The definition of the derivative

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We compute the instantaneous growth rate by computing the limit of average growth rates.

Given a function, it is often useful to know the rate at which the function changes. To give you a feeling why this is true, consider the following:

If $s(t)$ represents the position of an object on a line (with respect to time), the rate of change gives the velocity of the object.
If $v(t)$ represents the velocity of an object with respect to time, the rate of change gives the acceleration of the object.
If $R(x)$ represents the revenue generated by selling $x$ objects, the rate of change gives us the marginal revenue, meaning the additional revenue generated by selling one additional unit. Note, there is an implicit assumption that $x$ is quite large compared to $1$ .
If $C(x)$ represents the cost to produce $x$ objects, the rate of change gives us the marginal cost, meaning the additional cost generated by selling one additional unit. Again, there is an implicit assumption that $x$ is quite large compared to $1$ .
The rate of change of a function can help us approximate a complicated function with a simple function.
The rate of change of a function can be used to help us solve equations that we would not be able to solve via other methods.

From slopes of secant lines to slopes of tangent lines

We’ve been computing average rates of change for a while now,

${\text {average rate of change= }} \frac {\text {change in the function (output)}}{\text {change in the input }}.$

More precisely, the average rate of change of a function $f$ is given by ${\text {average rate of change of $f$ }} =\frac {f(x)-f(a)}{x-a},$ as the input changes from $a$ to $x$ .

What happens if we compute the average rate of change of $f$ for each value of $x$ as $x$ gets closer and closer to $a$ ? In other words, what is the meaning of the limit $\lim _{x\to a} \frac {f(x)-f(a)}{x-a},$ provided that the limit exists?

Naturally, we call this limit the instantaneous rate of change of the function $f$ at $a$ .

(In this example we will use abreviation AvRCh for the average rate of change, and InRCh for the instantaneous rate of change.)

Let $f(x)=2x^2+3$ .

(a): Find the expression for the average rate of change of $f$ between the points $a=2$ and $x=-1$ . $AvRCh=\frac {f(-1)-f(2)}{-1-\left (\answer [given]{2}\right )}$ Now evaluate the function, $AvRCh=\frac {2(-1)^2+3-(2(2)^2+3)}{-1-\left (\answer [given]{2}\right )}.$ Simplify, $AvRCh=\frac {5-11}{-3}=2.$
(b): Find the average rate of change of $f$ between the points $a=2$ and $x$ , $x\ne 2$ . $AvRCh=\frac {f(x)-f(2)}{x-\answer [given]{2}}$ Now substitute in for the function, $AvRCh=\frac {2x^2+3-11}{x-\answer [given]{2}}$ Simplify the top, $AvRCh=\frac {2x^2-8}{x-\answer [given]{2}}.$ Factor, $AvRCh=\frac {2(x^2-4)}{x-2}$ Factor and cancel, $AvRCh=\frac {2(x+2)\cancel {(x-2)}}{\cancel {x-2}}=2(x+2)$
(c): Find the instantaneous rate of change of $f$ at the point $x=2$ . $InRCh= \lim _{x\to 2} \frac {f(x)-f(2)}{x-2}$ So, the instantaneous rate of change is the limit, as $x\to 2$ , of average rates of change of $f$ between points $a=2$ and $x$ , for $x\ne 2$ .
We have already computed an expression for the average rate of change for all $x\ne 2$ . Therefore, $InRCh=\lim _{x\to 2}2(x+2)=8$

We can arrive at the same limit, $\lim _{x\to a} \frac {f(x)-f(a)}{x-a},$

within a completely different context.

Have a look at the figure below. The figure depicts a graph of the function $f$ , two points on the graph, $(a,f(a))$ and $(x,f(x))$ , and a secant line that passes through these two points.

The slope of this secant line is equal to $\frac {f(x)-f(a)}{x-a}$ !

This is exactly the expression for the average rate of change of $f$ as the input changes from $a$ to $x$ !

Therefore,

${\text {slope of secant line $=$}}{\text { average rate of change.}}$ As before, we can ask ourselves: What happens as $x$ gets closer and closer to $a$ ? In other words, what is the meaning of the limit of slopes of secant lines through the points $(a,f(a))$ and $(x,f(x))$ as $x$ gets closer and closer to $a$ ?

How can we interpret the limit $\lim _{x\to a} \frac {f(x)-f(a)}{x-a},$ provided that the limit exists?

This scenario is illustrated in the figure below. What happens as $x\to a$ ?

Notice that all the secant lines in the figure pass through the given point $(a,f(a))$ . What distinguishes these lines are their slopes; the slope of each line is determined by an additional point on the line, $(x,f(x))$ . The figure suggests that“the limit of slopes of secant lines” is the slope of a special line, called the tangent line to the curve $y=f(x)$ at the point $(a,f(a))$ . Notice how the tangent line and the curve are indistinguishable near the point $(a,f(a))$ . If one can “zoom in” on the graph at $(a, f(a))$ sufficiently so that it appears to be a straight line, then that line is the tangent line to $f(x)$ at the point $(a,f(a))$ . This is illustrated in the figure below.

Each of the following four figures depicts the graph of the same function $f$ and a line that passes through the point $(1,f(1))$ .

Only one of the figures depicts the line tangent to the curve $y=f(x)$ at the point $(1,f(1))$ . Which one?

Figure A Figure B Figure C Figure D

Only the line in figure $B$ looks indistinguishable from the curve $y=f(x)$ near the point $(1,f(1))$ .

Obviously, the limit $\lim _{x\to a} \frac {f(x)-f(a)}{x-a},$ has many different interpretations, depending on the context, such as the slope of the tangent line, the instantaneous rate of change, and the instantaneous velocity.

Therefore, this limit deserves a special name that could be used regardless of the context.

The derivative of $f$ at $a$ , denoted $f'(a)$ , is given by $f'(a) = \lim _{x\to a} \frac {f(x) - f(a)}{x-a},$ provided that the limit exists. We say that $f$ is differentiable at $a$ if this limit exists. Otherwise, we say that $f$ is non-differentiable at $a$ .

The definition of the derivative allows us to define a tangent line precisely.

Let $f$ be a function differentiable at $a$ . The line tangent to the curve $y=f(x)$ at the point $(a,f(a))$ is the line that passes through the point $(a,f(a))$ whose slope is equal to $f'(a)$ .

Naturally, by the point-slope equation of the line, it follows that the tangent line is given by the equation $y = f(a)+f'(a)(x-a).$

Let $f$ be a function given by $f(x) = 4x-3$ . What is the instantaneous rate of change of $f$ at $a$ , where $a$ is any real number?

The rate of change is given by $f'(a)$ , and so is the slope of the tangent line to the curve $y=f(x)$ at the point $(a,f(a))$ .

The line tangent to the curve $y=f(x) = 4x-3$ , is simply the curve itself!

The derivative is the slope of the line $y= 4x-3$ ! Therefore, $f'(a)=\answer {4}$ , for all real numbers $a$ .

Of course, we can verify this reasoning by simply computing $f'(a)$ , using the definition of the derivative. $f'(a) = \lim _{x\to a} \frac {f(x) - f(a)}{x-a}.$ Replace $f$ with its formula, $f'(a) = \lim _{x\to a} \frac {(4x-3)-(4a-3)}{x-a}.$ Simplify the numerator, $f'(a) = \lim _{x\to a} \frac {4x-4a}{x-a}.$

Factor the numerator, $f'(a) = \lim _{x\to a} \frac {4\cancel {(x-a)}}{\cancel {x-a}}.$ Compute the limit,

$f'(a) =4.$

But, most functions are not linear, and their graphs are not straight lines. Therefore, the computation of the derivative is not as simple as in the previous example.

Let $f(x) = 2x^2+3$ .

Find the slope of the tangent line to the curve $y=f(x)$ at the point $(2,f(2))$ .

Finding the slope of the tangent line at the point $(2,f(2))$ means finding $f'(2)$ $f'(2) = \lim _{x\to 2} \frac {f(x) - f(2)}{x-2}$ But, this limit gives us the instantaneous rate of change of $f$ at $a=2$ , which was computed in our first example! Therefore, the slope of the tangent line is $f'(2) = \lim _{x\to 2} \frac {f(x) - f(2)}{x-2}=8.$

We can confirm our results by looking at the graph of $y=f(x)$ and the line $y=8x-5$ .

Let $f(x) = \frac {1}{3-x}$ . Find an equation for the line tangent to the curve $y=f(x)$ at the point $(2, 1)$ .

To find an equation for a line, we need two pieces of information. We need to know a point on the line, and we need to know the slope of the line. In this question, we are given that the point $(2,1)$ is on the line. So, we need to find the slope of the tangent line. Finding the slope of the tangent line at the point $(2,1)$ means finding $f'(2)$ .

Start by writing out the definition of the derivative, $f'(2) = \lim _{x\to 2} \frac {f\left (\answer [given]{x}\right )-f\left (2\right )}{x-2} = \lim _{x\to 2} \frac {\frac {1}{3-x}-1}{x-2}.$ Multiply by $\frac {3-x}{3-x}$ to clear the fraction in the numerator, $f'(2) = \lim _{x\to 2} \frac {1 - \left (\answer [given]{3-x}\right )}{(x-2)(3-x)}.$ Combine like-terms in the numerator, $f'(2) = \lim _{x\to 2} \frac {\cancel {x-2}}{\cancel {(x-2)}(3-x)},$

Take the limit as $x$ goes to $2$ , $f'(2)= \lim _{x\to 2} \frac {1}{3-x} = \answer [given]{1}.$ We are looking for an equation of the line through the point $(2,1)$ with slope $m = f'(2) = 1$ . The point-slope formula tells us that the line has equation given by $y= 1+\answer [given]{x-2 },$ or $y=x-1$ . We can confirm our results by looking at the graph of $y=f(x)$ and the line $y=x-1$ .

An object moving along a straight line has position given by $s(t) = \sqrt {t+3}$ , where position is measured in meters and time in seconds. Find the (instantaneous) velocity of the object at time $t=6$ .

Velocity is the instantaneous rate of change of position with respect to time. We are being asked to find $v(6)=s'(6)$ . We have to adjust the definition of the the derivative, namely, replace $x$ with $t$ , $s'(6) = \lim _{t \to 6} \frac {s(t)-s(6)}{t-6} = \lim _{t\to 6} \frac {\sqrt {\answer [given]{t+3}}-3}{t-6}.$ Multiply by $\frac {\sqrt {t+3}+3}{\sqrt {t+3}+3}$ , in order to rationalize the expression, $s'(6) = \lim _{t\to 6} \left (\frac {\sqrt {t+3}-3}{t-6} \right )\left (\frac {\sqrt {t+3}+3}{\sqrt {t+3}+3} \right ).$ Now simplify the numerator, $s'(6) = \lim _{t \to 6} \frac {t+3 - 9}{(t-6)\left (\sqrt {t+3}+3 \right )}.$ Combine like-terms, $s'(6)= \lim _{t \to 6} \frac {\cancel {t-6}}{\cancel {(t-6)}\left (\sqrt {t+3}+3\right )}.$ Take the limit as $t$ goes to $6$ , $s'(6)= \lim _{t\to 6} \frac {1}{\sqrt {t+3}+3}=\answer [given]{\frac {1}{6}}.$

The object has velocity $\frac {1}{6}$ $\frac {m}{s}$ at time $t=6$ .

Below we can see the graph of $s=s(t)$ and the tangent line at $t=6$ , with a slope of $\frac {1}{6}$ . Notice, again, how the line fits the graph of the function $s$ near the point $(6,3)$ .