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 displacement (position relative to an origin) of an object 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 $1$ .
If $P(x)$ represents the profit gained by selling $x$ objects, the rate of change gives us the marginal profit, 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

You’ve been computing average rates of change for a while now, the computation is simply $\frac {\text {change in the function}}{\text {change in the input to the function}}.$ However, the question remains: Given a function that represents an amount, how exactly does one find the function that will give the instantaneous rate of change? Recall that the instantaneous rate of change of a line is the slope of the line. Hence the instantaneous rate of change of a function is the slope of the tangent line. For now, consider the following informal definition of a tangent line:

Given a function $f$ and a number $a$ in the domain of $f$ , 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 $(a,f(a))$ .

We illustrate this informal definition with the following diagram:

The derivative of a function $f$ at $a$ , is the instantaneous rate of change, and hence is the slope of the tangent line at $(a,f(a))$ .

What is the instantaneous rate of change of $f(x) = 4x-3$ ?

The rate of change is the slope of the tangent line.

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

The derivative is $\answer {4}$ .

Unfortunately, if $f$ is not a straight line we cannot use the slope formula to calculate this rate of change, since $(a,f(a))$ is the only point on this line that we know. In order to deal with this problem, we consider secant lines, lines that locally intersect the curve at two points. One of these points will be $(a, f(a))$ , the point at which we are trying to find the rate of change. If we call $h$ the difference between the $x$ -coordinates of the two points, then the second point for our secant line is $(a+h, f(a+h))$ . The slope of any secant line that passes through the points $(a,f(a))$ and $(a+h, f(a+h))$ is given by $\frac {\Delta y}{\Delta x}=\frac {f(x) - f(a)}{x - a}.$

If $f(x) = 2x^2+3$ , find the slope of the secant line through $(2,f(2))$ and $(x,f(x))$ in terms of $x$ .

Start with the slope formula, $\frac {\Delta y}{\Delta x} = \frac {f\left (\answer [given]{x}\right )-f\left (2\right )} {\answer [given]{x} - 2}.$ Replace $f$ with its formula: $\frac {\Delta y}{\Delta x} = \frac {\answer [given]{2x^2+3} - 11}{x-2}.$ Simplify the numerator: $\frac {\Delta y}{\Delta x} = \frac {2x^2-8}{x-2}.$ Factor the numerator, $\frac {\Delta y}{\Delta x} = \frac {2(x+2)\left (\answer [given]{x-2}\right )}{x-2}.$ Cancel $x-2$ from the numerator and denominator, $\frac {\Delta y}{\Delta x} = \answer [given]{2(x+2)}.$

Suppose the difference between $x$ and $a$ is some small number $h$ . That is $x-a=h$ or $x=a+h$ . Substituting this in to our slope formula gives us an alternate characterization of the slope of a secant line. $\frac {\Delta y}{\Delta x} = \frac {f(a+h)-f(a)}{(a+h)-a} = \frac {f(a+h)-f(a)}{h}.$

If $f(x) = x^2-2x$ , find the slope of the secant line through $x=2$ and $x=2+h$ , in terms of $h$ .

Start with the slope formula we just found, $\frac {\Delta y}{\Delta x} = \frac {f\left (\answer [given]{2+h}\right )-f\left (2\right )}{\answer [given]{h}}.$ Now substitute in for the function we know, $\frac {\Delta y}{\Delta x} = \frac {\answer [given]{(2+h)^2-2(2+h)} -0}{h}.$ Now expand the numerator of the fraction, $\frac {\Delta y}{\Delta x} = \frac {4+4h+h^2-4-2h }{h}.$ Now combine like-terms, $\frac {\Delta y}{\Delta x} = \frac {2h+h^2}{h}.$ Factor an $h$ from every term in the numerator, $\frac {\Delta y}{\Delta x} = \frac {h\left (\answer [given]{2+h}\right )}{h}.$ Cancel $h$ from the numerator and denominator, $\frac {\Delta y}{\Delta x} = \answer [given]{2+h}.$

The following diagram shows the secant lines for several values of $h$ , as well as the tangent line at $(a,f(a))$ .

Notice that as $a+h$ approaches $a$ , the slopes of the secant lines are approaching the slope of the tangent line. This leads to the definition of the derivative:

The derivative of $f$ at $a$ is $\eval {\ddx f(x)}_{x=a} = \lim _{h\to 0} \frac {f(a+h) - f(a)}{h}.$ If this limit exists, then we say that $f$ is differentiable at $a$ . If this limit does not exist for a given value of $a$ , then $f$ is non-differentiable at $a$ .

Which of the following computes the derivative, $\eval {\ddx f(x)}_{x=a}$ ?

$\lim _{h\to 0}\frac {(f(a)+h) - f(a)}{(a+h)-a}$ $\lim _{h\to 0}\frac {f(a+h) - f(a)}{(a+h)-a}$ $\lim _{h\to 0}\frac {(f(a)-h) - f(a)}{(a-h)-a}$ $\lim _{h\to 0}\frac {f(a-h) - f(a)}{(a-h)-a}$ $\lim _{h\to 0}\frac {f(a) - (f(a)+h)}{a-(a+h)}$ $\lim _{h\to 0}\frac {f(a) - f(a+h)}{a-(a+h)}$ $\lim _{h\to 0}\frac {f(a) - (f(a)-h)}{a-(a-h)}$ $\lim _{h\to 0}\frac {f(a) - f(a-h)}{a-(a-h)}$

Recalling our original characterization of the slope of the secant line between $a$ and $x$ which is near by to $a$ . We see that we could also calculate the derivative of a function by letting $x$ go to $a$ . This gives us the following definition:

The derivative of $f$ at $a$ is $\eval {\ddx f(x)}_{x=a} = \lim _{x\to a} \frac {f(x) - f(a)}{x-a}.$

There are several different notations for the derivative. The two we’ll mainly be using are $\eval {\ddx f(x)}_{x=a} = f'(a).$

Now we will give a number of examples.

If $f(x) = x^2-2x$ , find the derivative of $f$ at $2$ using the $h\to 0$ characterization.

Start with the definition of the derivative, $\eval {\ddx f(x)}_{x=2} = \lim _{h\to 0} \frac {f\left (\answer [given]{2+h}\right )-f\left (2\right )}{h}.$ Now substitute in for the function we know, $\eval {\ddx f(x)}_{x=2} = \lim _{h\to 0}\frac {\answer [given]{(2+h)^2-2(2+h)} -0}{h}.$ Now expand the numerator of the fraction, $\eval {\ddx f(x)}_{x=2} = \lim _{h\to 0} = \frac {4+4h+h^2-4-2h }{h}.$ Now combine like-terms, $\eval {\ddx f(x)}_{x=2} = \lim _{h\to 0}\frac {2h+h^2}{h}.$ Factor an $h$ from every term in the numerator, $\eval {\ddx f(x)}_{x=2} = \lim _{h\to 0}\frac {h\left (\answer [given]{2+h}\right )}{h}.$ Cancel $h$ from the numerator and denominator, $\eval {\ddx f(x)}_{x=2} = \lim _{h\to 0}\answer [given]{2+h}.$ Take the limit as $h$ goes to $0$ , $\eval {\ddx f(x)}_{x=2} = \answer [given]{2}.$ For your viewing pleasure, we have below the graph of $y=f(x)$ and the graph of the tangent line at the point $x=2$ . $\graph [xmin=-2,xmax=6,ymin=-6,ymax=6]{x^2-2x, 2(x-2)}$

Find the derivative of $f(x) = x^2+x+1$ at $x=-1$ using the $x\to a$ characterization.

Start with the definition of the derivative, $f'(-1) = \lim _{x\to -1}\frac {f\left (x\right )-f\left (\answer [given]{-1}\right )}{x-(-1)}.$ Replace $f$ with its formula, $f'(-1) = \lim _{x\to -1}\frac {x^2+x+1 - 1}{x+1}.$ Simplify the top, $f'(-1) = \lim _{x\to -1}\frac {x^2+x}{x+1}.$ Factor, $f'(-1) = \lim _{x\to -1}\frac {x\left (\answer [given]{x+1}\right )}{x+1}.$ Cancel the $x+1$ in the numerator and denominator, $f'(-1) = \lim _{x\to -1}x.$ The expression inside the limit is continuous everywhere, so evaluate the limit by plugging in, $f'(-1) = \answer [given]{-1}.$

Find an equation for the line tangent to $f(x) = \frac {1}{3-x}$ 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. In this question, we are given that $(2,1)$ is on the line. That means 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, $f'(2) = \lim _{h\to 0} \frac {f(\answer [given]{2+h})-f(2)}{h} = \lim _{h\to 0} \frac {\frac {1}{1-h} - 1}{h}.$ Multiply by $\frac {1-h}{1-h}$ to clear the fraction in the numerator, $f'(2) = \lim _{h\to 0} \frac {\answer [given]{1 - (1-h)}}{h(1-h)}.$ Combine like-terms in the numerator, $f'(2) = \lim _{h\to 0} \frac {h}{h(1-h)},$ Cancel $h$ from the numerator and denominator, $f'(2) = \lim _{h\to 0} \frac {1}{1-h},$ Take the limit as $h$ goes to $0$ , $f'(2) = \answer [given]{1}.$ 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 $y= \answer [given]{(x-2) + 1}.$ We can confirm our results by looking at the graph of $y=f(x)$ and the line $y=(x-2)+1$ . $\graph [xmin=-2,xmax=6,ymin=-6,ymax=6]{\frac {1}{3-x}, (x-2)+1}$

An object moving along a straight line has displacement given by $s(t) = \sqrt {t+3}$ . Find the velocity of the object at time $t=6$ .

Velocity is the rate of change of displacement with respect to time. We are being asked to find $\eval {\dd {t} s(t)}_{t=6}$ . The definition of the derivative gives $\eval {\dd {t} s(t)}_{t=6} = \lim _{h \to 0} \frac {s(\answer [given]{6+h})-s(6)}{h} = \lim _{h\to 0} \frac {\sqrt {\answer [given]{9+h}}-3}{h}.$ Multiply by $\frac {\sqrt {9+h}+3}{\sqrt {9+h}+3}$ , $\eval {\dd {t} s(t)}_{t=6} = \lim _{h\to 0} \left (\frac {\sqrt {9+h}-3}{h} \right )\left ( \frac {\sqrt {9+h}+3}{\sqrt {9+h}+3} \right ).$ Now expand the numerator, $\eval {\dd {t} s(t)}_{t=6} = \lim _{h \to 0} \frac {\answer [given]{9+h} - 9}{h\left (\sqrt {9+h}+3 \right )}.$ Combine like-terms, $\eval {\dd {t} s(t)}_{t=6} = \lim _{h \to 0} \frac {\answer [given]{h}}{h\left (\sqrt {9+h}+3\right )}.$ Cancel $h$ from the numerator and denominator, $\eval {\dd {t} s(t)}_{t=6} = \lim _{h\to 0} \frac {1}{\sqrt {9+h}+3}.$ Take the limit as $h$ tends to $0$ , $\eval {\dd {t} s(t)}_{t=6} = \answer [given]{\frac {1}{6}}.$ The object has velocity $\answer [given]{\frac {1}{6}}$ at time $t=6$ . Below we can see the graph of $y=s(t)$ and the tangent line at $t=6$ , with a slope of $\frac {1}{6}$ . $\graph [xmin=-1,xmax=15,ymin=-8,ymax=8]{\sqrt {x+3}, \frac {1}{6}(x-6)+3}$