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3. Derivatives

3.2 The Derivative equally a Function

Learning Objectives

Define the derivative function of a given role.
Graph a derivative office from the graph of a given office.
State the connection between derivatives and continuity.
Describe 3 conditions for when a part does not have a derivative.
Explain the meaning of a college-order derivative.

As we have seen, the derivative of a office at a given point gives us the rate of change or slope of the tangent line to the function at that point. If we differentiate a position function at a given time, nosotros obtain the velocity at that time. Information technology seems reasonable to conclude that knowing the derivative of the part at every point would produce valuable information almost the behavior of the role. However, the process of finding the derivative at even a handful of values using the techniques of the preceding section would chop-chop become quite wearisome. In this section we define the derivative role and learn a process for finding it.

Derivative Functions

The derivative function gives the derivative of a function at each betoken in the domain of the original function for which the derivative is defined. We can formally define a derivative function every bit follows.

Definition

Let $f$ be a function. The derivative function, denoted past $f^{\prime}$ , is the office whose domain consists of those values of $x$ such that the following limit exists:

$f^{\prime}(x)=\underset{h\to 0}{\lim}\frac{f(x+h)-f(x)}{h}$ .

A function $f(x)$ is said to exist differentiable at $a$ if
$f^{\prime}(a)$ exists. More by and large, a function is said to exist differentiable on $S$ if it is differentiable at every bespeak in an open gear up $S$ , and a differentiable part is ane in which $f^{\prime}(x)$ exists on its domain.

In the next few examples we use (Figure) to notice the derivative of a function.

Finding the Derivative of a Square-Root Part

Find the derivative of $f(x)=\sqrt{x}$ .

Solution

Start directly with the definition of the derivative role. Employ (Figure).

$\begin{array}{lllll}f^{\prime}(x)& =\underset{h\to 0}{\lim}\frac{\sqrt{x+h}-\sqrt{x}}{h} & & & \begin{array}{l}\text{Substitute} \, f(x+h)=\sqrt{x+h} \, \text{and} \, f(x)=\sqrt{x} \\ \text{into} \, f^{\prime}(x)=\underset{h\to 0}{\lim}\frac{f(x+h)-f(x)}{h}. \end{array} \\ & =\underset{h\to 0}{\lim}\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} & & & \begin{array}{l}\text{Multiply numerator and denominator by} \\ \sqrt{x+h}+\sqrt{x} \, \text{without distributing in the} \\ \text{denominator.} \end{array} \\ & =\underset{h\to 0}{\lim}\frac{h}{h(\sqrt{x+h}+\sqrt{x})} & & & \text{Multiply the numerators and simplify.} \\ & =\underset{h\to 0}{\lim}\frac{1}{(\sqrt{x+h}+\sqrt{x})} & & & \text{Cancel the} \, h. \\ & =\frac{1}{2\sqrt{x}} & & & \text{Evaluate the limit.} \end{array}$

Finding the Derivative of a Quadratic Function

Discover the derivative of the function $f(x)=x^2-2x$ .

Solution

Follow the same process here, but without having to multiply by the cohabit.

$\begin{array}{lllll}f^{\prime}(x) & =\underset{h\to 0}{\lim}\frac{((x+h)^2-2(x+h))-(x^2-2x)}{h} & & & \begin{array}{l}\text{Substitute} \, f(x+h)=(x+h)^2-2(x+h) \, \text{and} \\ f(x)=x^2-2x \, \text{into} \\ f^{\prime}(x)=\underset{h\to 0}{\lim}\frac{f(x+h)-f(x)}{h}. \end{array} \\ & =\underset{h\to 0}{\lim}\frac{x^2+2xh+h^2-2x-2h-x^2+2x}{h} & & & \text{Expand} \, (x+h)^2-2(x+h). \\ & =\underset{h\to 0}{\lim}\frac{2xh-2h+h^2}{h} & & & \text{Simplify.} \\ & =\underset{h\to 0}{\lim}\frac{h(2x-2+h)}{h} & & & \text{Factor out} \, h \, \text{from the numerator.} \\ & =\underset{h\to 0}{\lim}(2x-2+h) & & & \text{Cancel the common factor of} \, h. \\ & =2x-2 & & & \text{Evaluate the limit.} \end{array}$

Find the derivative of $f(x)=x^2$ .

Solution

$f^{\prime}(x)=2x$

We utilize a variety of unlike notations to express the derivative of a function. In (Effigy) we showed that if $f(x)=x^2-2x$ , and then $f^{\prime}(x)=2x-2$ . If we had expressed this function in the form $y=x^2-2x$ , we could have expressed the derivative as $y^{\prime}=2x-2$ or $\frac{dy}{dx}=2x-2$ . We could accept conveyed the aforementioned information past writing $\frac{d}{dx}(x^2-2x)=2x-2$ . Thus, for the function $y=f(x)$ , each of the post-obit notations represents the derivative of $f(x)$ :

$f^{\prime}(x), \, \frac{dy}{dx}, \, y^{\prime}, \, \frac{d}{dx}(f(x))$ .

In place of $f^{\prime}(a)$ we may also use $\frac{dy}{dx}\Big|_{x=a}$ Utilize of the $\frac{dy}{dx}$ notation (called Leibniz notation) is quite common in engineering and physics. To empathize this notation better, recall that the derivative of a office at a point is the limit of the slopes of secant lines as the secant lines approach the tangent line. The slopes of these secant lines are frequently expressed in the form $\frac{\Delta y}{\Delta x}$ where $\Delta y$ is the difference in the $y$ values corresponding to the departure in the $x$ values, which are expressed equally $\Delta x$ ((Figure)). Thus the derivative, which can exist idea of every bit the instantaneous rate of change of $y$ with respect to $x$ , is expressed as

$\frac{dy}{dx}=\underset{\Delta x\to 0}{\lim}\frac{\Delta y}{\Delta x}$ .

Graphing a Derivative

We accept already discussed how to graph a function, so given the equation of a function or the equation of a derivative function, we could graph it. Given both, we would expect to see a correspondence between the graphs of these two functions, since $f^{\prime}(x)$ gives the rate of change of a function $f(x)$ (or gradient of the tangent line to $f(x)$ ).

In (Effigy) we found that for $f(x)=\sqrt{x}, \, f^{\prime}(x)=1/2\sqrt{x}$ . If we graph these functions on the same axes, every bit in (Figure), we tin use the graphs to empathise the relationship betwixt these ii functions. Outset, we detect that $f(x)$ is increasing over its entire domain, which ways that the slopes of its tangent lines at all points are positive. Consequently, nosotros expect $f^{\prime}(x)>0$ for all values of $x$ in its domain. Furthermore, as $x$ increases, the slopes of the tangent lines to $f(x)$ are decreasing and nosotros expect to see a corresponding decrease in $f^{\prime}(x)$ . Nosotros too observe that $f(0)$ is undefined and that $\underset{x\to 0^+}{\lim}f^{\prime}(x)=+\infty$ , corresponding to a vertical tangent to $f(x)$ at 0.

In (Figure) we found that for $f(x)=x^2-2x, \, f^{\prime}(x)=2x-2$ . The graphs of these functions are shown in (Effigy). Notice that $f(x)$ is decreasing for . For these aforementioned values of $x, \, f^{\prime}(10)<0$ . For values of $x>1, \, f(x)$ is increasing and $f^{\prime}(x)>0$ . Also, $f(x)$ has a horizontal tangent at $x=1$ and $f^{\prime}(1)=0$ .

Sketching a Derivative Using a Function

Use the post-obit graph of $f(x)$ to sketch a graph of $f^{\prime}(x)$ .

The function f(x) is roughly sinusoidal, starting at (−4, 3), decreasing to a local minimum at (−2, 2), then increasing to a local maximum at (3, 6), and getting cut off at (7, 2).

Sketch the graph of $f(x)=x^2-4$ . On what interval is the graph of $f^{\prime}(x)$ above the $x$ -centrality?

Solution

$(0,+\infty)$

Derivatives and Continuity

Now that we can graph a derivative, permit's examine the behavior of the graphs. First, we consider the human relationship between differentiability and continuity. We will see that if a part is differentiable at a point, it must be continuous at that place; however, a function that is continuous at a signal need non exist differentiable at that point. In fact, a function may be continuous at a indicate and fail to be differentiable at the point for one of several reasons.

Proof

If $f(x)$ is differentiable at $a$ , then $f^{\prime}(a)$ exists and

$f^{\prime}(a)=\underset{x\to a}{\lim}\frac{f(x)-f(a)}{x-a}$ .

We want to prove that $f(x)$ is continuous at $a$ past showing that $\underset{x\to a}{\lim}f(x)=f(a)$ . Thus,

$\begin{array}{lllll} \underset{x\to a}{\lim}f(x) & =\underset{x\to a}{\lim}(f(x)-f(a)+f(a)) & & & \\ & =\underset{x\to a}{\lim}(\frac{f(x)-f(a)}{x-a}\cdot (x-a)+f(a)) & & & \text{Multiply and divide} \, f(x)-f(a) \, \text{by} \, x-a. \\ & =(\underset{x\to a}{\lim}\frac{f(x)-f(a)}{x-a}) \cdot (\underset{x\to a}{\lim}(x-a))+\underset{x\to a}{\lim}f(a) & & & \\ & =f^{\prime}(a) \cdot 0+f(a) & & & \\ & =f(a). & & & \end{array}$

Therefore, since $f(a)$ is defined and $\underset{x\to a}{\lim}f(x)=f(a)$ , nosotros conclude that $f$ is continuous at $a$ . $_\blacksquare$

We have merely proven that differentiability implies continuity, but now we consider whether continuity implies differentiability. To determine an answer to this question, we examine the part $f(x)=|x|$ . This function is continuous everywhere; even so, $f^{\prime}(0)$ is undefined. This observation leads the states to believe that continuity does non imply differentiability. Permit'south explore farther. For $f(x)=|x|$ ,

$f^{\prime}(0)=\underset{x\to 0}{\lim}\frac{f(x)-f(0)}{x-0}=\underset{x\to 0}{\lim}\frac{|x|-|0|}{x-0}=\underset{x\to 0}{\lim}\frac{|x|}{x}$ .

This limit does not exist because

$\underset{x\to 0^-}{\lim}\frac{|x|}{x}=-1 \, \text{and} \, \underset{x\to 0^+}{\lim}\frac{|x|}{x}=1$ .

Run into (Figure).

Allow's consider some additional situations in which a continuous function fails to be differentiable. Consider the function $f(x)=\sqrt[3]{x}$ :

$f^{\prime}(0)=\underset{x\to 0}{\lim}\frac{\sqrt[3]{x}-0}{x-0}=\underset{x\to 0}{\lim}\frac{1}{\sqrt[3]{x^2}}=+\infty$ .

Thus $f^{\prime}(0)$ does not exist. A quick wait at the graph of $f(x)=\sqrt[3]{x}$ clarifies the situation. The function has a vertical tangent line at 0 ((Figure)).

The function $f(x)=\begin{cases} x \sin(\frac{1}{x}) & \text{if} \, x \ne 0 \\ 0 & \text{if} \, x = 0 \end{cases}$ also has a derivative that exhibits interesting behavior at 0. We come across that

$f^{\prime}(0)=\underset{x\to 0}{\lim}\frac{x \sin(1/x)-0}{x-0}=\underset{x\to 0}{\lim} \sin(\frac{1}{x})$ .

This limit does not exist, substantially because the slopes of the secant lines continuously change management equally they arroyo zero ((Effigy)).

In summary:

We notice that if a function is non continuous, it cannot be differentiable, since every differentiable function must be continuous. Nevertheless, if a function is continuous, it may nonetheless neglect to be differentiable.
We saw that $f(x)=|x|$ failed to be differentiable at 0 because the limit of the slopes of the tangent lines on the left and right were not the same. Visually, this resulted in a sharp corner on the graph of the function at 0. From this we conclude that in order to be differentiable at a point, a function must be "polish" at that point.
Every bit we saw in the example of $f(x)=\sqrt[3]{x}$ , a office fails to be differentiable at a point where in that location is a vertical tangent line.
As we saw with $f(x)=\begin{cases} x \sin(\frac{1}{x}) & \text{if} \, x \ne 0 \\ 0 & \text{if} \, x = 0 \end{cases}$ a function may neglect to be differentiable at a point in more than complicated ways besides.

A Piecewise Function that is Continuous and Differentiable

Solution

For the function to be continuous at $x=-10, \, \underset{x\to 10^-}{\lim}f(x)=f(-10)$ . Thus, since

$\underset{x\to −10^-}{\lim}f(x)=\frac{1}{10}(-10)^2-10b+c=10-10b+c$

and $f(-10)=5$ , we must accept $10-10b+c=5$ . Equivalently, we have $c=10b-5$ .

For the function to exist differentiable at -x,

$f^{\prime}(10)=\underset{x\to −10}{\lim}\frac{f(x)-f(-10)}{x+10}$

must exist. Since $f(x)$ is defined using different rules on the right and the left, we must evaluate this limit from the right and the left and then set them equal to each other:

$\begin{array}{lllll} \underset{x\to −10^-}{\lim}\frac{f(x)-f(-10)}{x+10} & =\underset{x\to −10^-}{\lim}\frac{\frac{1}{10}x^2+bx+c-5}{x+10} & & & \\ & =\underset{x\to −10^-}{\lim}\frac{\frac{1}{10}x^2+bx+(10b-5)-5}{x+10} & & & \text{Substitute} \, c=10b-5. \\ & =\underset{x\to −10^-}{\lim}\frac{x^2-100+10bx+100b}{10(x+10)} & & & \\ & =\underset{x\to −10^-}{\lim}\frac{(x+10)(x-10+10b)}{10(x+10)} & & & \text{Factor by grouping.} \\ & =b-2 & & & \end{array}$

Nosotros also have

$\begin{array}{ll} \underset{x\to −10^+}{\lim}\frac{f(x)-f(-10)}{x+10} & =\underset{x\to −10^+}{\lim}\frac{-\frac{1}{4}x+\frac{5}{2}-5}{x+10} \\ & =\underset{x\to −10^+}{\lim}\frac{−(x+10)}{4(x+10)} \\ & =-\frac{1}{4} \end{array}$

This gives us $b-2=-\frac{1}{4}$ . Thus $b=\frac{7}{4}$ and $c=10(\frac{7}{4})-5=\frac{25}{2}$ .

Higher-Order Derivatives

The derivative of a part is itself a function, so nosotros can find the derivative of a derivative. For example, the derivative of a position role is the rate of change of position, or velocity. The derivative of velocity is the rate of modify of velocity, which is acceleration. The new function obtained past differentiating the derivative is called the 2nd derivative. Furthermore, nosotros can continue to take derivatives to obtain the third derivative, fourth derivative, and so on. Collectively, these are referred to every bit higher-order derivatives. The notation for the college-order derivatives of $y=f(x)$ tin can be expressed in whatsoever of the following forms:

$f''(x), \, f'''(x), \, f^{(4)}(x), \cdots ,f^{(n)}(x)$

$y'', \, y''', \, y^{(4)}, \cdots ,y^{(n)}$

$\frac{d^2y}{dx^2}, \, \frac{d^3y}{dx^3}, \, \frac{d^4y}{dx^4}, \cdots,\frac{d^ny}{dx^n}$ .

It is interesting to note that the notation for $\frac{d^2y}{dx^2}$ may be viewed as an attempt to express $\frac{d}{dx}(\frac{dy}{dx})$ more compactly. Analogously, $\frac{d}{dx}(\frac{d}{dx}(\frac{dy}{dx}))=\frac{d}{dx}(\frac{d^2y}{dx^2})=\frac{d^3y}{dx^3}$ .

Finding a Second Derivative

For $f(x)=2x^2-3x+1$ , find $f''(x)$ .

Solution

Showtime find $f^{\prime}(x)$ .

$\begin{array}{lllll}f^{\prime}(x) & =\underset{h\to 0}{\lim}\frac{(2(x+h)^2-3(x+h)+1)-(2x^2-3x+1)}{h} & & & \begin{array}{l}\text{Substitute} \, f(x)=2x^2-3x+1 \\ \text{and} \\ f(x+h)=2(x+h)^2-3(x+h)+1 \\ \text{into} \, f^{\prime}(x)=\underset{h\to 0}{\lim}\frac{f(x+h)-f(x)}{h}. \end{array} \\ & =\underset{h\to 0}{\lim}\frac{4xh+h^2-3h}{h} & & & \text{Simplify the numerator.} \\ & =\underset{h\to 0}{\lim}(4x+h-3) & & & \begin{array}{l}\text{Factor out the} \, h \, \text{in the numerator} \\ \text{and cancel with the} \, h \, \text{in the} \\ \text{denominator.} \end{array} \\ & =4x-3 & & & \text{Take the limit.} \end{array}$

Side by side, notice $f''(x)$ past taking the derivative of $f^{\prime}(x)=4x-3$ .

$\begin{array}{lllll} f''(x)& =\underset{h\to 0}{\lim}\frac{f^{\prime}(x+h)-f^{\prime}(x)}{h} & & & \begin{array}{l}\text{Use} \, f^{\prime}(x)=\underset{h\to 0}{\lim}\frac{f(x+h)-f(x)}{h} \, \text{with} \, f^{\prime}(x) \, \text{in} \\ \text{place of} \, f(x). \end{array} \\ & =\underset{h\to 0}{\lim}\frac{(4(x+h)-3)-(4x-3)}{h} & & & \begin{array}{l}\text{Substitute} \, f^{\prime}(x+h)=4(x+h)-3 \, \text{and} \\ f^{\prime}(x)=4x-3. \end{array} \\ & =\underset{h\to 0}{\lim}4 & & & \text{Simplify.} \\ & =4 & & & \text{Take the limit.} \end{array}$

Finding Acceleration

The position of a particle along a coordinate axis at time $t$ (in seconds) is given by $s(t)=3t^2-4t+1$ (in meters). Find the function that describes its acceleration at time $t$ .

Solution

Since $v(t)=s^{\prime}(t)$ and $a(t)=v^{\prime}(t)=s''(t)$ , we begin past finding the derivative of $s(t)$ :

$\begin{array}{ll}s^{\prime}(t) & =\underset{h\to 0}{\lim}\frac{s(t+h)-s(t)}{h} \\ & =\underset{h\to 0}{\lim}\frac{3(t+h)^2-4(t+h)+1-(3t^2-4t+1)}{h} \\ & =6t-4 \end{array}$

Next,

$\begin{array}{ll} s''(t) & =\underset{h\to 0}{\lim}\frac{s^{\prime}(t+h)-s^{\prime}(t)}{h} \\ & =\underset{h\to 0}{\lim}\frac{6(t+h)-4-(6t-4)}{h} \\ & =6 \end{array}$

Thus, $a=6 \, \text{m/s}^2$ .

Central Concepts

Cardinal Equations

The derivative function
$f^{\prime}(x)=\underset{h\to 0}{\lim}\frac{f(x+h)-f(x)}{h}$

For the following exercises, utilise the definition of a derivative to find $f^{\prime}(x)$ .

1. $f(x)=6$

2. $f(x)=2-3x$

3. $f(x)=\frac{2x}{7}+1$

four. $f(x)=4x^2$

Solution

$8x$

5. $f(x)=5x-x^2$

6. $f(x)=\sqrt{2x}$

Solution

$\frac{1}{\sqrt{2x}}$

7. $f(x)=\sqrt{x-6}$

8. $f(x)=\frac{9}{x}$

Solution

$\frac{-9}{x^2}$

nine. $f(x)=x+\frac{1}{x}$

10. $f(x)=\frac{1}{\sqrt{x}}$

Solution

$\frac{-1}{2x^{3/2}}$

For the post-obit exercises, use the graph of $y=f(x)$ to sketch the graph of its derivative $f^{\prime}(x)$ .

11. The function f(x) starts at (−2, 20) and decreases to pass through the origin and achieve a local minimum at roughly (0.5, −1). Then, it increases and passes through (1, 0) and achieves a local maximum at (2.25, 2) before decreasing again through (3, 0) to (4, −20).

12. The function f(x) starts at (−1.5, 20) and decreases to pass through (0, 10), where it appears to have a derivative of 0. Then it further decreases, passing through (1.7, 0) and achieving a minimum at (3, −17), at which point it increases rapidly through (3.8, 0) to (4, 20).

Solution

The function starts in the third quadrant and increases to touch the origin, then decreases to a minimum at (2, −16), before increasing through the x axis at x = 3, after which it continues increasing.

13. The function f(x) starts at (−2.25, −20) and increases rapidly to pass through (−2, 0) before achieving a local maximum at (−1.4, 8). Then the function decreases to the origin. The graph is symmetric about the y-axis, so the graph increases to (1.4, 8) before decreasing through (2, 0) and heading on down to (2.25, −20).

14. The function f(x) starts at (−3, −1) and increases to pass through (−1.5, 0) and achieve a local minimum at (1, 0). Then, it decreases and passes through (1.5, 0) and continues decreasing to (3, −1).

Solution

The function starts at (−3, 0), increases to a maximum at (−1.5, 1), decreases through the origin and to a minimum at (1.5, −1), and then increases to the x axis at x = 3.

For the following exercises, the given limit represents the derivative of a part $y=f(x)$ at $x=a$ . Find $f(x)$ and $a$ .

fifteen. $\underset{h\to 0}{\lim}\frac{(1+h)^{2/3}-1}{h}$

16. $\underset{h\to 0}{\lim}\frac{[3(2+h)^2+2]-14}{h}$

Solution

$f(x)=3x^2+2, \, a=2$

17. $\underset{h\to 0}{\lim}\frac{\cos(\pi+h)+1}{h}$

18. $\underset{h\to 0}{\lim}\frac{(2+h)^4-16}{h}$

Solution

$f(x)=x^4, \, a=2$

nineteen. $\underset{h\to 0}{\lim}\frac{[2(3+h)^2-(3+h)]-15}{h}$

20. $\underset{h\to 0}{\lim}\frac{e^h-1}{h}$

Solution

$f(x)=e^x, \, a=0$

For the following functions,

sketch the graph and
utilize the definition of a derivative to show that the function is not differentiable at $x=1$ .

21. $f(x)=\begin{cases} 2\sqrt{x} & \text{if} \, 0 \le x \le 1 \\ 3x-1 & \text{if} \, x>1 \end{cases}$

23. $f(x)=\begin{cases} -x^2+2 & \text{if} \, x \le 1 \\ x & \text{if} \, x>1 \end{cases}$

For the post-obit graphs,

decide for which values of $x=a$ the $\underset{x\to a}{\lim}f(x)$ exists merely $f$ is non continuous at $x=a$ , and
determine for which values of $x=a$ the role is continuous simply not differentiable at $x=a$ .

25.

For the post-obit functions, use $f''(x)=\underset{h\to 0}{\lim}\frac{f^{\prime}(x+h)-f^{\prime}(x)}{h}$ to find $f''(x)$ .

28. $f(x)=2-3x$

29. $f(x)=4x^2$

30. $f(x)=x+\frac{1}{x}$

Solution

$\frac{2}{x^3}$

For the following exercises, use a reckoner to graph $f(x)$ . Determine the part $f^{\prime}(x)$ , then utilize a calculator to graph $f^{\prime}(x)$ .

31. [T] $f(x)=-\frac{5}{x}$

33. [T] $f(x)=\sqrt{x}+3x$

35. [T] $f(x)=1+x+\frac{1}{x}$

For the following exercises, describe what the 2 expressions represent in terms of each of the given situations. Be sure to include units.

$\frac{f(x+h)-f(x)}{h}$
$f^{\prime}(x)=\underset{h\to 0}{\lim}\frac{f(x+h)-f(x)}{h}$

37. $P(x)$ denotes the population of a city at time $x$ in years.

38. $C(x)$ denotes the total amount of coin (in thousands of dollars) spent on concessions by $x$ customers at an amusement park.

Solution

a. Boilerplate charge per unit at which customers spent on concessions in thousands per customer.
b. Rate (in thousands per customer) at which $x$ customers spent coin on concessions in thousands per customer.

39. $R(x)$ denotes the total cost (in thousands of dollars) of manufacturing $x$ clock radios.

40. $g(x)$ denotes the grade (in per centum points) received on a exam, given $x$ hours of studying.

Solution

a. Average grade received on the test with an average study fourth dimension between two amounts.
b. Rate (in percentage points per hour) at which the course on the test increased or decreased for a given boilerplate study time of $x$ hours.

41. $B(x)$ denotes the cost (in dollars) of a sociology textbook at university bookstores in the United States in $x$ years since 1990.

42. $p(x)$ denotes atmospheric pressure at an altitude of $x$ feet.

Solution

a. Average alter of atmospheric pressure betwixt ii different altitudes.
b. Rate (torr per foot) at which atmospheric pressure is increasing or decreasing at $x$ feet.

Solution

a. The rate (in degrees per foot) at which temperature is increasing or decreasing for a given acme $x$ .
b. The rate of change of temperature as altitude changes at 1000 feet is -0.1 degrees per foot.

Solution

a. The rate at which the number of people who have come down with the flu is changing $t$ weeks afterward the initial outbreak.
b. The rate is increasing sharply up to the third calendar week, at which betoken it slows downwards and then becomes constant.

For the post-obit exercises, use the following tabular array, which shows the height $h$ of the Saturn V rocket for the Apollo xi mission $t$ seconds later launch.

Time (seconds)	Height (meters)
0	0
1	2
ii	4
3	13
four	25
5	32

47.What is the concrete meaning of $h^{\prime}(t)$ ? What are the units?