Again Considering at the Point Where C Passes Through the Origin

iii. Derivatives

3.two The Derivative as a Role

Learning Objectives

Define the derivative function of a given function.
Graph a derivative office from the graph of a given function.
Land the connectedness between derivatives and continuity.
Draw three conditions for when a function does not take a derivative.
Explain the significant of a higher-order derivative.

As we take seen, the derivative of a function at a given point gives us the rate of change or slope of the tangent line to the role at that bespeak. If we differentiate a position function at a given fourth dimension, nosotros obtain the velocity at that time. Information technology seems reasonable to conclude that knowing the derivative of the role at every bespeak would produce valuable data near the behavior of the function. Nevertheless, the process of finding the derivative at even a handful of values using the techniques of the preceding section would quickly become quite boring. In this department we ascertain the derivative function and learn a procedure for finding it.

Derivative Functions

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

Definition

Allow $f$ exist a role. The derivative function, denoted by $f^{\prime}$ , is the role 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 be differentiable at $a$ if
$f^{\prime}(a)$ exists. More than generally, a office is said to be differentiable on $S$ if information technology is differentiable at every point in an open up fix $S$ , and a differentiable function is 1 in which $f^{\prime}(x)$ exists on its domain.

In the side by side few examples we utilise (Figure) to discover the derivative of a function.

Finding the Derivative of a Foursquare-Root Function

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

Solution

Offset direct with the definition of the derivative function. 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, 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 utilise a variety of dissimilar notations to express the derivative of a function. In (Effigy) we showed that if $f(x)=x^2-2x$ , and so $f^{\prime}(x)=2x-2$ . If we had expressed this function in the form $y=x^2-2x$ , we could have expressed the derivative every bit $y^{\prime}=2x-2$ or $\frac{dy}{dx}=2x-2$ . Nosotros could have conveyed the same information by writing $\frac{d}{dx}(x^2-2x)=2x-2$ . Thus, for the function $y=f(x)$ , 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 $f^{\prime}(a)$ we may likewise apply $\frac{dy}{dx}\Big|_{x=a}$ Employ of the $\frac{dy}{dx}$ notation (called Leibniz notation) is quite mutual in engineering and physics. To sympathize this notation improve, recall that the derivative of a role at a point 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 ofttimes expressed in the course $\frac{\Delta y}{\Delta x}$ where $\Delta y$ is the difference in the $y$ values corresponding to the difference in the $x$ values, which are expressed every bit $\Delta x$ ((Figure)). Thus the derivative, which can be idea of as 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

Nosotros have already discussed how to graph a function, and then given the equation of a function or the equation of a derivative office, we could graph it. Given both, nosotros 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 slope of the tangent line to $f(x)$ ).

In (Effigy) nosotros found that for $f(x)=\sqrt{x}, \, f^{\prime}(x)=1/2\sqrt{x}$ . If we graph these functions on the same axes, as in (Figure), we tin can employ the graphs to sympathize the human relationship between these two functions. First, we observe that $f(x)$ is increasing over its unabridged domain, which means that the slopes of its tangent lines at all points are positive. Consequently, we wait $f^{\prime}(x)>0$ 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 respective subtract in $f^{\prime}(x)$ . We as well detect 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). Detect that $f(x)$ is decreasing for . 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 Part

Utilize 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 $f^{\prime}(x)$ above the $x$ -axis?

Solution

$(0,+\infty)$

Derivatives and Continuity

Now that nosotros can graph a derivative, let's examine the beliefs of the graphs. Commencement, we consider the relationship betwixt differentiability and continuity. We volition see that if a function is differentiable at a betoken, it must be continuous there; however, a function that is continuous at a point need not be differentiable at that point. In fact, a function may be continuous at a signal 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

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

Nosotros want to show 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 have merely proven that differentiability implies continuity, only now we consider whether continuity implies differentiability. To determine an answer to this question, nosotros examine the function $f(x)=|x|$ . This function is continuous everywhere; however, $f^{\prime}(0)$ is undefined. This observation leads u.s. to believe that continuity does not imply differentiability. Let'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 considering

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

See (Effigy).

Allow'southward consider some boosted situations in which a continuous part 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 be. A quick look at the graph of $f(x)=\sqrt[3]{x}$ clarifies the situation. The part has a vertical tangent line at 0 ((Effigy)).

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 beliefs at 0. Nosotros see 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, essentially because the slopes of the secant lines continuously alter direction as they arroyo nada ((Effigy)).

In summary:

We detect that if a function is not continuous, information technology cannot be differentiable, since every differentiable office must exist continuous. However, if a office is continuous, it may yet fail to exist 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 "shine" at that point.
As we saw in the instance of $f(x)=\sqrt[3]{x}$ , a role fails to exist differentiable at a point where there is a vertical tangent line.
Equally 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 neglect to be differentiable at a betoken in more complicated ways every bit well.

A Piecewise Role that is Continuous and Differentiable

Solution

For the office 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 be differentiable at -ten,

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

must exist. Since $f(x)$ is defined using different rules on the right and the left, we must evaluate this limit from the correct and the left then fix 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 besides accept

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

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

College-Society Derivatives

The derivative of a function is itself a function, and so we tin can find the derivative of a derivative. For case, the derivative of a position part is the rate of change of position, or velocity. The derivative of velocity is the rate of change of velocity, which is acceleration. The new role obtained by differentiating the derivative is called the second derivative. Furthermore, we tin keep to take derivatives to obtain the third derivative, 4th derivative, and so on. Collectively, these are referred to as higher-club derivatives. The note for the higher-gild derivatives of $y=f(x)$ can be expressed in any of the following forms:

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

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

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

It is interesting to annotation that the notation for $\frac{d^2y}{dx^2}$ may be viewed as an attempt to limited $\frac{d}{dx}(\frac{dy}{dx})$ more than 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 2d Derivative

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

Solution

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

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

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

Fundamental Concepts

Key Equations

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

For the post-obit exercises, use 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

$8x$

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

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

Solution

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

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

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

Solution

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

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

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 $x=a$ . Detect $f(x)$ 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. $\underset{h\to 0}{\lim}\frac{\cos(\pi+h)+1}{h}$

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

Solution

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

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

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
utilise the definition of a derivative to show 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 not continuous at $x=a$ , and
determine for which values of $x=a$ the function is continuous but not differentiable at $x=a$ .

25.

For the following functions, employ $f''(x)=\underset{h\to 0}{\lim}\frac{f^{\prime}(x+h)-f^{\prime}(x)}{h}$ 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 computer to graph $f(x)$ . Decide the function $f^{\prime}(x)$ , 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 post-obit exercises, describe what the ii expressions stand for 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 full amount of coin (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 client.
b. Charge per unit (in thousands per customer) at which $x$ customers spent money on concessions in thousands per customer.

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

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

Solution

a. Average grade received on the test with an average written report time betwixt two amounts.
b. Charge per unit (in percentage points per hour) 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 university bookstores in the United states in $x$ years since 1990.

42. $p(x)$ denotes atmospheric pressure level at an distance of $x$ anxiety.

Solution

a. Boilerplate alter of atmospheric pressure between 2 different altitudes.
b. Rate (torr per foot) at which atmospheric pressure is increasing or decreasing at $x$ feet.

Solution

a. The charge per unit (in degrees per foot) at which temperature is increasing or decreasing for a given acme $x$ .
b. The rate of modify of temperature every bit distance changes at 1000 feet is -0.1 degrees per foot.

Solution

a. The rate at which the number of people who accept come downwardly with the flu is changing $t$ weeks after the initial outbreak.
b. The rate is increasing sharply upward to the third week, at which point it slows downward and and so becomes constant.

For the following exercises, use the following table, which shows the superlative $h$ of the Saturn V rocket for the Apollo 11 mission $t$ seconds later launch.

Time (seconds)	Height (meters)
0	0
i	2
ii	four
3	13
4	25
v	32

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

alvareztriessir.blogspot.com

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

Again Considering at the Point Where C Passes Through the Origin

3.two The Derivative as a Role

Learning Objectives

Derivative Functions

Definition

Finding the Derivative of a Foursquare-Root Function

Solution

Finding the Derivative of a Quadratic Function

Solution

Solution

Graphing a Derivative

Sketching a Derivative Using a Part

Solution

Derivatives and Continuity

Proof

A Piecewise Role that is Continuous and Differentiable

Solution

College-Society Derivatives

Finding a 2d Derivative

Solution

Finding Acceleration

Solution

Fundamental Concepts

Key Equations

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

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