A common amusement park ride lifts riders to a height then allows them to free fall a certain distance before safely stopping them. Suppose such a ride drops riders from a height of 150 feet. Students of physics may recall that the height (in feet) of the riders, $t$ seconds after free fall (and ignoring air resistance, etc.) can be accurately modeled by $f(t)=-16{t}^{2}+150$.
Using this formula, it is easy to verify that, without intervention, the riders will hit the ground at $t=2.5\sqrt{1.5}\approx 3.06$ seconds. Suppose the designers of the ride decide to begin slowing the riders’ fall after 2 seconds (corresponding to a height of 86 ft.). How fast will the riders be traveling at that time?
We have been given a position function, but what we want to compute is a velocity at a specific point in time, i.e., we want an instantaneous velocity. We do not currently know how to calculate this.
However, we do know from common experience how to calculate an average velocity. (If we travel 60 miles in 2 hours, we know we had an average velocity of 30 mph.) We looked at this concept in Section 1.1 when we introduced the difference quotient. We have
$$\frac{\text{change in distance}}{\text{change in time}}=\frac{\text{\u201crise\u201d}}{\text{run}}=\text{average velocity}.$$ |
We can approximate the instantaneous velocity at $t=2$ by considering the average velocity over some time period containing $t=2$. If we make the time interval small, we will get a good approximation. (This fact is commonly used. For instance, high speed cameras are used to track fast moving objects. Distances are measured over a fixed number of frames to generate an accurate approximation of the velocity.)
Consider the interval from $t=2$ to $t=3$ (just before the riders hit the ground). On that interval, the average velocity is
$$\frac{f(3)-f(2)}{3-2}=\frac{f(3)-f(2)}{1}=-80\text{ft/s},$$ |
where the minus sign indicates that the riders are moving down. By narrowing the interval we consider, we will likely get a better approximation of the instantaneous velocity. On $[2,2.5]$ we have
$$\frac{f(2.5)-f(2)}{2.5-2}=\frac{f(2.5)-f(2)}{0.5}=-72\text{ft/s}.$$ |
We can do this for smaller and smaller intervals of time. For instance, over a time span of 1/10${}^{\text{th}}$ of a second, i.e., on $[2,2.1]$, we have
$$\frac{f(2.1)-f(2)}{2.1-2}=\frac{f(2.1)-f(2)}{0.1}=-65.6\text{ft/s}.$$ |
Over a time span of 1/100${}^{\text{th}}$ of a second, on $[2,2.01]$, the average velocity is
$$\frac{f(2.01)-f(2)}{2.01-2}=\frac{f(2.01)-f(2)}{0.01}=-64.16\text{ft/s}.$$ |
What we are really computing is the average velocity on the interval $[2,2+h]$ for small values of $h$. That is, we are computing
$$\frac{f(2+h)-f(2)}{h}$$ |
where $h$ is the change in time after $2$ seconds.
What we really want is for $h=0$, but this, of course, returns the familiar “$0/0$” indeterminate form. So we employ a limit, as we did in Section 1.1.
We can approximate the value of this limit numerically with small values of $h$ as seen in Figure 2.1.1. It looks as though the velocity is approaching $-64$ ft/s. Computing the limit directly gives
$\underset{h\to 0}{lim}{\displaystyle \frac{f(2+h)-f(2)}{h}}$ | $=\underset{h\to 0}{lim}{\displaystyle \frac{-16{(2+h)}^{2}+150-(-16{(2)}^{2}+150)}{h}}$ | ||
$=\underset{h\to 0}{lim}{\displaystyle \frac{-64h-16{h}^{2}}{h}}$ | |||
$=\underset{h\to 0}{lim}(-64-16h)$ | |||
$=-64.$ |
Graphically, we can view the average velocities we computed numerically as the slopes of secant lines on the graph of $f$ going through the points $(2,f(2))$ and $(2+h,f(2+h))$, as in Figure 2.1.2. In Figure 2.1.3, the secant line corresponding to $h=1$ is shown in three contexts. Figure 2.1.3(a) shows a “zoomed out” version of $f$ with its secant line. In (b), we zoom in around the points of intersection between $f$ and the secant line. Notice how well this secant line approximates $f$ between those two points — it is a common practice to approximate functions with straight lines.
As $h\to 0$, these secant lines approach the tangent line, a line that goes through the point $(2,f(2))$ with the special slope of $-64$. In parts (c) and (d) of Figure 2.1.3, we zoom in around the point $(2,86)$. In (c) we see the secant line, which approximates $f$ well, but not as well the tangent line shown in (d).
(a) | (b) |
(c) | (d) |
We have just introduced a number of important concepts that we will flesh out more within this section. First, we formally define two of them.
Let $f$ be a continuous function on an open interval $I$ and let $c$ be in $I$. The derivative of $f$ at $c$, denoted ${f}^{\prime}(c)$, is
$$\underset{h\to 0}{lim}\frac{f(c+h)-f(c)}{h},$$ |
provided the limit exists.
If the limit exists, we say that $f$ is differentiable at $c$; if the limit does not exist, then $f$ is not differentiable at $c$. If $f$ is differentiable at every point in $I$, then $f$ is differentiable on $I$.
Let $f$ be continuous on an open interval $I$ and differentiable at $c$, for some $c$ in $I$. The line with equation $\mathrm{\ell}(x)={f}^{\prime}(c)(x-c)+f(c)$ is the tangent line to the graph of $f$ at $c$; that is, it is the line through $(c,f(c))$ whose slope is the derivative of $f$ at $c$.
Some examples will help us understand these definitions.
Let $f(x)=3{x}^{2}+5x-7$. Find:
${f}^{\prime}(1)$
The equation of the tangent line to the graph of $f$ at $x=1$.
${f}^{\prime}(3)$
The equation of the tangent line to the graph $f$ at $x=3$.
Solution
We compute this directly using Definition 2.1.1.
${f}^{\prime}(1)$ | $=\underset{h\to 0}{lim}{\displaystyle \frac{f(1+h)-f(1)}{h}}$ | ||
$=\underset{h\to 0}{lim}{\displaystyle \frac{3{(1+h)}^{2}+5(1+h)-7-(3{(1)}^{2}+5(1)-7)}{h}}$ | |||
$=\underset{h\to 0}{lim}{\displaystyle \frac{3{h}^{2}+11h}{h}}$ | |||
$=\underset{h\to 0}{lim}(3h+11)=11.$ |
The tangent line at $x=1$ has slope ${f}^{\prime}(1)$ and goes through the point $(1,f(1))=(1,1)$. Thus the tangent line has equation, in point-slope form, $y=11(x-1)+1$. In slope-intercept form we have $y=11x-10$.
Again, using the definition,
${f}^{\prime}(3)$ | $=\underset{h\to 0}{lim}{\displaystyle \frac{f(3+h)-f(3)}{h}}$ | ||
$=\underset{h\to 0}{lim}{\displaystyle \frac{3{(3+h)}^{2}+5(3+h)-7-(3{(3)}^{2}+5(3)-7)}{h}}$ | |||
$=\underset{h\to 0}{lim}{\displaystyle \frac{3{h}^{2}+23h}{h}}$ | |||
$=\underset{h\to 0}{lim}(3h+23)$ | |||
$=23.$ |
The tangent line at $x=3$ has slope $23$ and goes through the point $(3,f(3))$, which is $(3,35)$. Thus the tangent line has equation $y=23(x-3)+35=23x-34$.
A graph of $f$ is given in Figure 2.1.4 along with the tangent lines at $x=1$ and $x=3$.
Linear functions are easy to work with; many functions that arise in the course of solving real problems are not easy to work with. A common practice in mathematical problem solving is to approximate difficult functions with not-so-difficult functions. Lines are a common choice. It turns out that at any given point on the graph of a differentiable function $f$, the best linear approximation to $f$ is its tangent line. That is one reason we’ll spend considerable time finding tangent lines to functions.
One type of function that does not benefit from a tangent-line approximation is a line; it is rather simple to recognize that the tangent line to a line is the line itself. We look at this in the following example.
Consider $f(x)=3x+5$. Find the equation of the tangent line to $f$ at $x=1$ and $x=7$.
SolutionWe find the slope of the tangent line with Definition 2.1.1.
${f}^{\prime}(1)$ | $=\underset{h\to 0}{lim}{\displaystyle \frac{f(1+h)-f(1)}{h}}$ | ||
$=\underset{h\to 0}{lim}{\displaystyle \frac{3(1+h)+5-(3+5)}{h}}$ | |||
$=\underset{h\to 0}{lim}{\displaystyle \frac{3h}{h}}$ | |||
$=\underset{h\to 0}{lim}3$ | |||
$=3.$ |
We just found that ${f}^{\prime}(1)=3$. That is, we found the instantaneous rate of change of $f(x)=3x+5$ is $3$. This is not surprising; lines are characterized by being the only functions with a constant rate of change. That rate of change is called the slope of the line. Since their rates of change are constant, their instantaneous rates of change are always the same; they are all the slope.
So given a line $f(x)=ax+b$, the derivative at any point $x$ will be $a$; that is, ${f}^{\prime}(x)=a$.
It is now easy to see that the tangent line to the graph of $f$ at $x=1$ is just $f$, with the same being true for $x=7$.
We often desire to find the tangent line to the graph of a function without knowing the actual derivative of the function. In these cases, the best we may be able to do is approximate the tangent line. We demonstrate this in the next example.
Approximate the equation of the tangent line to the graph of $f(x)=\mathrm{sin}x$ at $x=0$.
SolutionIn order to find the equation of the tangent line, we need a slope and a point. The point is given to us: $(0,\mathrm{sin}0)=(0,0)$. To compute the slope, we need the derivative. This is where we will make an approximation. Recall that
$${f}^{\prime}(0)\approx \frac{\mathrm{sin}(0+h)-\mathrm{sin}0}{h}$$ |
for a small value of $h$. We choose (somewhat arbitrarily) to let $h=0.1$. Thus ^{†}^{†}margin: Λ
$${f}^{\prime}(0)\approx \frac{\mathrm{sin}(0.1)-\mathrm{sin}0}{0.1}\approx 0.9983.$$ |
Thus our approximation of the equation of the tangent line is $y=0.9983(x-0)+0=0.9983x$; it is graphed in Figure 2.1.5. The graph seems to imply the approximation is rather good.
Recall from Section 1.3 that $\underset{x\to 0}{lim}{\displaystyle \frac{\mathrm{sin}x}{x}}=1$, meaning for values of $x$ near 0, $\mathrm{sin}x\approx x$. Since the slope of the line $y=x$ is 1 at $x=0$, it should seem reasonable that “the slope of $f(x)=\mathrm{sin}x$” is near 1 at $x=0$. In fact, since we approximated the value of the slope to be $0.9983$, we might guess the actual value is 1. We’ll come back to this later.
Consider again Example 2.1.1. To find the derivative of $f$ at $x=1$, we needed to evaluate a limit. To find the derivative of $f$ at $x=3$, we needed to again evaluate a limit. We have this process:
$$\begin{array}{c}\hfill \text{input specific}\hfill \\ \hfill \text{number}c\hfill \end{array}\u27f6\begin{array}{c}\hfill \text{do something}\hfill \\ \hfill \text{to}f\text{and}c\hfill \end{array}\u27f6\begin{array}{c}\hfill \text{return}\hfill \\ \hfill \text{number}f\phantom{\rule{1.66666666666667pt}{0ex}}{}^{\prime}(c)\hfill \end{array}$$ |
This process describes a function; given one input (the value of $c$), we return exactly one output (the value of ${f}^{\prime}(c)$). The “do something” box is where the tedious work (taking limits) of this function occurs.
Instead of applying this function repeatedly for different values of $c$, let us apply it just once to the variable $x$. We then take a limit just once. The process now looks like:
$$\text{input variable x}\u27f6\begin{array}{c}\hfill \text{do something}\hfill \\ \hfill \text{to}f\text{and}x\hfill \end{array}\u27f6\begin{array}{c}\hfill \text{return}\hfill \\ \hfill \text{function}f\phantom{\rule{1.66666666666667pt}{0ex}}{}^{\prime}(x)\hfill \end{array}$$ |
The output is the “derivative function,” ${f}^{\prime}(x)$. The ${f}^{\prime}(x)$ function will take a number $c$ as input and return the derivative of $f$ at $c$. This calls for a definition.
Let $f$ be a differentiable function on an open interval $I$. The function
$${f}^{\prime}(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$$ |
is the derivative of $f$.
Notation:
Let $y=f(x)$. The following notations all represent the derivative:
$${f}^{\prime}(x)={y}^{\prime}=\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}f}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(f)=\frac{\mathrm{d}}{\mathrm{d}x}(y).$$ |
Important: The notation $\frac{\mathrm{d}y}{\mathrm{d}x}$ is one symbol; it is not the fraction “$\mathrm{d}y/\mathrm{d}x$”. The notation, while somewhat confusing at first, was chosen with care. A fraction-looking symbol was chosen because the derivative has many fraction-like properties. Among other places, we see these properties at work when we talk about the units of the derivative, when we discuss the Chain Rule, and when we learn about integration (topics that appear in later sections and chapters).
Examples will help us understand this definition.
Let $f(x)=3{x}^{2}+5x-7$ as in Example 2.1.1. Find ${f}^{\prime}(x)$.
SolutionWe apply Definition 2.1.3.
${f}^{\prime}(x)$ | $=\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)-f(x)}{h}}$ | ||
$=\underset{h\to 0}{lim}{\displaystyle \frac{3{(x+h)}^{2}+5(x+h)-7-(3{x}^{2}+5x-7)}{h}}$ | |||
$=\underset{h\to 0}{lim}{\displaystyle \frac{3{h}^{2}+6xh+5h}{h}}$ | |||
$=\underset{h\to 0}{lim}(3h+6x+5)$ | |||
$=6x+5$ |
So ${f}^{\prime}(x)=6x+5$. Recall earlier we found that ${f}^{\prime}(1)=11$ and ${f}^{\prime}(3)=23$. Note our new computation of ${f}^{\prime}(x)$ affirm these facts.
Let $f(x)={\displaystyle \frac{1}{x+1}}$. Find ${f}^{\prime}(x)$.
SolutionWe apply Definition 2.1.3.
${f}^{\prime}(x)$ | $=\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)-f(x)}{h}}$ | |||
$=\underset{h\to 0}{lim}{\displaystyle \frac{\frac{1}{x+h+1}-\frac{1}{x+1}}{h}}$ | ||||
Now find a common denominator and subtract; factor $1/h$ out front to facilitate reading. | ||||
${f}^{\prime}(x)$ | $=\underset{h\to 0}{lim}{\displaystyle \frac{1}{h}}\cdot \left({\displaystyle \frac{x+1}{(x+1)(x+h+1)}}-{\displaystyle \frac{x+h+1}{(x+1)(x+h+1)}}\right)$ | |||
$=\underset{h\to 0}{lim}{\displaystyle \frac{1}{h}}\cdot \left({\displaystyle \frac{x+1-(x+h+1)}{(x+1)(x+h+1)}}\right)$ | ||||
$=\underset{h\to 0}{lim}{\displaystyle \frac{1}{h}}\cdot \left({\displaystyle \frac{-h}{(x+1)(x+h+1)}}\right)$ | ||||
$=\underset{h\to 0}{lim}{\displaystyle \frac{-1}{(x+1)(x+h+1)}}$ | ||||
$={\displaystyle \frac{-1}{(x+1)(x+1)}}$ | ||||
$={\displaystyle \frac{-1}{{(x+1)}^{2}}}$ |
So ${f}^{\prime}(x)={\displaystyle \frac{-1}{{(x+1)}^{2}}}$. To practice using our notation, we could also state
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{x+1}\right)=\frac{-1}{{(x+1)}^{2}}.$$ |
Find the derivative of $f(x)=\mathrm{sin}x$.
SolutionBefore applying Definition 2.1.3, note that once the derivative is found, we can find the actual tangent line to $f(x)=\mathrm{sin}x$ at $x=0$, whereas we settled for an approximation in Example 2.1.3.
${f}^{\prime}\left(x\right)$ | |||
$=\underset{h\to 0}{lim}{\displaystyle \frac{\mathrm{sin}\left(x+h\right)-\mathrm{sin}x}{h}}$ | $\left(\text{Use trig identity}\mathrm{sin}\left(x+h\right)=\mathrm{sin}x\mathrm{cos}h+\mathrm{cos}x\mathrm{sin}h\right)$ | ||
$=\underset{h\to 0}{lim}{\displaystyle \frac{\mathrm{sin}x\mathrm{cos}h+\mathrm{cos}x\mathrm{sin}h-\mathrm{sin}x}{h}}$ | (regroup) | ||
$=\underset{h\to 0}{lim}{\displaystyle \frac{\mathrm{sin}x\left(\mathrm{cos}h-1\right)+\mathrm{cos}x\mathrm{sin}h}{h}}$ | (split into two fractions) | ||
$=\underset{h\to 0}{lim}\left({\displaystyle \frac{\mathrm{sin}x\left(\mathrm{cos}h-1\right)}{h}}+{\displaystyle \frac{\mathrm{cos}x\mathrm{sin}h}{h}}\right)$ | $\left(\text{use}\underset{h\to 0}{lim}{\displaystyle \frac{\mathrm{cos}h-1}{h}}=0\text{and}\underset{h\to 0}{lim}{\displaystyle \frac{\mathrm{sin}h}{h}}=1\right)$ | ||
$=\mathrm{sin}x\cdot 0+\mathrm{cos}x\cdot 1$ | |||
$=\mathrm{cos}x.$ |
We have found that when $f(x)=\mathrm{sin}x$, ${f}^{\prime}(x)=\mathrm{cos}x$ (see Figure 2.1.6).
Initially, this might be somewhat surprising; the result of a tedious limit process and the sine function is a nice function. Then again, perhaps this is not entirely surprising. The sine function is periodic — it repeats itself on regular intervals. Therefore its rate of change also repeats itself on the same regular intervals. In fact, if we think about ${f}^{\prime}(x)$ as the slope of the tangent to the sine curve we notice the following
when the slope of tangent lines is 0 then ${f}^{\prime}(x)=\mathrm{cos}x$ crosses the $x-$axis;
when the slopes of the tangent lines are positive then ${f}^{\prime}$ lies above the $x-$axis; and
when the slopes of the tangent lines are negative then ${f}^{\prime}$ lies below the $x-$axis.
We should have known the derivative would be periodic; we now know exactly which periodic function it is.
Thinking back to Example 2.1.3, we can find the slope of the tangent line to $f(x)=\mathrm{sin}x$ at $x=0$ using our derivative. We approximated the slope as $0.9983$; we now know the slope is exactly $\mathrm{cos}0=1$.
SolutionWe need to evaluate $\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)-f(x)}{h}}.$ As $f$ is piecewise-defined, we need to consider separately the limits when $$ and when $x>0$.
When $$:
$\frac{\mathrm{d}}{\mathrm{d}x}}\left(-x\right)$ | $=\underset{h\to 0}{lim}{\displaystyle \frac{-(x+h)-(-x)}{h}}$ | ||
$=\underset{h\to 0}{lim}{\displaystyle \frac{-h}{h}}$ | |||
$=\underset{h\to 0}{lim}-1$ | |||
$=-1.$ |
When $x>0$, a similar computation shows that $\frac{\mathrm{d}}{\mathrm{d}x}}\left(x\right)=1$.
We need to also find the derivative at $x=0$. By the definition of the derivative at a point, we have
$${f}^{\prime}(0)=\underset{h\to 0}{lim}\frac{f(0+h)-f(0)}{h}.$$ |
Since $x=0$ is the point where our function’s definition switches from one piece to the other, we need to consider left and right-hand limits. Consider the following, where we compute the left and right hand limits side by side.
$\underset{h\to {0}^{-}}{lim}{\displaystyle \frac{f(0+h)-f(0)}{h}}$
$=$
$\underset{h\to {0}^{-}}{lim}{\displaystyle \frac{-h-0}{h}}$
$=$
$\underset{h\to {0}^{-}}{lim}-1$
$=-1$
$\underset{h\to {0}^{+}}{lim}{\displaystyle \frac{f(0+h)-f(0)}{h}}$
$=$
$\underset{h\to {0}^{+}}{lim}{\displaystyle \frac{h-0}{h}}$
$=$
$\underset{h\to {0}^{+}}{lim}1$
$=1$
^{†}^{†}margin:
Λ
The last lines of each column tell the story: the left and right hand limits are not equal. Therefore the limit does not exist at 0, and $f$ is not differentiable at 0; see Figure 2.1.8.
So we have
$$ |
At $x=0$, ${f}^{\prime}(x)$ does not exist; there is a jump discontinuity at 0. So $f(x)=\left|x\right|$ is differentiable everywhere except at 0.
The point of non-differentiability came where the piecewise defined function switched from one piece to the other. Our next example shows that this does not always cause trouble.
Find the derivative of $f(x)$, where $f(x)=\{\begin{array}{cc}\mathrm{sin}x\hfill & x\le \pi /2\hfill \\ 1\hfill & x>\pi /2\hfill \end{array}.$ See Figure 2.1.9.
SolutionUsing Example 2.1.6, we know that when $$, ${f}^{\prime}(x)=\mathrm{cos}x$. It is easy to verify that when $x>\pi /2$, ${f}^{\prime}(x)=0$; consider:
$$\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}=\underset{h\to 0}{lim}\frac{1-1}{h}=\underset{h\to 0}{lim}0=0.$$ |
So far we have
$$ |
We still need to find ${f}^{\prime}(\pi /2)$. Notice at $x=\pi /2$ that both pieces of ${f}^{\prime}$ are 0, meaning we can state that ${f}^{\prime}(\pi /2)=0$.
Being more rigorous, we can again evaluate the difference quotient limit at $x=\pi /2$, utilizing again left and right-hand limits:
$\underset{h\to {0}^{-}}{lim}{\displaystyle \frac{f\left(\pi /2+h\right)-f\left(\pi /2\right)}{h}}$
$=$
$\underset{h\to {0}^{-}}{lim}{\displaystyle \frac{\mathrm{sin}\left(\pi /2+h\right)-\mathrm{sin}\left(\pi /2\right)}{h}}$
$=$
$\underset{h\to {0}^{-}}{lim}{\displaystyle \frac{\mathrm{sin}\left(\frac{\pi}{2}\right)\mathrm{cos}\left(h\right)+\mathrm{sin}\left(h\right)\mathrm{cos}\left(\frac{\pi}{2}\right)-\mathrm{sin}\left(\frac{\pi}{2}\right)}{h}}$
$=$
$\underset{h\to {0}^{-}}{lim}{\displaystyle \frac{1\cdot \mathrm{cos}\left(h\right)+\mathrm{sin}\left(h\right)\cdot 0-1}{h}}$
$=$
$0$
$\underset{h\to {0}^{+}}{lim}{\displaystyle \frac{f\left(\pi /2+h\right)-f\left(\pi /2\right)}{h}}$
$=$
$\underset{h\to {0}^{+}}{lim}{\displaystyle \frac{1-1}{h}}$
$=$
$\underset{h\to {0}^{+}}{lim}{\displaystyle \frac{0}{h}}$
$=$
$0$
Since both the left and right hand limits are 0 at $x=\pi /2$, the limit exists and ${f}^{\prime}(\pi /2)$ exists (and is 0). Therefore we can fully write ${f}^{\prime}$ as
$${f}^{\prime}(x)=\{\begin{array}{cc}\mathrm{cos}x\hfill & x\le \pi /2\hfill \\ 0\hfill & x>\pi /2\hfill \end{array}.$$ |
See Figure 2.1.10 for a graph of this function.
Recall we pseudo-defined a continuous function as one in which we could sketch its graph without lifting our pencil. We can give a pseudo-definition for differentiability as well: it is a continuous function that does not have any “sharp corners.” One such sharp corner is shown in Figure 2.1.7. Even though the function $f$ in Example 2.1.8 is piecewise-defined, the transition is “smooth” hence it is differentiable. Note how in the graph of $f$ in Figure 2.1.9 it is difficult to tell when $f$ switches from one piece to the other; there is no “corner.”
To better understand the definition of a derivative, experiment with the Geogebra app at http://mathinsight.org/applet/secant_line_slope.
This section defined the derivative; in some sense, it answers the question of “What is the derivative?” The next section addresses the question “What does the derivative mean?”
T/F: Let $f$ be a position function. The average rate of change on $[a,b]$ is the slope of the line through the points $(a,f\left(a\right))$ and $(b,f\left(b\right))$.
In your own words, explain the difference between the average rate of change and instantaneous rate of change.
Let $y=f\left(x\right)$. Give three different notations equivalent to “${f}^{\prime}\left(x\right)$.”
In Exercises 5–14.,
use the definition of the derivative to compute the derivative of the given function.
Find the tangent line to the graph of the given function at $x=c$.
$f\left(x\right)=6$ at $x=-2$
$f\left(x\right)=2x$ at $x=3$
$f\left(x\right)=4-3x$ at $x=7$
$g\left(x\right)={x}^{2}$ at $x=-2$
$h\left(x\right)=2x-{x}^{2}$ at $x=1$
$f\left(x\right)=3{x}^{2}-x+4$ at $x=-1$
$g\left(x\right)=\sqrt{x+3}$ at $x=1$
$r\left(x\right)={\displaystyle \frac{1}{x}}$ at $x=-2$
$h\left(x\right)={\displaystyle \frac{3}{\sqrt{x}}}$ at $x=4$
$f\left(x\right)={\displaystyle \frac{1}{x-2}}$ at $x=3$
In Exercises 15–18., each limit represents the derivative of some function, $f$, at some number $c$. State an appropriate $f$ and $c$ for each.
$\underset{h\to 0}{lim}{\displaystyle \frac{\sqrt{16+h}-4}{h}}$
$\underset{h\to 0}{lim}{\displaystyle \frac{{\left(3+h\right)}^{4}-81}{h}}$
$\underset{h\to 0}{lim}{\displaystyle \frac{\frac{1}{2+h}-\frac{1}{2}}{h}}$
$\underset{h\to 0}{lim}{\displaystyle \frac{\mathrm{cos}\left(-\pi +h\right)+1}{h}}$
In Exercises 19–24., a function $f$ and an $x$-value $a$ are given. Approximate the equation of the tangent line to the graph of $f$ at $x=a$ by numerically approximating ${f}^{\prime}\left(a\right)$, using $h=0.1$.
$f\left(x\right)={x}^{2}+2x+1$, $x=3$
$f\left(x\right)=\sqrt{x}$, $x=4$
$f\left(x\right)={\displaystyle \frac{10}{x+1}}$, $x=9$
$f\left(x\right)={e}^{x}$, $x=2$
$f\left(x\right)=\mathrm{ln}x$, $x=2$
$f\left(x\right)=\mathrm{cos}x$, $x=0$
In Exercises 27–32., a graph of a function $f\left(x\right)$ is given. Using the graph, sketch ${f}^{\prime}\left(x\right)$.
In Exercises 33–34., a graph of $g\left(x\right)$ is given. Using the graph, answer the following questions.
Where is $g\left(x\right)>0$?
Where is $$?
Where is $g\left(x\right)=0$?
Where is ${g}^{\prime}\left(x\right)>0$?
Where is $$?
Where is ${g}^{\prime}\left(x\right)=0$?
Approximate $\underset{x\to 5}{lim}{\displaystyle \frac{{x}^{2}+2x-35}{{x}^{2}-10.5x+27.5}}$.
Use the Bisection Method to approximate, accurate to two decimal places, the root of $g\left(x\right)={x}^{3}+{x}^{2}+x-1$ on $[0.5,0.6]$.