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Defining the Derivative

Recognize the meaning of the tangent to a curve at a point.

Tangent Lines

We begin our study of calculus by revisiting the notion of secant lines and tangent lines. Recall that we used the slope of a secant line to a function at a point $(a,f(a))$ to estimate the rate of change, or the rate at which one variable changes in relation to another variable. We can obtain the slope of the secant by choosing a value of $x$ near $a$ and drawing a line through the points $(a,f(a))$ and $(x,f(x)),$ as shown in . The slope of this line is given by an equation in the form of a difference quotient:

\[{m}_{\text{sec}}=\frac{f(x)-f(a)}{x-a}.\]

We can also calculate the slope of a secant line to a function at a value a by using this equation and replacing $x$ with $a+h,$ where $h$ is a value close to 0. We can then calculate the slope of the line through the points $(a,f(a))$ and $(a+h,f(a+h)).$ In this case, we find the secant line has a slope given by the following difference quotient with increment $h\text{:}$

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

These two expressions for calculating the slope of a secant line are illustrated in . We will see that each of these two methods for finding the slope of a secant line is of value. Depending on the setting, we can choose one or the other. The primary consideration in our choice usually depends on ease of calculation.

In (a) we see that, as the values of $x$ approach $a,$ the slopes of the secant lines provide better estimates of the rate of change of the function at $a.$ Furthermore, the secant lines themselves approach the tangent line to the function at $a,$ which represents the limit of the secant lines. Similarly, (b) shows that as the values of $h$ get closer to $0,$ the secant lines also approach the tangent line. The slope of the tangent line at $a$ is the rate of change of the function at $a,$ as shown in (c).

Formally we may define the tangent line to the graph of a function as follows.

Condensed — the full section is in OpenStax Calculus Volume 1.

The Derivative of a Function at a Point

The type of limit we compute in order to find the slope of the line tangent to a function at a point occurs in many applications across many disciplines. These applications include velocity and acceleration in physics, marginal profit functions in business, and growth rates in biology. This limit occurs so frequently that we give this value a special name: the derivative. The process of finding a derivative is called differentiation.

Example

Try it.

For $f(x)={x}^{2},$ use a table to estimate ${f}^{'}(3)$ using .

Solution

Create a table using values of $x$ just below $3$ and just above $3.$

$x$	$\frac{{x}^{2}-9}{x-3}$
$2.9$	$5.9$
$2.99$	$5.99$
$2.999$	$5.999$
$3.001$	$6.001$
$3.01$	$6.01$
$3.1$	$6.1$

After examining the table, we see that a good estimate is ${f}^{'}(3)=6.$

Example

Try it.

For $f(x)=3{x}^{2}-4x+1,$ find ${f}^{'}(2)$ by using .

Solution

Substitute the given function and value directly into the equation.

\[\begin{array}{lllll}{f}^{'}(2) & =\underset{x\to 2}{\text{lim}}\frac{f(x)-f(2)}{x-2} & & & \text{Apply the definition.} \\ & =\underset{x\to 2}{\text{lim}}\frac{(3{x}^{2}-4x+1)-5}{x-2} & & & \text{Substitute}\ f(x)=3{x}^{2}-4x+1\ \text{and}\ f(2)=5. \\ & =\underset{x\to 2}{\text{lim}}\frac{(x-2)(3x+2)}{x-2} & & & \text{Simplify and factor the numerator.} \\ & =\underset{x\to 2}{\text{lim}}(3x+2) & & & \text{Cancel the common factor.} \\ & =8 & & & \text{Evaluate the limit.}\end{array}\]

Condensed — the full section is in OpenStax Calculus Volume 1.

Velocities and Rates of Change

Now that we can evaluate a derivative, we can use it in velocity applications. Recall that if $s(t)$ is the position of an object moving along a coordinate axis, the average velocity of the object over a time interval $[a,t]$ if $t>a$ or $[t,a]$ if \(t\[{v}_{\text{ave}}=\frac{s(t)-s(a)}{t-a}.\]

As the values of $t$ approach $a,$ the values of ${v}_{\text{ave}}$ approach the value we call the instantaneous velocity at $a.$ That is, instantaneous velocity at $a,$ denoted $v(a),$ is given by

\[v(a)={s}^{'}(a)=\underset{t\to a}{\text{lim}}\frac{s(t)-s(a)}{t-a}.\]

To better understand the relationship between average velocity and instantaneous velocity, see . In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time $t=a$ whose position at time $t$ is given by the function $s(t).$ The slope of the secant line (shown in green) is the average velocity of the object over the time interval $[a,t].$

We can use to calculate the instantaneous velocity, or we can estimate the velocity of a moving object by using a table of values. We can then confirm the estimate by using .

Example

Try it.

A lead weight on a spring is oscillating up and down. Its position at time $t$ with respect to a fixed horizontal line is given by $s(t)=\text{sin}\ t$ (). Use a table of values to estimate $v(0).$ Check the estimate by using .

Solution

We can estimate the instantaneous velocity at $t=0$ by computing a table of average velocities using values of $t$ approaching $0,$ as shown in .

$t$	$\frac{\text{sin}\ t-\text{sin}\ 0}{t-0}=\frac{\text{sin}\ t}{t}$
$-0.1$	$0.998334166$
$-0.01$	$0.9999833333$
$-0.001$	$0.999999833$
$0.001$	$0.999999833$
$0.01$	$0.9999833333$
$0.1$	$0.998334166$

From the table we see that the average velocity over the time interval $[-0.1,0]$ is $0.998334166,$ the average velocity over the time interval $[-0.01,0]$ is $0.9999833333,$ and so forth. Using this table of values, it appears that a good estimate is $v(0)=1.$

By using , we can see that

\[v(0)={s}^{'}(0)=\underset{t\to 0}{\text{lim}}\frac{\text{sin}\ t-\text{sin}\ 0}{t-0}=\underset{t\to 0}{\text{lim}}\frac{\text{sin}\ t}{t}=1.\]

Thus, in fact, $v(0)=1.$

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

The slope of the tangent line to a curve measures the instantaneous rate of change of a curve. We can calculate it by finding the limit of the difference quotient or the difference quotient with increment $h.$
The derivative of a function $f(x)$ at a value $a$ is found using either of the definitions for the slope of the tangent line.
Velocity is the rate of change of position. As such, the velocity $v(t)$ at time $t$ is the derivative of the position $s(t)$ at time $t.$ Average velocity is given by
\[{v}_{\text{ave}}=\frac{s(t)-s(a)}{t-a}.\]
Instantaneous velocity is given by
\[v(a)={s}^{'}(a)=\underset{t\to a}{\text{lim}}\frac{s(t)-s(a)}{t-a}.\]
We may estimate a derivative by using a table of values.

Key Equations

Difference quotient	$Q=\frac{f(x)-f(a)}{x-a}$
Difference quotient with increment $h$	$Q=\frac{f(a+h)-f(a)}{a+h-a}=\frac{f(a+h)-f(a)}{h}$
Slope of tangent line	${m}_{\text{tan}}=\underset{x\to a}{\text{lim}}\frac{f(x)-f(a)}{x-a}$ ${m}_{\text{tan}}=\underset{h\to 0}{\text{lim}}\frac{f(a+h)-f(a)}{h}$
Derivative of $f(x)$ at $a$	${f}^{'}(a)=\underset{x\to a}{\text{lim}}\frac{f(x)-f(a)}{x-a}$ ${f}^{'}(a)=\underset{h\to 0}{\text{lim}}\frac{f(a+h)-f(a)}{h}$
Average velocity	${v}_{a\text{ve}}=\frac{s(t)-s(a)}{t-a}$
Instantaneous velocity	$v(a)={s}^{'}(a)=\underset{t\to a}{\text{lim}}\frac{s(t)-s(a)}{t-a}$

Defining the Derivative

For the following exercises, use to find the slope of the secant line between the values ${x}_{1}$ and ${x}_{2}$ for each function $y=f(x).$

For the following functions,

use to find the slope of the tangent line ${m}_{\text{tan}}={f}^{'}(a),$ and
find an equation of the tangent line to $f$ at $x=a.$

For the following functions $y=f(x),$ find ${f}^{'}(a)$ using .

For the following exercises, given the function $y=f(x),$

find the slope of the secant line $PQ$ for each point $Q(x,f(x))$ with $x$ value given in the table.
Use the answers from a. to estimate the value of the slope of the tangent line at $P.$
Use the answer from b. to find an equation of the tangent line to $f$ at point $P.$

[T] For the following position functions $y=s(t),$ an object is moving along a straight line, where $t$ is in seconds and $s$ is in meters. Find

the simplified expression for the average velocity from $t=2$ to $t=2+h;$
the average velocity between $t=2$ and $t=2+h,$ where $\text{(i)}\ h=0.1,$ $\text{(ii)}\ h=0.01,$ $\text{(iii)}\ h=0.001,$ and $\text{(iv)}\ h=0.0001;$ and
use the answer from a. to estimate the instantaneous velocity at $t=2$ second.

For the following exercises, use the limit definition of derivative to show that the derivative does not exist at $x=a$ for each of the given functions.

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 an equation of the line tangent to the graph of $f(x)={x}^{2}$ at $x=3.$

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First find the slope of the tangent line. In this example, use .
\[\begin{array}{lllll}{m}_{\text{tan}} & =\underset{x\to 3}{\text{lim}}\frac{f(x)-f(3)}{x-3} & & & \text{Apply the definition.} \\ & =\underset{x\to 3}{\text{lim}}\frac{{x}^{2}-9}{x-3} & & & \text{Substitute}\ f(x)={x}^{2}\ \text{and}\ f(3)=9. \\ & =\underset{x\to 3}{\text{lim}}\frac{(x-3)(x+3)}{x-3}=\underset{x\to 3}{\text{lim}}(x+3)=6 & & & \text{Factor the numerator to evaluate the limit.}\end{array}\]
Next, find a point on the tangent line. Since the line is tangent to the graph of $f(x)$ at $x=3,$ it passes through the point $(3,f(3)).$ We have $f(3)=9,$ so the tangent line passes through the point $(3,9).$
Using the point-slope equation of the line with the slope $m=6$ and the point $(3,9),$ we obtain the line $y-9=6(x-3).$ Simplifying, we have $y=6x-9.$ The graph of $f(x)={x}^{2}$ and its tangent line at $3$ are shown in .
Use to find the slope of the line tangent to the graph of $f(x)={x}^{2}$ at $x=3.$

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The steps are very similar to . See for the definition.
\[\begin{array}{lllll}{m}_{\text{tan}} & =\underset{h\to 0}{\text{lim}}\frac{f(3+h)-f(3)}{h} & & & \text{Apply the definition.} \\ & =\underset{h\to 0}{\text{lim}}\frac{{(3+h)}^{2}-9}{h} & & & \text{Substitute}\ f(3+h)={(3+h)}^{2}\ \text{and}\ f(3)=9. \\ & =\underset{h\to 0}{\text{lim}}\frac{9+6h+{h}^{2}-9}{h} & & & \text{Expand and simplify to evaluate the limit.} \\ & =\underset{h\to 0}{\text{lim}}\frac{h(6+h)}{h}=\underset{h\to 0}{\text{lim}}(6+h)=6\end{array}\]
We obtained the same value for the slope of the tangent line by using the other definition, demonstrating that the formulas can be interchanged.
Find an equation of the line tangent to the graph of $f(x)=1\text{/}x$ at $x=2.$

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We can use , but as we have seen, the results are the same if we use .
\[\begin{array}{lllll}{m}_{\text{tan}} & =\underset{x\to 2}{\text{lim}}\frac{f(x)-f(2)}{x-2} & & & \text{Apply the definition.} \\ & =\underset{x\to 2}{\text{lim}}\frac{\frac{1}{x}-\frac{1}{2}}{x-2} & & & \text{Substitute}\ f(x)=\frac{1}{x}\ \text{and}\ f(2)=\frac{1}{2}. \\ & =\underset{x\to 2}{\text{lim}}\frac{\frac{1}{x}-\frac{1}{2}}{x-2}\cdot \frac{2x}{2x} & & & \begin{array}{l}\text{Multiply numerator and denominator by}\ 2x\ \text{to} \\ \text{simplify fractions.}\end{array} \\ & =\underset{x\to 2}{\text{lim}}\frac{(2-x)}{(x-2)(2x)} & & & \text{Simplify.} \\ & =\underset{x\to 2}{\text{lim}}\frac{-1}{2x} & & & \text{Simplify using}\ \frac{2-x}{x-2}=-1,\text{for}\ x\ne 2. \\ & =-\frac{1}{4} & & & \text{Evaluate the limit.}\end{array}\]
We now know that the slope of the tangent line is $-\frac{1}{4}.$ To find an equation of the tangent line, we also need a point on the line. We know that $f(2)=\frac{1}{2}.$ Since the tangent line passes through the point $(2,\frac{1}{2})$ we can use the point-slope equation of a line to find an equation of the tangent line. Thus the tangent line has the equation $y=-\frac{1}{4}x+1.$ The graphs of $f(x)=\frac{1}{x}$ and $y=-\frac{1}{4}x+1$ are shown in .
Find the slope of the line tangent to the graph of $f(x)=\sqrt{x}$ at $x=4.$

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$\frac{1}{4}$
For $f(x)={x}^{2},$ use a table to estimate ${f}^{'}(3)$ using .

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Create a table using values of $x$ just below $3$ and just above $3.$
$x$ $\frac{{x}^{2}-9}{x-3}$
$2.9$ $5.9$
$2.99$ $5.99$
$2.999$ $5.999$
$3.001$ $6.001$
$3.01$ $6.01$
$3.1$ $6.1$
After examining the table, we see that a good estimate is ${f}^{'}(3)=6.$
For $f(x)={x}^{2},$ use a table to estimate ${f}^{'}(3)$ using .

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$6$
For $f(x)=3{x}^{2}-4x+1,$ find ${f}^{'}(2)$ by using .

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Substitute the given function and value directly into the equation.
\[\begin{array}{lllll}{f}^{'}(2) & =\underset{x\to 2}{\text{lim}}\frac{f(x)-f(2)}{x-2} & & & \text{Apply the definition.} \\ & =\underset{x\to 2}{\text{lim}}\frac{(3{x}^{2}-4x+1)-5}{x-2} & & & \text{Substitute}\ f(x)=3{x}^{2}-4x+1\ \text{and}\ f(2)=5. \\ & =\underset{x\to 2}{\text{lim}}\frac{(x-2)(3x+2)}{x-2} & & & \text{Simplify and factor the numerator.} \\ & =\underset{x\to 2}{\text{lim}}(3x+2) & & & \text{Cancel the common factor.} \\ & =8 & & & \text{Evaluate the limit.}\end{array}\]
For $f(x)=3{x}^{2}-4x+1,$ find ${f}^{'}(2)$ by using .

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Using this equation, we can substitute two values of the function into the equation, and we should get the same value as in .
\[\begin{array}{lllll}{f}^{'}(2) & =\underset{h\to 0}{\text{lim}}\frac{f(2+h)-f(2)}{h} & & & \text{Apply the definition.} \\ & =\underset{h\to 0}{\text{lim}}\frac{(3{(2+h)}^{2}-4(2+h)+1)-5}{h} & & & \begin{array}{l}\text{Substitute}\ f(2)=5\ \text{and} \\ f(2+h)=3{(2+h)}^{2}-4(2+h)+1.\end{array} \\ & =\underset{h\to 0}{\text{lim}}\frac{3{h}^{2}+8h}{h} & & & \text{Simplify the numerator.} \\ & =\underset{h\to 0}{\text{lim}}\frac{h(3h+8)}{h} & & & \text{Factor the numerator.} \\ & =\underset{h\to 0}{\text{lim}}(3h+8) & & & \text{Cancel the common factor.} \\ & =8 & & & \text{Evaluate the limit.}\end{array}\]
The results are the same whether we use or .
For $f(x)={x}^{2}+3x+2,$ find ${f}^{'}(1).$

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${f}^{'}(1)=5$

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We can estimate the instantaneous velocity at $t=0$ by computing a table of average velocities using values of $t$ approaching $0,$ as shown in .

$t$	$\frac{\text{sin}\ t-\text{sin}\ 0}{t-0}=\frac{\text{sin}\ t}{t}$
$-0.1$	$0.998334166$
$-0.01$	$0.9999833333$
$-0.001$	$0.999999833$
$0.001$	$0.999999833$
$0.01$	$0.9999833333$
$0.1$	$0.998334166$

By using , we can see that

\[v(0)={s}^{'}(0)=\underset{t\to 0}{\text{lim}}\frac{\text{sin}\ t-\text{sin}\ 0}{t-0}=\underset{t\to 0}{\text{lim}}\frac{\text{sin}\ t}{t}=1.\]

Thus, in fact, $v(0)=1.$

A rock is dropped from a height of $64$ feet. Its height above ground at time $t$ seconds later is given by $s(t)=-16{t}^{2}+64,0\le t\le 2.$ Find its instantaneous velocity $1$ second after it is dropped, using .

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$-32$ ft/s
Reaching a top speed of $270.49$ mph, the Hennessey Venom GT is one of the fastest cars in the world. In tests it went from $0$ to $60$ mph in $3.05$ seconds, from $0\ \text{to}\ 100$ mph in $5.88$ seconds, from $0\ \text{to}\ 200$ mph in $14.51$ seconds, and from $0\ \text{to}\ 229.9$ mph in $19.96$ seconds. Use this data to draw a conclusion about the rate of change of velocity (that is, its acceleration) as it approaches $229.9$ mph. Does the rate at which the car is accelerating appear to be increasing, decreasing, or constant?

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First observe that $60$ mph = $88$ ft/s, $100$ mph $\approx 146.67$ ft/s, $200$ mph $\approx 293.33$ ft/s, and $229.9$ mph $\approx 337.19$ ft/s. We can summarize the information in a table.
$t$ $v(t)$
$0$ $0$
$3.05$ $88$
$5.88$ $146.67$
$14.51$ $293.33$
$19.96$ $337.19$
Now compute the average acceleration of the car in feet per second per second on intervals of the form $[t,19.96]$ as $t$ approaches $19.96,$ as shown in the following table.
$t$ $\frac{v(t)-v(19.96)}{t-19.96}=\frac{v(t)-337.19}{t-19.96}$
$0.0$ $16.89$
$3.05$ $14.74$
$5.88$ $13.53$
$14.51$ $8.05$
The rate at which the car is accelerating is decreasing as its velocity approaches $229.9$ mph $\text{(}337.19$ ft/s).
A homeowner sets the thermostat so that the temperature in the house begins to drop from $70\text{^{\circ}}\text{F}$ at $9$ p.m., reaches a low of $60\text{^{\circ}}$ during the night, and rises back to $70\text{^{\circ}}$ by $7$ a.m. the next morning. Suppose that the temperature in the house is given by $T(t)=0.4{t}^{2}-4t+70$ for $0\le t\le 10,$ where $t$ is the number of hours past $9$ p.m. Find the instantaneous rate of change of the temperature at midnight.

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Since midnight is $3$ hours past $9$ p.m., we want to compute ${T}^{'}(3).$ Refer to .
\[\begin{array}{lllll}{T}^{'}(3) & =\underset{t\to 3}{\text{lim}}\frac{T(t)-T(3)}{t-3} & & & \text{Apply the definition.} \\ & =\underset{t\to 3}{\text{lim}}\frac{0.4{t}^{2}-4t+70-61.6}{t-3} & & & \begin{array}{l}\text{Substitute}\ T(t)=0.4{t}^{2}-4t+70\ \text{and} \\ T(3)=61.6.\end{array} \\ & =\underset{t\to 3}{\text{lim}}\frac{0.4{t}^{2}-4t+8.4}{t-3} & & & \text{Simplify.} \\ & =\underset{t\to 3}{\text{lim}}\frac{0.4(t-3)(t-7)}{t-3} & & & =\underset{t\to 3}{\text{lim}}\frac{0.4(t-3)(t-7)}{t-3} \\ & =\underset{t\to 3}{\text{lim}}0.4(t-7) & & & \text{Cancel.} \\ & =-1.6 & & & \text{Evaluate the limit.}\end{array}\]
The instantaneous rate of change of the temperature at midnight is $-1.6\text{^{\circ}}\text{F}$ per hour.
A toy company can sell $x$ electronic gaming systems at a price of $p=-0.01x+400$ dollars per gaming system. The cost of manufacturing $x$ systems is given by $C(x)=100x+10,000$ dollars. Find the rate of change of profit when $10,000$ games are produced. Should the toy company increase or decrease production?

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The profit $P(x)$ earned by producing $x$ gaming systems is $R(x)-C(x),$ where $R(x)$ is the revenue obtained from the sale of $x$ games. Since the company can sell $x$ games at $p=-0.01x+400$ per game,
\[R(x)=xp=x(-0.01x+400)=-0.01{x}^{2}+400x.\]
Consequently,
\[P(x)=-0.01{x}^{2}+300x-10,000.\]
Therefore, evaluating the rate of change of profit gives
\[\begin{array}{ll}{P}^{'}(10000) & =\underset{x\to 10000}{\text{lim}}\frac{P(x)-P(10000)}{x-10000} \\ & =\underset{x\to 10000}{\text{lim}}\frac{-0.01{x}^{2}+300x-10000-1990000}{x-10000} \\ & =\underset{x\to 10000}{\text{lim}}\frac{-0.01{x}^{2}+300x-2000000}{x-10000} \\ & =100.\end{array}\]
Since the rate of change of profit ${P}^{'}(10,000)>0$ and $P(10,000)>0,$ the company should increase production.
A coffee shop determines that the daily profit on scones obtained by charging $s$ dollars per scone is $P(s)=-20{s}^{2}+150s-10.$ The coffee shop currently charges $\text{\$}3.25$ per scone. Find ${P}^{'}(3.25),$ the rate of change of profit when the price is $\text{\$}3.25$ and decide whether or not the coffee shop should consider raising or lowering its prices on scones.

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${P}^{'}(3.25)=20>0;$ raise prices
$f(x)=4x+7;{x}_{1}=2,{x}_{2}=5$

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$4$
$f(x)=8x-3;{x}_{1}=-1,{x}_{2}=3$
$f(x)={x}^{2}+2x+1;{x}_{1}=3,{x}_{2}=3.5$

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$8.5$
$f(x)=\text{-}{x}^{2}+x+2;{x}_{1}=0.5,{x}_{2}=1.5$
$f(x)=\frac{4}{3x-1};{x}_{1}=1,{x}_{2}=3$

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$-\frac{3}{4}$
$f(x)=\frac{x-7}{2x+1};{x}_{1}=0,{x}_{2}=2$
$f(x)=\sqrt{x};{x}_{1}=1,{x}_{2}=16$

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$0.2$
$f(x)=\sqrt{x-9};{x}_{1}=10,{x}_{2}=13$
$f(x)={x}^{1\text{/}3}+1;{x}_{1}=0,{x}_{2}=8$

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$0.25$
$f(x)=6{x}^{2\text{/}3}+2{x}^{1\text{/}3};{x}_{1}=1,{x}_{2}=27$
$f(x)=3-4x,a=2$

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a. $-4$ b. $y=3-4x$
$f(x)=\frac{x}{5}+6,a=-1$
$f(x)={x}^{2}+x,a=1$

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a. $3$ b. $y=3x-1$
$f(x)=1-x-{x}^{2},a=0$
$f(x)=\frac{7}{x},a=3$

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a. $\frac{-7}{9}$ b. $y=\frac{-7}{9}x+\frac{14}{3}$
$f(x)=\sqrt{x+8},a=1$
$f(x)=2-3{x}^{2},a=-2$

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a. $12$ b. $y=12x+14$
$f(x)=\frac{-3}{x-1},a=4$
$f(x)=\frac{2}{x+3},a=-4$

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a. $-2$ b. $y=-2x-10$
$f(x)=\frac{3}{{x}^{2}},a=3$
$f(x)=5x+4,a=-1$

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$5$
$f(x)=-7x+1,a=3$
$f(x)={x}^{2}+9x,a=2$

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$13$
$f(x)=3{x}^{2}-x+2,a=1$
$f(x)=\sqrt{x},a=4$

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$\frac{1}{4}$

Symbols used here

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

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

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

imaginary unit
i² = −1.

^\circ

degrees
1/360 of a full turn. 180° = π radians.

\leq,\ \geq

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

\neq

not equal
The two sides are different.

\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.

\sin,\ \cos,\ \tan

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

\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.

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

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

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

derivative
Instantaneous rate of change; slope of the graph.

\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).

How to: Defining the Derivative

Recognize the meaning of the tangent to a curve at a point.
Calculate the slope of a tangent line.
Identify the derivative as the limit of a difference quotient.
Calculate the derivative of a given function at a point.
Describe the velocity as a rate of change.
Explain the difference between average velocity and instantaneous velocity.
Estimate the derivative from a table of values.
The slope of the tangent line to a curve measures the instantaneous rate of change of a curve. We can calculate it by finding the limit of the difference quotient or the difference quotient with increment

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.

ลองดูสิ

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

เพิ่มเติมใน Calculus

Limits Derivatives 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

\(x\)	\(\frac{{x}^{2}-9}{x-3}\)
\(2.9\)	\(5.9\)
\(2.99\)	\(5.99\)
\(2.999\)	\(5.999\)
\(3.001\)	\(6.001\)
\(3.01\)	\(6.01\)
\(3.1\)	\(6.1\)

\(t\)	\(\frac{\text{sin}\ t-\text{sin}\ 0}{t-0}=\frac{\text{sin}\ t}{t}\)
\(-0.1\)	\(0.998334166\)
\(-0.01\)	\(0.9999833333\)
\(-0.001\)	\(0.999999833\)
\(0.001\)	\(0.999999833\)
\(0.01\)	\(0.9999833333\)
\(0.1\)	\(0.998334166\)

Difference quotient	\(Q=\frac{f(x)-f(a)}{x-a}\)
Difference quotient with increment \(h\)	\(Q=\frac{f(a+h)-f(a)}{a+h-a}=\frac{f(a+h)-f(a)}{h}\)
Slope of tangent line	\({m}_{\text{tan}}=\underset{x\to a}{\text{lim}}\frac{f(x)-f(a)}{x-a}\) \({m}_{\text{tan}}=\underset{h\to 0}{\text{lim}}\frac{f(a+h)-f(a)}{h}\)
Derivative of \(f(x)\) at \(a\)	\({f}^{'}(a)=\underset{x\to a}{\text{lim}}\frac{f(x)-f(a)}{x-a}\) \({f}^{'}(a)=\underset{h\to 0}{\text{lim}}\frac{f(a+h)-f(a)}{h}\)
Average velocity	\({v}_{a\text{ve}}=\frac{s(t)-s(a)}{t-a}\)
Instantaneous velocity	\(v(a)={s}^{'}(a)=\underset{t\to a}{\text{lim}}\frac{s(t)-s(a)}{t-a}\)

\(x\)	\(\frac{{x}^{2}-9}{x-3}\)
\(2.9\)	\(5.9\)
\(2.99\)	\(5.99\)
\(2.999\)	\(5.999\)
\(3.001\)	\(6.001\)
\(3.01\)	\(6.01\)
\(3.1\)	\(6.1\)

\(t\)	\(\frac{\text{sin}\ t-\text{sin}\ 0}{t-0}=\frac{\text{sin}\ t}{t}\)
\(-0.1\)	\(0.998334166\)
\(-0.01\)	\(0.9999833333\)
\(-0.001\)	\(0.999999833\)
\(0.001\)	\(0.999999833\)
\(0.01\)	\(0.9999833333\)
\(0.1\)	\(0.998334166\)

\(t\)	\(v(t)\)
\(0\)	\(0\)
\(3.05\)	\(88\)
\(5.88\)	\(146.67\)
\(14.51\)	\(293.33\)
\(19.96\)	\(337.19\)

\(t\)	\(\frac{v(t)-v(19.96)}{t-19.96}=\frac{v(t)-337.19}{t-19.96}\)
\(0.0\)	\(16.89\)
\(3.05\)	\(14.74\)
\(5.88\)	\(13.53\)
\(14.51\)	\(8.05\)