3.1 Defining the Derivative

Big idea

Two questions that look unrelated turn out to be the same question. How steep is a curve at a single point, and how fast is a moving object at a single instant? Both ask for a rate at a place where nothing changes, and a rate needs two readings. The way out is to compute the rate you can compute, over an interval, and then shrink the interval.

Over an interval you have the average rate of change: the change in output divided by the change in input. Geometrically that number is the slope of the line through the two points on the graph, called a secant line. It is an honest answer to a slightly different question, and it is exact. What it hides is everything that happened between the endpoints, which is why it cannot by itself say how steep the curve is at one end.

Now fix one endpoint and slide the other toward it. The secant lines pivot around the fixed point, and if the slopes settle on a single number, that number is the derivative at the point and the limiting line is the tangent line. The whole definition is one limit, and the limit is essential: at the moment the two points coincide, the quotient reads zero over zero and says nothing.

This is where the limit machinery from the previous chapter earns its keep. Every difference quotient is an indeterminate form, so you never evaluate it by substitution. You do algebra first - expand, cancel a common factor of the increment, rationalize - until the quotient becomes something continuous, and only then substitute.

Decoder

The derivative of $f$ at $a$ is the limit of the slopes of the secant lines through the point $(a, f(a))$, and when that limit exists it is the slope of the tangent line to the graph there.

Two phrases carry the content. “The limit of the slopes” says the derivative is not a slope you measure; it is a number that slopes approach. “When that limit exists” says the derivative can fail to exist, and it does for functions with a jump or a sharp point, because the secants approaching from the left and from the right settle on different numbers.

The definition comes in two spellings. Naming the second point $a + h$ makes the increment explicit and the algebra mechanical. Naming it $x$ and letting $x \to a$ makes the cancellation easier when the function is a quotient or a root. They are the same limit after the substitution $x = a + h$, so use whichever leaves you with a factor you can cancel.

The classic mistake is to stop at the simplification and forget the limit, or to plug $h = 0$ into the quotient before cancelling. Substituting first gives $0/0$, which is not a number and not an error message; it is a signal that the cancellation has not been done yet.

Definitions and results

Average rate of change. For $f$ on the interval from $a$ to $b$ with $a \ne b$, the average rate of change is

$$ \frac{f(b) - f(a)}{b - a} $$

Its units are output units per input unit, which is how you check that an interpretation is sensible.

Secant line. The line through $(a, f(a))$ and $(b, f(b))$. Its slope is exactly the average rate of change over that interval.

The derivative at a point. If the limit exists,

$$ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} $$

and $f$ is called differentiable at $a$. The quotient inside is the difference quotient.

The alternative form. The same number, written with the moving point named directly:

$$ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} $$

Tangent line. If $f'(a)$ exists, the tangent line to the graph at $x = a$ is

$$ y = f(a) + f'(a)(x - a) $$

It is the line through the point with the derivative as its slope, and near $x = a$ it is the best straight-line stand-in for the curve.

Normal line. The line through the same point perpendicular to the tangent. When $f'(a) \ne 0$ its slope is $-1/f'(a)$, so its equation is $y = f(a) - \frac{1}{f'(a)}(x - a)$. If $f'(a) = 0$ the normal line is vertical.

Instantaneous velocity. If $s(t)$ gives position at time $t$, the average velocity over an interval is the average rate of change of $s$, and the instantaneous velocity at time $t$ is $s'(t)$. Speed is the absolute value of velocity.

Why the limit is needed. At $h = 0$ both numerator and denominator of the difference quotient are zero. The quotient is undefined at that point but can be simplified everywhere else, and the simplified form is what you take the limit of.

Worked examples

Average rate against instantaneous rate

Take $f(x) = x^2 + 1$ on the interval from $x = 1$ to $x = 4$. Since $f(1) = 2$ and $f(4) = 17$, the average rate of change is

$$ \frac{17 - 2}{4 - 1} = \frac{15}{3} = 5 $$

For the instantaneous rate at $x = 1$, build the difference quotient and cancel:

$$ \frac{f(1+h) - f(1)}{h} = \frac{(1 + 2h + h^2 + 1) - 2}{h} = \frac{2h + h^2}{h} = 2 + h $$

The last expression is defined at $h = 0$, so the limit is $f'(1) = 2$. The same computation at $x = 4$ gives $8 + h$, so $f'(4) = 8$. The average rate $5$ sits between $2$ and $8$, as it must for a curve whose steepness grows across the interval.

A quadratic from the definition

Find $f'(2)$ for $f(x) = 3x^2 - 2x$. Here $f(2) = 12 - 4 = 8$, and

$$ \frac{f(2+h) - f(2)}{h} = \frac{3(4 + 4h + h^2) - 2(2+h) - 8}{h} = \frac{10h + 3h^2}{h} = 10 + 3h $$

Letting $h \to 0$ gives $f'(2) = 10$. Notice that the constant terms cancelled before the division; if they do not, check the arithmetic, because a leftover constant makes the quotient blow up rather than settle.

A reciprocal, using the other form

Find $f'(3)$ for $f(x) = 1/x$. The $x \to a$ form avoids expanding anything:

$$ \frac{\frac{1}{x} - \frac{1}{3}}{x - 3} = \frac{\frac{3 - x}{3x}}{x - 3} = \frac{-(x-3)}{3x(x-3)} = \frac{-1}{3x} $$

As $x \to 3$ this tends to $-1/9$, so $f'(3) = -1/9$. The sign is right: $1/x$ decreases for positive $x$.

Tangent and normal lines to a square root

Let $f(x) = \sqrt{x}$ and work at $a = 4$, where $f(4) = 2$. Rationalizing turns the quotient into something that survives substitution:

$$ \frac{\sqrt{x} - 2}{x - 4} = \frac{\sqrt{x} - 2}{(\sqrt{x} - 2)(\sqrt{x} + 2)} = \frac{1}{\sqrt{x} + 2} $$

So $f'(4) = 1/4$. The tangent line is $y = 2 + \frac{1}{4}(x - 4)$, which simplifies to $y = \frac{x}{4} + 1$. The normal line has slope $-4$:

$$ y = 2 - 4(x - 4) = -4x + 18 $$

Check both at $x = 4$: the tangent gives $1 + 1 = 2$ and the normal gives $-16 + 18 = 2$. Both pass through $(4, 2)$, and their slopes multiply to $-1$.

Practice

Begin with intervals, where no limits are involved. Compute the change in output over the change in input and attach units when the problem supplies them.

Next, the definition itself. Build the difference quotient, cancel the increment, and take the limit. Both spellings of the definition appear, so choose the one that makes the algebra shorter.

Last, lines. Once you have the point and the derivative there, the tangent line is immediate and the normal line is the negative reciprocal slope through the same point.

Quiz

Six items on average rates of change, the limit definition of the derivative at a point, and tangent and normal lines.