maths.free › Calculus › 12. Introduction to Calculus › Derivatives

Derivatives

Power, product, quotient and chain rules, applied one at a time.

The derivative is the slope of the graph at each point. It is computed by rules — power, sum, constant multiple, product, quotient, chain — and the solver applies exactly one rule per step so you can see which rule fires where. The dashed curve is the derivative plotted against the function.

Finding the Average Rate of Change of a Function

The functions describing the examples above involve a change over time. Change divided by time is one example of a rate. The rates of change in the previous examples are each different. In other words, some changed faster than others. If we were to graph the functions, we could compare the rates by determining the slopes of the graphs.

A tangent line to a curve is a line that intersects the curve at only a single point but does not cross it there. (The tangent line may intersect the curve at another point away from the point of interest.) If we zoom in on a curve at that point, the curve appears linear, and the slope of the curve at that point is close to the slope of the tangent line at that point.

represents the function $f(x)={x}^{3}-4x.$ We can see the slope at various points along the curve.

slope at $x=-2$ is 8
slope at $x=-1$ is –1
slope at $x=2$ is 8

Let’s imagine a point on the curve of function $f$ at $x=a$ as shown in . The coordinates of the point are $(a,f(a)).$ Connect this point with a second point on the curve a little to the right of $x=a,$ with an x-value increased by some small real number $h.$ The coordinates of this second point are $(a+h,f(a+h))$ for some positive-value $h.$

We can calculate the slope of the line connecting the two points $(a,f(a))$ and $(a+h,f(a+h)),$ called a secant line, by applying the slope formula,

\[\text{slope = }\frac{\text{change in }y}{\text{change in }x}\]

We use the notation ${m}_{\text{sec}}$ to represent the slope of the secant line connecting two points.

\[\begin{array}{l}{m}_{\text{sec}}=\frac{f(a+h)-f(a)}{(a+h)-(a)} \\ =\frac{f(a+h)-f(a)}{a+h-a}\end{array}\]\[{m}_{\text{sec}}=\frac{f(a+h)-f(a)}{h}\]

Example

Try it.

Find the average rate of change connecting the points $(2,-6)$ and $(-1,5).$

Solution

We know the average rate of change connecting two points may be given by

\[\text{AROC}=\frac{f(a+h)-f(a)}{h}.\]

If one point is $(2,-6),$ or $(2,f(2)),$ then $f(2)=-6.$

The value $h$ is the displacement from $2$ to $-1,$ which equals $-1-2=-3.$

For the other point, $f(a+h)$ is the y-coordinate at $a+h,$ which is $2+(-3)$ or $-1,$ so $f(a+h)=f(-1)=5.$

\[\begin{array}{l}\text{AROC}=\frac{f(a+h)-f(a)}{h} \\ =\frac{5-(-6)}{-3} \\ =\frac{11}{-3} \\ =-\frac{11}{3}\end{array}\]

Condensed — the full section is in OpenStax Precalculus 2e.

Understanding the Instantaneous Rate of Change

Now that we can find the average rate of change, suppose we make $h$ in smaller and smaller. Then $a+h$ will approach $a$ as $h$ gets smaller, getting closer and closer to 0. Likewise, the second point $(a+h,f(a+h))$ will approach the first point, $(a,f(a)).$ As a consequence, the connecting line between the two points, called the secant line, will get closer and closer to being a tangent to the function at $x=a,$ and the slope of the secant line will get closer and closer to the slope of the tangent at $x=a.$ See .

Because we are looking for the slope of the tangent at $x=a,$ we can think of the measure of the slope of the curve of a function $f$ at a given point as the rate of change at a particular instant. We call this slope the instantaneous rate of change, or the derivative of the function at $x=a.$ Both can be found by finding the limit of the slope of a line connecting the point at $x=a$ with a second point infinitesimally close along the curve. For a function $f$ both the instantaneous rate of change of the function and the derivative of the function at $x=a$ are written as $f'(a),$ and we can define them as a two-sided limit that has the same value whether approached from the left or the right.

\[{f}^{'}(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}\]

The expression by which the limit is found is known as the difference quotient.

Condensed — the full section is in OpenStax Precalculus 2e.

Finding Instantaneous Rates of Change

Many applications of the derivative involve determining the rate of change at a given instant of a function with the independent variable time—which is why the term instantaneous is used. Consider the height of a ball tossed upward with an initial velocity of 64 feet per second, given by $s(t)=-16{t}^{2}+64t+6,$ where $t$ is measured in seconds and $s(t)$ is measured in feet. We know the path is that of a parabola. The derivative will tell us how the height is changing at any given point in time. The height of the ball is shown in as a function of time. In physics, we call this the “s-t graph.”

Example

Try it.

Using the function above, $s(t)=-16{t}^{2}+64t+6,$ what is the instantaneous velocity of the ball at 1 second and 3 seconds into its flight?

Solution

The velocity at $t=1$ and $t=3$ is the instantaneous rate of change of distance per time, or velocity. Notice that the initial height is 6 feet. To find the instantaneous velocity, we find the derivative and evaluate it at $t=1$ and $t=3:$

\[\begin{array}{ll}{f}^{'}(a)=\underset{h\ \to \ 0}{\lim }\ \frac{f(a+h)-f(a)}{h} & \\ {s}^{'}(t)=\underset{h\ \to \ 0}{\lim }\ \frac{-16{(t+h)}^{2}+64(t+h)+6-(-16{t}^{2}+64t+6)}{h} & \text{Substitute }s(t+h)\text{ and }s(t). \\ =\underset{h\ \to \ 0}{\lim \ }\frac{-16{t}^{2}-32ht-{h}^{2}+64t+64h+6+16{t}^{2}-64t-6}{h} & \text{Distribute}. \\ =\underset{h\ \to \ 0}{\lim }\ \frac{-32ht-{h}^{2}+64h}{h} & \text{Simplify}. \\ =\underset{h\ \to \ 0}{\lim }\ \frac{h(-32t-h+64)}{h} & \text{Factor the numerator}. \\ =\underset{h\ \to \ 0}{\lim }-32t-h+64 & \text{Cancel out the common factor }h. \\ {s}^{'}(t)=-32t+64 & \text{Evaluate the limit by letting }h=0.\end{array}\]

For any value of $t$, ${s}^{'}(t)$ tells us the velocity at that value of $t.$

Evaluate $t=1$ and $t=3.$

\[\begin{array}{l}{s}^{'}(1)=-32(1)+64=32 \\ {s}^{'}(3)=-32(3)+64=-32\end{array}\]

The velocity of the ball after 1 second is 32 feet per second, as it is on the way up.

The velocity of the ball after 3 seconds is $-32$ feet per second, as it is on the way down.

Condensed — the full section is in OpenStax Precalculus 2e.

Finding an Equation of a Line Tangent to the Graph of a Function

The equation of a tangent line to a curve of the function $f(x)$ at $x=a$ is derived from the point-slope form of a line, $y=m(x-{x}_{1})+{y}_{1}.$ The slope of the line is the slope of the curve at $x=a$ and is therefore equal to ${f}^{'}(a),$ the derivative of $f(x)$ at $x=a.$ The coordinate pair of the point on the line at $x=a$ is $(a,f(a)).$

If we substitute into the point-slope form, we have

The equation of the tangent line is

\[y=f'(a)(x-a)+f(a)\]

Example

Try it.

Find the equation of a line tangent to the curve $f(x)={x}^{2}-4x$ at $x=3.$

Solution

Using:

\[f'(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}\]

Substitute $f(a+h)={(a+h)}^{2}-4(a+h)$ and $f(a)={a}^{2}-4a.$

\[\begin{array}{ll}{f}^{'}(a)=\underset{h\to 0}{\lim }\frac{(a+h)(a+h)-4(a+h)-({a}^{2}-4a)}{h} & \\ =\underset{h\to 0}{\lim }\frac{{a}^{2}+2ah+{h}^{2}-4a-4h-{a}^{2}+4a}{h} & \text{Remove parentheses}. \\ =\underset{h\to 0}{\lim }\frac{{a}^{2}+2ah+{h}^{2}-4a-4h-{a}^{2}+4a}{h} & \text{Combine like terms}. \\ =\underset{h\to 0}{\lim }\frac{2ah+{h}^{2}-4h}{h} & \\ =\underset{h\to 0}{\lim }\frac{h(2a+h-4)}{h} & \text{Factor out }h. \\ =2a+0-4 & \\ {f}^{'}(a)=2a-4 & \text{Evaluate the limit}. \\ {f}^{'}(3)=2(3)-4=2 & \end{array}\]

Equation of tangent line at $x=3:$

\[\begin{array}{l}y=f'(a)(x-a)+f(a) \\ y=f'(3)(x-3)+f(3) \\ y=2(x-3)+(-3) \\ y=2x-9\end{array}\]

Finding the Instantaneous Speed of a Particle

If a function measures position versus time, the derivative measures displacement versus time, or the speed of the object. A change in speed or direction relative to a change in time is known as velocity. The velocity at a given instant is known as instantaneous velocity.

In trying to find the speed or velocity of an object at a given instant, we seem to encounter a contradiction. We normally define speed as the distance traveled divided by the elapsed time. But in an instant, no distance is traveled, and no time elapses. How will we divide zero by zero? The use of a derivative solves this problem. A derivative allows us to say that even while the object’s velocity is constantly changing, it has a certain velocity at a given instant. That means that if the object traveled at that exact velocity for a unit of time, it would travel the specified distance.

Example

Try it.

A ball is tossed upward from a height of 200 feet with an initial velocity of 36 ft/sec. If the height of the ball in feet after $t$ seconds is given by $s(t)=-16{t}^{2}+36t+200,$ find the instantaneous velocity of the ball at $t=2.$

Solution

First, we must find the derivative ${s}^{'}(t)$. Then we evaluate the derivative at $t=2,$ using $s(a+h)=-16{(a+h)}^{2}+36(a+h)+200$ and $s(a)=-16{a}^{2}+36a+200.$

\[\begin{array}{l}{s}^{'}(a)=\underset{h\to 0}{\lim }\frac{s(a+h)-s(a)}{h} \\ =\underset{h\to 0}{\lim }\frac{-16{(a+h)}^{2}+36(a+h)+200-(-16{a}^{2}+36a+200)}{h} \\ =\underset{h\to 0}{\lim }\frac{-16({a}^{2}+2ah+{h}^{2})+36(a+h)+200-(-16{a}^{2}+36a+200)}{h} \\ =\underset{h\to 0}{\lim }\frac{-16{a}^{2}-32ah-16{h}^{2}+36a+36h+200+16{a}^{2}-36a-200}{h} \\ =\underset{h\to 0}{\lim }\frac{-16{a}^{2}-32ah-16{h}^{2}+36a+36h+200+16{a}^{2}-36a-200}{h} \\ =\underset{h\to 0}{\lim }\frac{-32ah-16{h}^{2}+36h}{h} \\ =\underset{h\to 0}{\lim }\frac{h(-32a-16h+36)}{h} \\ =\underset{h\to 0}{\lim }(-32a-16h+36) \\ =-32a-16⋅0+36 \\ {s}^{'}(a)=-32a+36 \\ {s}^{'}(2)=-32(2)+36 \\ =-28\end{array}\]

Condensed — the full section is in OpenStax Precalculus 2e.

Key Equations

average rate of change	$\text{AROC}=\frac{f(a+h)-f(a)}{h}$
derivative of a function	${f}^{'}(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}$

Key Concepts

The slope of the secant line connecting two points is the average rate of change of the function between those points. See .
The derivative, or instantaneous rate of change, is a measure of the slope of the curve of a function at a given point, or the slope of the line tangent to the curve at that point. See , , and .
The difference quotient is the quotient in the formula for the instantaneous rate of change:
$\frac{f(a+h)-f(a)}{h}$
Instantaneous rates of change can be used to find solutions to many real-world problems. See .
The instantaneous rate of change can be found by observing the slope of a function at a point on a graph by drawing a line tangent to the function at that point. See .
Instantaneous rates of change can be interpreted to describe real-world situations. See and .
Some functions are not differentiable at a point or points. See .
The point-slope form of a line can be used to find the equation of a line tangent to the curve of a function. See .
Velocity is a change in position relative to time. Instantaneous velocity describes the velocity of an object at a given instant. Average velocity describes the velocity maintained over an interval of time.
Using the derivative makes it possible to calculate instantaneous velocity even though there is no elapsed time. See .

કામ કરેલ ઉદાહરણ: `derivative of sin(x)*x^2`

Calculus · Derivatives

Differentiate x^2·sin(x)

\frac{d}{dx}\left[x^{2} \sin{\left(x \right)}\right]

પગલું દ્વારા પગલું

\frac{d}{dx}\left[x^{2} \sin{\left(x \right)}\right]
Start from the derivative to compute.
x^{2} \frac{d}{d x} \sin{\left(x \right)} + \sin{\left(x \right)} \frac{d}{d x} x^{2}
Product rule with u = x^{2} and v = \sin{\left(x \right)}: (uv)′ = u′v + uv′.
x^{2} \cos{\left(x \right)} + \sin{\left(x \right)} \frac{d}{d x} x^{2}
(sin u)′ = cos u.
x^{2} \cos{\left(x \right)} + 2 x \sin{\left(x \right)}
Power rule: (xⁿ)′ = n·xⁿ⁻¹ with n = 2.
x \left(x \cos{\left(x \right)} + 2 \sin{\left(x \right)}\right)
Simplify.

જવાબ બતાવો

f'(x) = x \left(x \cos{\left(x \right)} + 2 \sin{\left(x \right)}\right)

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Find the average rate of change connecting the points $(2,-6)$ and $(-1,5).$

જવાબ બતાવો

We know the average rate of change connecting two points may be given by
\[\text{AROC}=\frac{f(a+h)-f(a)}{h}.\]
If one point is $(2,-6),$ or $(2,f(2)),$ then $f(2)=-6.$
The value $h$ is the displacement from $2$ to $-1,$ which equals $-1-2=-3.$
For the other point, $f(a+h)$ is the y-coordinate at $a+h,$ which is $2+(-3)$ or $-1,$ so $f(a+h)=f(-1)=5.$
\[\begin{array}{l}\text{AROC}=\frac{f(a+h)-f(a)}{h} \\ =\frac{5-(-6)}{-3} \\ =\frac{11}{-3} \\ =-\frac{11}{3}\end{array}\]
Find the average rate of change connecting the points $(-5,1.5)$ and $(-2.5,9).$

જવાબ બતાવો

3
Find the derivative of the function $f(x)={x}^{2}-3x+5$ at $x=a.$

જવાબ બતાવો

We have:
\[\begin{array}{ll}{f}^{'}(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h} & \text{ Definition of a derivative }\end{array}\]
Substitute $f(a+h)={(a+h)}^{2}-3(a+h)+5$ and $f(a)={a}^{2}-3a+5.$
\[\begin{array}{ll}{f}^{'}(a)=\underset{h\to 0}{\lim }\frac{(a+h)(a+h)-3(a+h)+5-({a}^{2}-3a+5)}{h} & \\ =\underset{h\to 0}{\lim }\frac{{a}^{2}+2ah+{h}^{2}-3a-3h+5-{a}^{2}+3a-5}{h} & \text{Evaluate to remove parentheses}. \\ =\underset{h\to 0}{\lim }\frac{{a}^{2}+2ah+{h}^{2}-3a-3h+5-{a}^{2}+3a-5}{h} & \text{Simplify}. \\ =\underset{h\to 0}{\lim }\frac{2ah+{h}^{2}-3h}{h} & \\ =\underset{h\to 0}{\lim }\frac{h(2a+h-3)}{h} & \text{Factor out an }h. \\ =2a+0-3 & \text{Evaluate the limit}. \\ =2a-3 & \end{array}\]
Find the derivative of the function $f(x)=3{x}^{2}+7x$ at $x=a.$

જવાબ બતાવો

${f}^{'}(a)=6a+7$
Find the derivative of the function $f(x)=\frac{3+x}{2-x}$ at $x=a.$

જવાબ બતાવો

$\begin{array}{ll}{f}^{'}(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h} & \\ =\underset{h\to 0}{\lim }\frac{\frac{3+(a+h)}{2-(a+h)}-(\frac{3+a}{2-a})}{h} & \text{Substitute }f(a+h)\text{ and }f(a) \\ =\underset{h\to 0}{\lim }\frac{(2-(a+h))(2-a)[\frac{3+(a+h)}{2-(a+h)}-(\frac{3+a}{2-a})]}{(2-(a+h))(2-a)(h)} & \text{Multiply numerator and denominator by }(2-(a+h))(2-a) \\ =\underset{h\to 0}{\lim }\frac{(2-(a+h))(2-a)(\frac{3+(a+h)}{(2-(a+h))})-(2-(a+h))(2-a)(\frac{3+a}{2-a})}{(2-(a+h))(2-a)(h)} & \text{Distribute} \\ =\underset{h\to 0}{\lim }\frac{6-3a+2a-{a}^{2}+2h-ah-6+3a+3h-2a+{a}^{2}+ah}{(2-(a+h))(2-a)(h)} & \text{Multiply} \\ =\underset{h\to 0}{\lim }\frac{5h}{(2-(a+h))(2-a)(h)} & \text{Combine like terms} \\ =\underset{h\to 0}{\lim }\frac{5}{(2-(a+h))(2-a)} & \text{Cancel like factors} \\ =\frac{5}{(2-(a+0))(2-a)}=\frac{5}{(2-a)(2-a)}=\frac{5}{{(2-a)}^{2}} & \text{Evaluate the limit}\end{array}$
Find the derivative of the function $f(x)=\frac{10x+11}{5x+4}$ at $x=a.$

જવાબ બતાવો

${f}^{'}(a)=\frac{-15}{{(5a+4)}^{2}}$
Find the derivative of the function $f(x)=4\sqrt{x}$ at $x=36.$

જવાબ બતાવો

We have
\[\begin{array}{ll}{f}^{'}(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h} & \\ =\underset{h\to 0}{\lim }\frac{4\sqrt{a+h}-4\sqrt{a}}{h} & \text{Substitute }f(a+h)\text{ and }f(a)\end{array}\]
Multiply the numerator and denominator by the conjugate: $\frac{4\sqrt{a+h}+4\sqrt{a}}{4\sqrt{a+h}+4\sqrt{a}}.$
\[\begin{array}{ll}\ {f}^{'}(a)=\underset{h\to 0}{\lim }(\frac{4\sqrt{a+h}-4\sqrt{a}}{h})⋅(\frac{4\sqrt{a+h}+4\sqrt{a}}{4\sqrt{a+h}+4\sqrt{a}}) & \\ =\underset{h\to 0}{\lim }(\frac{16(a+h)-16a}{h4(\sqrt{a+h}+\sqrt{a})}) & \text{Multiply}. \\ =\underset{h\to 0}{\lim }(\frac{16a+16h-16a}{h4(\sqrt{a+h}+\sqrt{a})}) & \text{Distribute and combine like terms}. \\ =\underset{h\to 0}{\lim }(\frac{16h}{h(4\sqrt{a+h}+4\sqrt{a})}) & \text{Simplify}. \\ =\underset{h\to 0}{\lim }(\frac{16}{4\sqrt{a+h}+4\sqrt{a}}) & \text{Evaluate the limit by letting }h=0. \\ =\frac{16}{8\sqrt{a}}=\frac{2}{\sqrt{a}} & \\ {f}^{'}(36)=\frac{2}{\sqrt{36}} & \text{Evaluate the derivative at }x=36. \\ =\frac{2}{6} & \\ =\frac{1}{3} & \end{array}\]
Find the derivative of the function $f(x)=9\sqrt{x}$ at $x=9.$

જવાબ બતાવો

$\frac{3}{2}$
Using the function above, $s(t)=-16{t}^{2}+64t+6,$ what is the instantaneous velocity of the ball at 1 second and 3 seconds into its flight?

જવાબ બતાવો

The velocity at $t=1$ and $t=3$ is the instantaneous rate of change of distance per time, or velocity. Notice that the initial height is 6 feet. To find the instantaneous velocity, we find the derivative and evaluate it at $t=1$ and $t=3:$
\[\begin{array}{ll}{f}^{'}(a)=\underset{h\ \to \ 0}{\lim }\ \frac{f(a+h)-f(a)}{h} & \\ {s}^{'}(t)=\underset{h\ \to \ 0}{\lim }\ \frac{-16{(t+h)}^{2}+64(t+h)+6-(-16{t}^{2}+64t+6)}{h} & \text{Substitute }s(t+h)\text{ and }s(t). \\ =\underset{h\ \to \ 0}{\lim \ }\frac{-16{t}^{2}-32ht-{h}^{2}+64t+64h+6+16{t}^{2}-64t-6}{h} & \text{Distribute}. \\ =\underset{h\ \to \ 0}{\lim }\ \frac{-32ht-{h}^{2}+64h}{h} & \text{Simplify}. \\ =\underset{h\ \to \ 0}{\lim }\ \frac{h(-32t-h+64)}{h} & \text{Factor the numerator}. \\ =\underset{h\ \to \ 0}{\lim }-32t-h+64 & \text{Cancel out the common factor }h. \\ {s}^{'}(t)=-32t+64 & \text{Evaluate the limit by letting }h=0.\end{array}\]
For any value of $t$, ${s}^{'}(t)$ tells us the velocity at that value of $t.$
Evaluate $t=1$ and $t=3.$
\[\begin{array}{l}{s}^{'}(1)=-32(1)+64=32 \\ {s}^{'}(3)=-32(3)+64=-32\end{array}\]
The velocity of the ball after 1 second is 32 feet per second, as it is on the way up.
The velocity of the ball after 3 seconds is $-32$ feet per second, as it is on the way down.
The position of the ball is given by $s(t)=-16{t}^{2}+64t+6.$ What is its velocity 2 seconds into flight?

જવાબ બતાવો

0
From the graph of the function $y=f(x)$ presented in , estimate each of the following:
1. ⓐ $f(0)$
2. ⓑ $f(2)$
3. ⓒ $f'(0)$
4. ⓓ $f'(2)$
જવાબ બતાવો
To find the functional value, $f(a),$ find the y-coordinate at $x=a.$
To find the derivative at $x=a,$ ${f}^{'}(a),$ draw a tangent line at $x=a,$ and estimate the slope of that tangent line. See .
1. ⓐ $f(0)$ is the y-coordinate at $x=0.$ The point has coordinates $(0,1),$ thus $f(0)=1.$
2. ⓑ $f(2)$ is the y-coordinate at $x=2.$ The point has coordinates $(2,1),$ thus $f(2)=1.$
3. ⓒ ${f}^{'}(0)$ is found by estimating the slope of the tangent line to the curve at $x=0.$ The tangent line to the curve at $x=0$ appears horizontal. Horizontal lines have a slope of 0, thus ${f}^{'}(0)=0.$
4. ⓓ ${f}^{'}(2)$ is found by estimating the slope of the tangent line to the curve at $x=2.$ Observe the path of the tangent line to the curve at $x=2.$ As the $x$ value moves one unit to the right, the $y$ value moves up four units to another point on the line. Thus, the slope is 4, so ${f}^{'}(2)=4.$
Using the graph of the function $f(x)={x}^{3}-3x$ shown in , estimate: $f(1),$ ${f}^{'}(1),$ $f(0),$ and ${f}^{'}(0).$

જવાબ બતાવો

$-2,$ 0, 0, $-3$
The cost in dollars of producing $x$ laptop computers in dollars is $f(x)={x}^{2}-100x.$ At the point where 200 computers have been produced, what is the approximate cost of producing the 201^st unit?

જવાબ બતાવો

If $f(x)={x}^{2}-100x$ describes the cost of producing $x$ computers, ${f}^{'}(x)$ will describe the marginal cost. We need to find the derivative. For purposes of calculating the derivative, we can use the following functions:
\[\begin{array}{l}f(x+h)={(x+h)}^{2}-100(x+h) \\ f(x)={x}^{2}-100x\end{array}\]
$\begin{array}{ll}\ {f}^{'}(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h} & \text{Formula for a derivative} \\ =\underset{h\to 0}{\lim }\frac{{(x+h)}^{2}-100(x+h)-({x}^{2}-100x)}{h} & \text{Substitute }f(x+h)\text{ and }f(x). \\ =\underset{h\to 0}{\lim }\frac{{x}^{2}+2xh+{h}^{2}-100x-100h-{x}^{2}+100x}{h} & \text{Multiply polynomials, distribute}. \\ =\underset{h\to 0}{\lim }\frac{2xh+{h}^{2}-100h}{h} & \text{Collect like terms}. \\ =\underset{h\to 0}{\lim }\frac{h(2x+h-100)}{h} & \text{Factor and cancel like terms}. \\ =\underset{h\to 0}{\lim }\ 2x+h-100 & \text{Simplify}. \\ =(2x-100) & \text{Evaluate when }h=0. \\ {f}^{'}(x)=2x-100 & \text{Formula for marginal cost} \\ {f}^{'}(200)=2(200)-100=300 & \text{Evaluate for 200 units}.\end{array}$
The marginal cost of producing the 201^st unit will be approximately $300.
A car leaves an intersection. The distance it travels in miles is given by the function $f(t),$ where $t$ represents hours. Explain the following notations:
1. ⓐ $f(0)=0$
2. ⓑ ${f}^{'}(1)=60$
3. ⓒ $f(1)=70$
4. ⓓ $f(2.5)=150$
જવાબ બતાવો
First we need to evaluate the function $f(t)$ and the derivative of the function ${f}^{'}(t),$ and distinguish between the two. When we evaluate the function $f(t),$ we are finding the distance the car has traveled in $t$ hours. When we evaluate the derivative ${f}^{'}(t),$ we are finding the speed of the car after $t$ hours.
1. ⓐ $f(0)=0$ means that in zero hours, the car has traveled zero miles.
2. ⓑ ${f}^{'}(1)=60$ means that one hour into the trip, the car is traveling 60 miles per hour.
3. ⓒ $f(1)=70$ means that one hour into the trip, the car has traveled 70 miles. At some point during the first hour, then, the car must have been traveling faster than it was at the 1-hour mark.
4. ⓓ $f(2.5)=150$ means that two hours and thirty minutes into the trip, the car has traveled 150 miles.
A runner runs along a straight east-west road. The function $f(t)$ gives how many feet eastward of her starting point she is after $t$ seconds. Interpret each of the following as it relates to the runner.
1. ⓐ $f(0)=0$
2. ⓑ $f(10)=150$
3. ⓒ ${f}^{'}(10)=15$
4. ⓓ ${f}^{'}(20)=-10$
5. ⓔ $f(40)=-100$
જવાબ બતાવો
1. ⓐ After zero seconds, she has traveled 0 feet.
2. ⓑ After 10 seconds, she has traveled 150 feet east.
3. ⓒ After 10 seconds, she is moving eastward at a rate of 15 ft/sec.
4. ⓓ After 20 seconds, she is moving westward at a rate of 10 ft/sec.
5. ⓔAfter 40 seconds, she is 100 feet westward of her starting point.
Using , determine where the function is
1. continuous
2. discontinuous
3. differentiable
4. not differentiable
At the points where the graph is discontinuous or not differentiable, state why.
જવાબ બતાવો

The graph of $f(x)$ is continuous on $(-\infty ,-2)\cup (-2, 1)\cup (1,\infty ).$ The graph of $f(x)$ has a removable discontinuity at $x=-2$ and a jump discontinuity at $x=1.$ See .
The graph of is differentiable on $(-\infty ,-2)\cup (-2,-1)\cup (-1,1)\cup (1,2)\cup (2,\infty ).$ The graph of $f(x)$ is not differentiable at $x=-2$ because it is a point of discontinuity, at $x=-1$ because of a sharp corner, at $x=1$ because it is a point of discontinuity, and at $x=2$ because of a sharp corner. See .
Determine where the function $y=f(x)$ shown in is continuous and differentiable from the graph.

જવાબ બતાવો

The graph of $f$ is continuous on $(-\infty ,1)\cup (1,3)\cup (3,\infty ).$ The graph of $f$ is discontinuous at $x=1$ and $x=3.$ The graph of $f$ is differentiable on $(-\infty ,1)\cup (1,3)\cup (3,\infty ).$ The graph of $f$ is not differentiable at $x=1$ and $x=3.$
Find the equation of a line tangent to the curve $f(x)={x}^{2}-4x$ at $x=3.$

જવાબ બતાવો

Using:
\[f'(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}\]
Substitute $f(a+h)={(a+h)}^{2}-4(a+h)$ and $f(a)={a}^{2}-4a.$
\[\begin{array}{ll}{f}^{'}(a)=\underset{h\to 0}{\lim }\frac{(a+h)(a+h)-4(a+h)-({a}^{2}-4a)}{h} & \\ =\underset{h\to 0}{\lim }\frac{{a}^{2}+2ah+{h}^{2}-4a-4h-{a}^{2}+4a}{h} & \text{Remove parentheses}. \\ =\underset{h\to 0}{\lim }\frac{{a}^{2}+2ah+{h}^{2}-4a-4h-{a}^{2}+4a}{h} & \text{Combine like terms}. \\ =\underset{h\to 0}{\lim }\frac{2ah+{h}^{2}-4h}{h} & \\ =\underset{h\to 0}{\lim }\frac{h(2a+h-4)}{h} & \text{Factor out }h. \\ =2a+0-4 & \\ {f}^{'}(a)=2a-4 & \text{Evaluate the limit}. \\ {f}^{'}(3)=2(3)-4=2 & \end{array}\]
Equation of tangent line at $x=3:$
\[\begin{array}{l}y=f'(a)(x-a)+f(a) \\ y=f'(3)(x-3)+f(3) \\ y=2(x-3)+(-3) \\ y=2x-9\end{array}\]
Find the equation of a tangent line to the curve of the function $f(x)=5{x}^{2}-x+4$ at $x=2.$

જવાબ બતાવો

$y=19x-16$
A ball is tossed upward from a height of 200 feet with an initial velocity of 36 ft/sec. If the height of the ball in feet after $t$ seconds is given by $s(t)=-16{t}^{2}+36t+200,$ find the instantaneous velocity of the ball at $t=2.$

જવાબ બતાવો

First, we must find the derivative ${s}^{'}(t)$. Then we evaluate the derivative at $t=2,$ using $s(a+h)=-16{(a+h)}^{2}+36(a+h)+200$ and $s(a)=-16{a}^{2}+36a+200.$
\[\begin{array}{l}{s}^{'}(a)=\underset{h\to 0}{\lim }\frac{s(a+h)-s(a)}{h} \\ =\underset{h\to 0}{\lim }\frac{-16{(a+h)}^{2}+36(a+h)+200-(-16{a}^{2}+36a+200)}{h} \\ =\underset{h\to 0}{\lim }\frac{-16({a}^{2}+2ah+{h}^{2})+36(a+h)+200-(-16{a}^{2}+36a+200)}{h} \\ =\underset{h\to 0}{\lim }\frac{-16{a}^{2}-32ah-16{h}^{2}+36a+36h+200+16{a}^{2}-36a-200}{h} \\ =\underset{h\to 0}{\lim }\frac{-16{a}^{2}-32ah-16{h}^{2}+36a+36h+200+16{a}^{2}-36a-200}{h} \\ =\underset{h\to 0}{\lim }\frac{-32ah-16{h}^{2}+36h}{h} \\ =\underset{h\to 0}{\lim }\frac{h(-32a-16h+36)}{h} \\ =\underset{h\to 0}{\lim }(-32a-16h+36) \\ =-32a-16⋅0+36 \\ {s}^{'}(a)=-32a+36 \\ {s}^{'}(2)=-32(2)+36 \\ =-28\end{array}\]
A fireworks rocket is shot upward out of a pit 12 ft below the ground at a velocity of 60 ft/sec. Its height in feet after $t$ seconds is given by $s=-16{t}^{2}+60t-12.$ What is its instantaneous velocity after 4 seconds?

જવાબ બતાવો

–68 ft/sec, it is dropping back to Earth at a rate of 68 ft/s.
How is the slope of a linear function similar to the derivative?

જવાબ બતાવો

The slope of a linear function stays the same. The derivative of a general function varies according to $x.$ Both the slope of a line and the derivative at a point measure the rate of change of the function.
What is the difference between the average rate of change of a function on the interval $[x,x+h]$ and the derivative of the function at $x?$
A car traveled 110 miles during the time period from 2:00 P.M. to 4:00 P.M. What was the car's average velocity? At exactly 2:30 P.M., the speed of the car registered exactly 62 miles per hour. What is another name for the speed of the car at 2:30 P.M.? Why does this speed differ from the average velocity?

જવાબ બતાવો

Average velocity is 55 miles per hour. The instantaneous velocity at 2:30 p.m. is 62 miles per hour. The instantaneous velocity measures the velocity of the car at an instant of time whereas the average velocity gives the velocity of the car over an interval.
Explain the concept of the slope of a curve at point $x.$
Suppose water is flowing into a tank at an average rate of 45 gallons per minute. Translate this statement into the language of mathematics.

જવાબ બતાવો

The average rate of change of the amount of water in the tank is 45 gallons per minute. If $f(x)$ is the function giving the amount of water in the tank at any time $t,$ then the average rate of change of $f(x)$ between $t=a$ and $t=b$ is $f(a)+45(b-a).$
$f(x)=3x-4$
$f(x)=-2x+1$

જવાબ બતાવો

${f}^{'}(x)=-2$
$f(x)={x}^{2}-2x+1$
$f(x)=2{x}^{2}+x-3$

જવાબ બતાવો

${f}^{'}(x)=4x+1$
$f(x)=2{x}^{2}+5$
$f(x)=\frac{-1}{x-2}$

જવાબ બતાવો

${f}^{'}(x)=\frac{1}{{(x-2)}^{2}}$
$f(x)=\frac{2+x}{1-x}$
$f(x)=\frac{5-2x}{3+2x}$

જવાબ બતાવો

$\frac{-16}{{(3+2x)}^{2}}$
$f(x)=\sqrt{1+3x}$
$f(x)=3{x}^{3}-{x}^{2}+2x+5$

જવાબ બતાવો

${f}^{'}(x)=9{x}^{2}-2x+2$
$f(x)=5$
$f(x)=5\pi$

જવાબ બતાવો

${f}^{'}(x)=0$
$(-2,0)$ and $(-4,5)$
$(4,-3)$ and $(-2,-1)$

જવાબ બતાવો

$-\frac{1}{3}$

Symbols used here

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Derivatives

Find the derivative of a function.
Find instantaneous rates of change.
Find an equation of the tangent line to the graph of a function at a point.
Find the instantaneous velocity of a particle.
slope at
slope at
slope at
The derivative of a function

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

તમારા પોતાના પ્રયત્ન કરો

Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

આમાં વધુ Calculus

Limits Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests

average rate of change	\(\text{AROC}=\frac{f(a+h)-f(a)}{h}\)
derivative of a function	\({f}^{'}(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}\)

Finding the Average Rate of Change of a Function

Understanding the Instantaneous Rate of Change

Finding Instantaneous Rates of Change

Finding an Equation of a Line Tangent to the Graph of a Function

Finding the Instantaneous Speed of a Particle

Key Equations

Key Concepts

કામ કરેલ ઉદાહરણ: derivative of sin(x)*x^2

Differentiate x^2·sin(x)

પગલું દ્વારા પગલું

Practice (40)

Symbols used here

How to: Derivatives

Questions people ask

તમારા પોતાના પ્રયત્ન કરો

આમાં વધુ Calculus

કામ કરેલ ઉદાહરણ: `derivative of sin(x)*x^2`