Again Considering at the Point Where C Passes Through the Origin

3. Derivatives

3.2 The Derivative as a Function

Learning Objectives

Ascertain the derivative function of a given function.
Graph a derivative function from the graph of a given office.
Land the connection between derivatives and continuity.
Depict three weather for when a function does non have a derivative.
Explain the meaning of a college-order derivative.

Equally we accept seen, the derivative of a part at a given point gives united states of america the rate of change or slope of the tangent line to the office at that point. If nosotros differentiate a position function at a given fourth dimension, we obtain the velocity at that time. It seems reasonable to conclude that knowing the derivative of the function at every bespeak would produce valuable information well-nigh the behavior of the function. Nonetheless, the process of finding the derivative at fifty-fifty a handful of values using the techniques of the preceding section would apace become quite tedious. In this department nosotros define the derivative function and learn a procedure for finding information technology.

Derivative Functions

The derivative function gives the derivative of a role at each point in the domain of the original function for which the derivative is divers. Nosotros tin formally define a derivative part as follows.

Definition

Permit $f$ be a function. The derivative function, denoted by $f^{\prime}$ , is the function whose domain consists of those values of $x$ such that the post-obit limit exists:

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

A function $f(x)$ is said to be differentiable at $a$ if
$f^{\prime}(a)$ exists. More generally, a function is said to exist differentiable on $S$ if information technology is differentiable at every point in an open set $S$ , and a differentiable office is one in which $f^{\prime}(x)$ exists on its domain.

In the next few examples we utilize (Effigy) to discover the derivative of a function.

Finding the Derivative of a Foursquare-Root Office

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

Solution

Beginning directly with the definition of the derivative part. Use (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

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

Solution

Follow the same procedure here, simply without having to multiply by the conjugate.

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

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

Solution

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

Nosotros use a variety of different notations to limited the derivative of a function. In (Figure) we showed that if $f(x)=x^2-2x$ , then $f^{\prime}(x)=2x-2$ . If we had expressed this part in the class $y=x^2-2x$ , nosotros could have expressed the derivative every bit $y^{\prime}=2x-2$ or $\frac{dy}{dx}=2x-2$ . Nosotros could take conveyed the same information by writing $\frac{d}{dx}(x^2-2x)=2x-2$ . Thus, for the role $y=f(x)$ , each of the following notations represents the derivative of $f(x)$ :

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

In place of $f^{\prime}(a)$ nosotros may also apply $\frac{dy}{dx}\Big|_{x=a}$ Use of the $\frac{dy}{dx}$ annotation (called Leibniz notation) is quite mutual in applied science and physics. To sympathise this notation improve, call back that the derivative of a role at a betoken is the limit of the slopes of secant lines every bit the secant lines approach the tangent line. The slopes of these secant lines are often expressed in the course $\frac{\Delta y}{\Delta x}$ where $\Delta y$ is the deviation in the $y$ values corresponding to the departure in the $x$ values, which are expressed equally $\Delta x$ ((Figure)). Thus the derivative, which tin can be thought of as the instantaneous charge per unit 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

Nosotros accept already discussed how to graph a function, so given the equation of a function or the equation of a derivative part, we could graph it. Given both, nosotros would expect to come across a correspondence between the graphs of these 2 functions, since $f^{\prime}(x)$ gives the rate of change of a role $f(x)$ (or slope 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, equally in (Figure), we can utilise the graphs to understand the relationship between these two functions. Kickoff, we notice that $f(x)$ is increasing over its entire domain, which ways that the slopes of its tangent lines at all points are positive. Consequently, we look $f^{\prime}(x)>0$ for all values of $x$ in its domain. Furthermore, as $x$ increases, the slopes of the tangent lines to $f(x)$ are decreasing and we expect to see a corresponding decrease in $f^{\prime}(x)$ . We besides discover 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 plant that for $f(x)=x^2-2x, \, f^{\prime}(x)=2x-2$ . The graphs of these functions are shown in (Effigy). Discover that $f(x)$ is decreasing for $x<1$ . For these same values of $x, \, f^{\prime}(x)<0$ . 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 following 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 $f^{\prime}(x)$ higher up the $x$ -centrality?

Solution

$(0,+\infty)$

Derivatives and Continuity

Now that we tin can graph a derivative, let's examine the behavior of the graphs. Outset, we consider the relationship between differentiability and continuity. We will see that if a part is differentiable at a point, information technology must be continuous there; however, a role that is continuous at a point need not be differentiable at that indicate. In fact, a function may be continuous at a indicate and fail to be differentiable at the signal 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 bear witness that $f(x)$ is continuous at $a$ by 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 $\underset{x\to a}{\lim}f(x)=f(a)$ , we conclude that $f$ is continuous at $a$ . $_\blacksquare$

We take just proven that differentiability implies continuity, only 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; however, $f^{\prime}(0)$ is undefined. This ascertainment leads usa to believe that continuity does not imply differentiability. Let'due south explore further. 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 considering

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

See (Figure).

Let'due south consider some additional situations in which a continuous role 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 look at the graph of $f(x)=\sqrt[3]{x}$ clarifies the situation. The function has a vertical tangent line at 0 ((Figure)).

The function $f(x)=\begin{cases} x \sin(\frac{1}{x}) & \text{if} \, x \ne 0 \\ 0 & \text{if} \, x = 0 \end{cases}$ too has a derivative that exhibits interesting behavior at 0. Nosotros 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 non exist, essentially because the slopes of the secant lines continuously change direction every bit they approach zero ((Figure)).

In summary:

We discover that if a function is not continuous, it cannot be differentiable, since every differentiable part must be continuous. However, if a function is continuous, it may still fail to be differentiable.
Nosotros saw that $f(x)=|x|$ failed to exist differentiable at 0 because the limit of the slopes of the tangent lines on the left and right were not the aforementioned. 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 bespeak, a role must exist "smooth" at that point.
As we saw in the example of $f(x)=\sqrt[3]{x}$ , a office fails to exist differentiable at a point where at that place 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 part may fail to be differentiable at a betoken in more complicated means besides.

A Piecewise Part that is Continuous and Differentiable

Solution

For the part 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 $f(-10)=5$ , nosotros must accept $10-10b+c=5$ . Equivalently, nosotros take $c=10b-5$ .

For the function to exist differentiable at -10,

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

must be. Since $f(x)$ is divers using different rules on the right and the left, we must evaluate this limit from the right and the left and then set up 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 usa $b-2=-\frac{1}{4}$ . Thus $b=\frac{7}{4}$ and $c=10(\frac{7}{4})-5=\frac{25}{2}$ .

Higher-Gild Derivatives

The derivative of a part is itself a function, and so nosotros can notice the derivative of a derivative. For example, the derivative of a position part is the rate of change of position, or velocity. The derivative of velocity is the charge per unit of change of velocity, which is acceleration. The new function obtained by differentiating the derivative is called the 2nd derivative. Furthermore, we tin continue to take derivatives to obtain the third derivative, fourth derivative, and so on. Collectively, these are referred to as higher-society derivatives. The annotation for the higher-lodge derivatives of $y=f(x)$ can exist expressed in any of the following forms:

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

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

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

Information technology is interesting to annotation 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

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

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

Adjacent, find $f''(x)$ by 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 forth a coordinate axis at fourth dimension $t$ (in seconds) is given past $s(t)=3t^2-4t+1$ (in meters). Detect the function that describes its dispatch at fourth dimension $t$ .

Solution

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

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

Side by side,

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

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

Key Concepts

Key Equations

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

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

i. $f(x)=6$

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

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

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

Solution

$8x$

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

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

Solution

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

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

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

Solution

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

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

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

Solution

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

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

xi. 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.

xiii. 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).

fourteen. 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 function $y=f(x)$ at $x=a$ . Find $f(x)$ and $a$ .

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

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

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

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

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

Solution

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

For the post-obit functions,

sketch the graph and
utilize the definition of a derivative to testify that the part 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 following graphs,

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

25.

For the following 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 post-obit exercises, use a calculator to graph $f(x)$ . Decide the role $f^{\prime}(x)$ , then use 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, draw what the two expressions correspond 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 metropolis at time $x$ in years.

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

Solution

a. Average rate at which customers spent on concessions in thousands per customer.
b. Rate (in thousands per client) 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.

xl. $g(x)$ denotes the class (in percentage points) received on a test, given $x$ hours of studying.

Solution

a. Average grade received on the exam with an average report time between two amounts.
b. Rate (in pct points per hr) at which the grade on the test increased or decreased for a given average study time of $x$ hours.

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

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

Solution

a. Boilerplate modify of atmospheric force per unit area between ii different altitudes.
b. Charge per unit (torr per foot) at which atmospheric pressure is increasing or decreasing at $x$ anxiety.

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 m feet is -0.1 degrees per human foot.

Solution

a. The charge per unit at which the number of people who have come up down with the flu is irresolute $t$ weeks later the initial outbreak.
b. The charge per unit is increasing sharply up to the third calendar week, at which betoken information technology slows downwardly and then becomes constant.

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

Time (seconds)	Height (meters)
0	0
one	ii
2	4
3	13
4	25
five	32

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

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Source: https://opentextbc.ca/calculusv1openstax/chapter/the-derivative-as-a-function/

Again Considering at the Point Where C Passes Through the Origin

3.2 The Derivative as a Function

Learning Objectives

Derivative Functions

Definition

Finding the Derivative of a Foursquare-Root Office

Solution

Finding the Derivative of a Quadratic Function

Solution

Solution

Graphing a Derivative

Sketching a Derivative Using a Function

Solution

Derivatives and Continuity

Proof

A Piecewise Part that is Continuous and Differentiable

Solution

Higher-Gild Derivatives

Finding a Second Derivative

Solution

Finding Acceleration

Solution

Key Concepts

Key Equations

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

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