Unit 2: Differentiation: Definition and Fundamental Properties

The Derivative as a Limit (Instantaneous Rate of Change)

When you first meet the derivative, it helps to remember the problem it was invented to solve: how do you measure change at a single instant? Average rate of change is familiar: you can compute “miles per hour” over an interval of time, or the slope of a secant line between two points on a graph. But “instantaneous” speed at exactly t=2 seconds or the slope of the tangent line at exactly x=5 can’t be found by a simple average over a nonzero interval.

The key idea is to shrink the interval until it becomes a point, using limits.

Average rate of change and secant slopes

There are two common ways of talking about “rate of change.”

1) Average rate of change (over an interval) uses the difference quotient. For a function f(x), the average rate of change from x=a to x=a+h (where h \neq 0) is

\frac{f(a+h)-f(a)}{(a+h)-a}

which simplifies to the difference quotient

\frac{f(a+h)-f(a)}{h}

A simpler way to say “average rate of change” between two points is the familiar slope formula

\frac{y_2-y_1}{x_2-x_1}

Geometrically, this is the slope of the secant line through the points \big(a,f(a)\big) and \big(a+h,f(a+h)\big). The closer the two points are, the more accurate this secant slope is at approximating the slope “at” a.

2) Instantaneous rate of change (at a specific point in time) uses the same difference quotient, but with a limit as the interval size goes to zero.

From secant to tangent: letting the interval shrink

For a linear line, slope is “rise over run,” but for a curved graph the slope changes as you move along it. To get the slope at a single input, you let the second point slide toward the first point, meaning h \to 0. If the slopes of the secant lines approach a single value, that value is the slope of the tangent line, which touches the curve at exactly one point (locally).

Definition of the derivative at a point

The derivative of f at a, written f'(a), is

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

If this limit exists (and is finite), f is differentiable at a.

Why this matters: it gives a precise way to compute tangent slopes and defines instantaneous rate of change in applied contexts (velocity, marginal cost, population growth rate at a moment, and more).

Equivalent limit form (sometimes more convenient)

You may also see

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

This is the same idea: a secant slope where the second input x approaches a.

Units and interpretation

If f has units (say, meters) and x has units (say, seconds), then the derivative has units

\frac{\text{units of }f}{\text{units of }x}

So if s(t) is position in meters, s'(t) is velocity in meters per second.

Worked example 1: derivative from the definition

Find f'(a) for f(x)=x^2 using the limit definition.

Start with

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

Compute the numerator:

f(a+h)=(a+h)^2=a^2+2ah+h^2

f(a)=a^2

So

\frac{f(a+h)-f(a)}{h}=\frac{(a^2+2ah+h^2)-a^2}{h}

Simplify:

\frac{2ah+h^2}{h}=2a+h

Now take the limit as h\to 0:

f'(a)=\lim_{h\to 0}(2a+h)=2a

So the derivative function is f'(x)=2x.

What to notice: if you try to plug in h=0 too early, you get \frac{0}{0}, which is indeterminate. The algebra step where you factor/cancel h is crucial.

Worked example 2: tangent line slope at a point

Let f(x)=x^2. Find the slope of the tangent line at x=3.

From above, f'(x)=2x, so

f'(3)=2\cdot 3=6

That means the tangent line slope at x=3 is 6.

Exam Focus

Typical question patterns:

• “Use the limit definition to find f'(x) for a given function (often a polynomial or radical).”
• “Compute f'(a) from the definition at a specific point.”
• “Interpret the derivative as a slope or instantaneous rate of change with units.”

Common mistakes:

• Plugging in h=0 before simplifying, leading to \frac{0}{0}.
• Algebra errors expanding expressions like (a+h)^2.
• Forgetting: if the limit does not exist, the derivative does not exist.

Derivative Notation and the Derivative as a Function

Once you can compute a derivative at a single point, the next step is realizing you can compute it at every input (where it exists). This produces a new function: the derivative function.

Many notations, same meaning

Derivative notation depends on context (geometry, physics, pure function notation). These are equivalent ways to express “the derivative of y=f(x) with respect to x.”

A common confusion is thinking \frac{dy}{dx} is a “fraction” you can treat like ordinary algebra in all situations. In AP Calculus, you’ll sometimes manipulate it informally later, but conceptually it represents a single quantity: the derivative.

First and second derivative notation

You will also see notation for second derivatives (the derivative of the derivative):

The derivative function

If f'(a) exists for many values of a, define

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

This new function takes an input x and outputs the slope of the tangent line to f at that input.

A key relationship you’ll use repeatedly is: where f is increasing, f' is positive; where f is decreasing, f' is negative.

Relating the derivative to tangent lines

The tangent line to y=f(x) at x=a has slope f'(a) and passes through \big(a,f(a)\big). Using point-slope form:

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

Worked example 1: writing a tangent line equation

Let f(x)=\sqrt{x}. Find the equation of the tangent line at x=4.

Compute f'(4) from the definition:

f'(4)=\lim_{h\to 0}\frac{\sqrt{4+h}-\sqrt{4}}{h}=\lim_{h\to 0}\frac{\sqrt{4+h}-2}{h}

Multiply by the conjugate:

\frac{\sqrt{4+h}-2}{h}\cdot\frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}=\frac{(4+h)-4}{h(\sqrt{4+h}+2)}

Simplify:

\frac{h}{h(\sqrt{4+h}+2)}=\frac{1}{\sqrt{4+h}+2}

Now take the limit:

f'(4)=\lim_{h\to 0}\frac{1}{\sqrt{4+h}+2}=\frac{1}{4}

So the slope is \frac{1}{4}. The point is \big(4,2\big), so

y-2=\frac{1}{4}(x-4)

Worked example 2: interpreting a derivative value

Suppose P(t) is the number of bacteria in a culture at time t (hours). If P'(3)=120, then at exactly t=3 hours the population is increasing at 120 bacteria per hour.

A common trap is to interpret it as “the population is 120 at 3 hours.” That would be P(3)=120, not the derivative.

Exam Focus

Typical question patterns:

• “Find the equation of the tangent line to f at x=a.”
• “Given a real-world context, interpret f'(a) with units.”
• “Write an expression for f'(x) using the limit definition.”

Common mistakes:

• Using f(a) as the slope instead of f'(a).
• Forgetting to use the point \big(a,f(a)\big) in point-slope form.
• Mixing up f(a) and f'(a) in context questions.

Estimating Derivatives from Tables and Graphs

In many AP problems, you are not given a formula for f(x). Instead, you might get a table of values, a graph, or a description. In those cases, you estimate the derivative by thinking the way the definition thinks: slope of nearby secant lines.

Estimating from a table using secant slopes

If you have values of f(x) near x=a, you can approximate f'(a) with a difference quotient. A common approximation uses a symmetric difference quotient, which tends to be more accurate when the function is reasonably smooth:

f'(a)\approx\frac{f(a+h)-f(a-h)}{2h}

Symmetric estimates often reduce “bias” from curvature because they balance the left and right behavior around a.

Estimating from a graph

From a graph, f'(a) is the slope of the tangent line at x=a. Since you can’t draw a perfect tangent line, you approximate by drawing the tangent line as best you can, choosing two clear points on that tangent line (not necessarily points on the curve), and computing rise over run. If the tangent line appears horizontal, the derivative is near 0.

Worked example 1: estimating from a table

You are given:

Estimate f'(2).

Left and right secant slopes:

\text{Left slope}=\frac{f(2.0)-f(1.9)}{2.0-1.9}=\frac{6.00-5.72}{0.1}=2.8

\text{Right slope}=\frac{f(2.1)-f(2.0)}{2.1-2.0}=\frac{6.31-6.00}{0.1}=3.1

Average them (matching the symmetric idea):

f'(2)\approx\frac{2.8+3.1}{2}=2.95

So the instantaneous rate of change near x=2 is about 2.95 units of f per unit of x.

Worked example 2: estimating from a graph description

If a graph shows the tangent line at x=1 passing through approximately (1,2) and (3,6), then

f'(1)\approx\frac{6-2}{3-1}=2

Even if the curve itself doesn’t pass through (3,6), it’s okay as long as those are points on the tangent line you drew.

Exam Focus

Typical question patterns:

• “Estimate f'(a) from a table of values near a.”
• “Use the graph of f to approximate f'(a) or identify where f'(x)=0.”
• “Compare left- and right-hand estimates to decide whether a derivative might exist.”

Common mistakes:

• Using points on the curve far from a instead of points on the tangent line near a.
• Mixing up \Delta y and \Delta x (computing run over rise).
• Using a one-sided difference when the problem expects a symmetric estimate.

Differentiability and Continuity

A major conceptual checkpoint is understanding when a derivative exists. Differentiability tells you whether a function is “smooth enough” to have a well-defined tangent slope at a point.

What it means to be differentiable

A function f is differentiable at x=a if the limit

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

exists and is finite. Intuitively, as you zoom in near x=a, the curve should look more and more like a straight line with a single slope.

Differentiability implies continuity

A crucial theorem:

• If f is differentiable at a, then f is continuous at a.

Equivalently:

• If f is not continuous at a, then f cannot be differentiable at a.

Important caution: the converse is false.

• A function can be continuous at a but not differentiable at a.

Common reasons a derivative fails to exist

1) Corner or cusp: the left-hand slope and right-hand slope approach different values (corner), or slope becomes infinite in opposite ways (cusp).

Example shape: |x| at x=0.

2) Vertical tangent: slopes grow without bound (infinite slope). In AP language, you typically say the derivative does not exist there.

Example shape: \sqrt[3]{x} at x=0 has a vertical tangent.

3) Discontinuity (hole, jump, asymptote): not continuous, so not differentiable.

One-sided derivatives and corners

Define one-sided derivatives:

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

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

For f'(a) to exist, both must exist and be equal.

Worked example 1: continuous but not differentiable

Consider f(x)=|x| at x=0.

Left side: if h