate the graph you expected?

3. Use the limit definition of the derivative to find a formula for $p'(x)$ where $p(x) = 5x^2 - 4x + 12\text{.}$

4. Compare and contrast the formulas for $g'(x)$ and $p'(x)$ you have found. How do the constants 5, 4, 12, and 3 affect the results?

Let $g$ be a continuous function (that is, one with no jumps or holes in the graph) and suppose that a graph of $y= g'(x)$ is given by the graph on the right in Figure1.67.

1. Observe that for every value of $x$ that satisfies $0 \lt x \lt 2\text{,}$ the value of $g'(x)$ is constant. What does this tell you about the behavior of the graph of $y = g(x)$ on this interval?

2. On what intervals other than $0 \lt x \lt 2$ do you expect $y = g(x)$ to be a linear function? Why?

3. At which values of $x$ is $g'(x)$ not defined? What behavior does this lead you to expect to see in the graph of $y=g(x)\text{?}$

4. Suppose that $g(0) = 1\text{.}$ On the axes provided at left in Figure1.67, sketch an accurate graph of $y = g(x)\text{.}$

For each graph that provides an original function $y = f(x)$ in Figure1.68, your task is to sketch an approximate graph of its derivative function, $y = f'(x)\text{,}$ on the axes immediately below. View the scale of the grid for the graph of $f$ as being $1 \times 1\text{,}$ and assume the horizontal scale of the grid for the graph of $f'$ is identical to that for $f\text{.}$ If you need to adjust the vertical scale on the axes for the graph of $f'\text{,}$ you should label that accordingly.