AP Calculus BC Unit 2 Notes: Understanding the Derivative from First Principles

Average and Instantaneous Rates of Change

What “rate of change” means

A rate of change describes how one quantity changes in response to another. In calculus, you usually study how an output f(x) changes when the input x changes. The key idea is that “change” is measured by differences:

• Change in input: \Delta x = x_2 - x_1
• Change in output: \Delta f = f(x_2) - f(x_1)

The ratio \Delta f/\Delta x tells you “output change per unit input change.” This ratio is the backbone of derivatives.

Why this matters: almost everything you do with derivatives is a refined version of the same question—how fast is something changing right now? That shows up in velocity, growth rates, marginal cost, slope of a curve, sensitivity to inputs, and more.

Average rate of change (secant slope)

The average rate of change of f from x=a to x=b is

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

Conceptually, this is the slope of the secant line through the two points \big(a,f(a)\big) and \big(b,f(b)\big).

• If the average rate of change is positive, the function increased overall on that interval.
• If it’s negative, the function decreased overall.
• Units matter: if f is in meters and x is in seconds, then \Delta f/\Delta x is meters per second.

Average rate of change is “global” across an interval—you’re blending all the behavior between a and b into one number.

Instantaneous rate of change (tangent slope)

The instantaneous rate of change at a single input value (say, at x=a) is meant to capture how f(x) is changing _right at_ a.

A common analogy: if you drive 60 miles in 1 hour, your average speed is 60 mph. But your speedometer reports your speed at that moment—that’s an instantaneous rate.

Mathematically, you get instantaneous rate by shrinking the interval used for the average rate of change. You look at secant slopes as one point approaches the other:

• Start with the secant slope between x=a and x=a+h:

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

• Then let h get closer and closer to 0. If those secant slopes approach a single number, that limiting value is the instantaneous rate of change at a.

This is the central motivation for the derivative.

Worked example: average rate of change

Suppose f(x)=x^2. Find the average rate of change from x=1 to x=3.

Compute the difference quotient:

\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4

Interpretation: on average, f(x) increased by 4 units of output per 1 unit of input over [1,3].

Worked example: secant slopes approaching a tangent slope

Let f(x)=x^2 and examine the secant slope from x=2 to x=2+h:

\frac{f(2+h)-f(2)}{h}=\frac{(2+h)^2-4}{h}=\frac{4+4h+h^2-4}{h}=\frac{4h+h^2}{h}=4+h

As h approaches 0, 4+h approaches 4. That suggests the instantaneous rate of change at x=2 is 4, which matches the derivative idea you’ll formalize next.

Exam Focus

• Typical question patterns:
  • Compute an average rate of change over an interval from a formula, table, or graph.
  • Interpret average rate of change in context (including correct units).
  • Compare average rates on different intervals to describe where a function changes faster.
• Common mistakes:
  • Swapping the order in the numerator (using f(a)-f(b)) while keeping b-a, which flips the sign.
  • Forgetting units or mixing them (for example, using minutes in one place and hours in another).
  • Treating average rate of change as if it describes the behavior at a point (confusing “average over an interval” with “instantaneous at a point”).

Defining the Derivative of a Function

What the derivative is

The derivative of a function at a point is the exact mathematical definition of “instantaneous rate of change” and “slope of the tangent line.”

More precisely, the derivative of f at x=a (if it exists) is the limit of the secant slopes as the second point approaches a.

Why the limit is necessary

You might wonder: why not just plug h=0 into the secant slope formula?

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

If you set h=0, you get

\frac{f(a)-f(a)}{0}

which is division by zero—undefined. The limit process is how calculus resolves this: you don’t ever divide by zero; instead you ask what value the quotient approaches as h approaches zero.

Formal definition (difference quotient limit)

The derivative of f at x=a is

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

If this limit exists (as a finite real number), f is **differentiable** at a.

Interpretations of f'(a):

• Slope interpretation: the slope of the tangent line to the graph y=f(x) at x=a.
• Rate interpretation: the instantaneous rate at which f(x) changes with respect to x at x=a.

Equivalent definition (two-point limit)

Another common form uses a second input value x approaching a:

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

Both definitions describe the same idea: secant slopes approaching a tangent slope.

Derivative as a new function

So far, f'(a) is “the derivative at a point.” If you compute the derivative at every point where it exists, you get a new function called the derivative function:

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

This is powerful because it lets you study how the original function behaves everywhere: where it’s increasing, decreasing, leveling off, changing quickly, and so on.

Notation you must recognize

In AP Calculus, derivative notation is intentionally flexible. You need to be fluent reading all common forms.

A key conceptual point: \frac{dy}{dx} is notation for a derivative, not an actual fraction in the usual algebraic sense (though later, it behaves fraction-like in useful ways such as related rates). In this section, treat it as a single symbol meaning “the derivative.”

Worked example: derivative from the definition

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

Start with the definition:

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

Substitute f(x)=x^2:

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

Expand and simplify:

f'(x)=\lim_{h\to 0}\frac{x^2+2xh+h^2-x^2}{h}=\lim_{h\to 0}\frac{2xh+h^2}{h}=\lim_{h\to 0}(2x+h)

Now take the limit:

f'(x)=2x

This example shows the typical mechanism: you simplify the difference quotient so the h in the denominator cancels before taking the limit.

Worked example: derivative at a point

Let f(x)=\sqrt{x}. Find f'(4) using the definition.

Use the two-point form (often convenient for numeric points):

f'(4)=\lim_{x\to 4}\frac{\sqrt{x}-\sqrt{4}}{x-4}=\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}

Direct substitution gives 0/0, so you simplify. A classic strategy is to multiply by the conjugate:

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

Simplify:

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

Now substitute x=4:

f'(4)=\frac{1}{4}

Interpretation: at x=4, the function \sqrt{x} is increasing at a rate of 0.25 units of output per 1 unit of input.

Exam Focus

• Typical question patterns:
  • Use the limit definition to compute f'(a) for a specific function and point.
  • Show algebraic steps that resolve an indeterminate form like 0/0 (factoring, conjugates).
  • Interpret f'(a) as a slope or an instantaneous rate in a word problem.
• Common mistakes:
  • Plugging in h=0 too early (you must simplify before evaluating the limit).
  • Dropping parentheses when substituting x+h (for example, writing f(x+h)=x+h^2 instead of f(x+h)=(x+h)^2).
  • Confusing f'(a) with f(a) (derivative value vs function value).

Estimating Derivatives at a Point

What it means to estimate a derivative

Not every problem gives you an explicit formula for f(x). On the AP exam, you may get a **graph** of a function or a **table** of values and still be asked for f'(a).

To estimate the derivative at x=a, you approximate the slope of the tangent line by computing slopes of secant lines using points close to a. The closer the points are (and the smoother the graph is near a), the better the approximation tends to be.

This matters because it connects the abstract limit definition to real data: in science, economics, and statistics, you often have measurements, not formulas.

Estimating from a table of values

Suppose you know values of f(x) near x=a. A basic approximation uses a one-sided difference quotient:

Right-hand estimate:

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

Left-hand estimate:

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

A typically better estimate (when the table includes symmetric points) is the symmetric difference quotient:

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

Why symmetric helps: it balances the error from one side with the error from the other side when the function is reasonably smooth.

Worked example: estimating from a table

Assume you have the following values near x=2:

Estimate f'(2).

• Right-hand estimate using h=0.1:

\frac{f(2.1)-f(2.0)}{0.1}=\frac{4.41-4.00}{0.1}=4.1

• Left-hand estimate using h=0.1:

\frac{f(2.0)-f(1.9)}{0.1}=\frac{4.00-3.61}{0.1}=3.9

• Symmetric estimate:

\frac{f(2.1)-f(1.9)}{0.2}=\frac{4.41-3.61}{0.2}=4.0

So f'(2)\approx 4.0. (In fact, these values match f(x)=x^2, whose exact derivative at 2 is 4.)

A common AP skill is explaining your choice: symmetric is usually preferred if the data are available and the function seems smooth.

Estimating from a graph

From a graph, f'(a) is the slope of the tangent line at x=a. Since you cannot draw a perfect tangent line, you approximate it:

1. Locate the point on the curve at x=a.
2. Sketch the tangent line that just “kisses” the curve there (locally matches its direction).
3. Pick two clear points on your tangent line (not necessarily on the curve) and compute slope.

Slope formula reminder (rise over run):

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

Practical tip: choose points far enough apart on the tangent line to reduce reading error, but still on the tangent line you sketched.

Estimating when a function is given implicitly or as data

Sometimes you’re given that f is differentiable and you’re provided nearby values. Your job is not to “find a formula,” but to approximate the derivative directly from differences. In applied contexts:

• If f(t) is position, then f'(t) is velocity.
• If f(t) is velocity, then f'(t) is acceleration.
• If C(q) is cost, then C'(q) is marginal cost (approximate extra cost for producing one more unit near q).

The same numerical difference quotients are used in all of these.

What can go wrong when estimating

Estimation is sensitive to two big issues:

• Step size too large: If the interval is wide, you’re closer to an average rate than an instantaneous rate.
• Function not smooth near the point: If there’s a corner, cusp, discontinuity, or vertical tangent, secant slopes may not settle toward a single finite value.

Exam Focus

• Typical question patterns:
  • Estimate f'(a) from a table using left, right, or symmetric differences.
  • Estimate f'(a) from a graph by approximating the tangent line slope.
  • Interpret the meaning of an estimated derivative in context (including units).
• Common mistakes:
  • Using points that are not close to a (producing an average slope that may differ from the tangent slope).
  • Mixing up f'(a) with f(a) when reading a graph (slope vs height).
  • Computing slope using two points on the curve rather than on the tangent line when the prompt specifically asks for a tangent-line estimate.

Differentiability and Continuity

Continuity: “no breaks” in the graph

A function f is **continuous** at x=a if three conditions hold:

1. f(a) is defined.
2. \lim_{x\to a} f(x) exists.
3. \lim_{x\to a} f(x)=f(a).

Intuitively, continuity at a means you can approach a from both sides and the function values settle to the function’s actual value at a—no holes, jumps, or asymptotic blow-ups at that point.

Continuity matters here because derivatives are defined using limits. If a function has a discontinuity at a, it cannot have a well-defined tangent line there in the usual sense.

Differentiability: “has a well-defined tangent slope”

A function f is **differentiable** at x=a if the derivative limit exists:

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

Geometrically, differentiability at a means the graph has a single, unambiguous tangent direction at that point.

The key relationship

A fundamental theorem-level fact you must know for AP Calculus:

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

In words: differentiability implies continuity.

But the converse is not always true:

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

Why? Continuity only guarantees no break in the graph; differentiability requires smoothness in how the graph changes direction.

Common reasons a function is not differentiable

Even if a function is continuous, the derivative can fail to exist at a point for several recognizable reasons.

1) Corner (sharp turn)

At a corner, the left-hand slope and right-hand slope are finite but unequal, so there is no single tangent slope.

Classic example: f(x)=|x| at x=0.

Using one-sided difference quotients:

Right-hand slope at 0:

\lim_{h\to 0^+}\frac{|h|-0}{h}=\lim_{h\to 0^+}\frac{h}{h}=1

Left-hand slope at 0:

\lim_{h\to 0^-}\frac{|h|-0}{h}=\lim_{h\to 0^-}\frac{-h}{h}=-1

Since the one-sided limits differ, f'(0) does not exist.

This is a common misconception point: the graph is perfectly continuous at 0, but not differentiable there.

2) Cusp (pointed tip)

A cusp is a point where slopes become unbounded in opposite directions (you get a “point” that is sharper than a corner). The function may still be continuous, but the derivative fails because the slope does not approach a single finite number.

You’ll often recognize cusps by a very pointy shape where the tangent would be “almost vertical” and changes direction abruptly.

3) Vertical tangent (infinite slope)

At a vertical tangent, the slope becomes infinite (or undefined as a finite number). In many AP contexts, this means the derivative does not exist as a real number.

Example: f(x)=x^{1/3} at x=0 has a vertical tangent; the function is continuous at 0, but the slope blows up near 0.

4) Discontinuity

If f is discontinuous at a (jump, hole, infinite discontinuity), then f'(a) cannot exist.

Reason: the difference quotient limit depends on f(a+h) approaching f(a) in a controlled way as h approaches 0. A break in the graph prevents that.

How to test differentiability in practice (AP mindset)

On the AP exam, differentiability is often assessed conceptually:

• If you see a jump/hole/asymptote at x=a, the function is not continuous and therefore not differentiable.
• If the function is continuous but has a corner/cusp/vertical tangent, it is not differentiable.
• If the curve looks smooth (no sharp features) at a, it’s reasonable to assume differentiability unless told otherwise.

If you’re given a piecewise function and asked about differentiability at the joining point, you usually need two checks:

1. Continuity at the join.
2. Equality of left and right derivatives at the join.

For the second check, you conceptually compare the left-hand and right-hand limits of the difference quotient (or compute one-sided derivatives).

Worked example: continuous but not differentiable

Define

f(x)=|x|

• Continuity at x=0: |x| has no break, and \lim_{x\to 0}|x|=0=f(0), so it is continuous.
• Differentiability at x=0: as shown earlier, left slope -1 and right slope 1 do not match, so not differentiable.

This example is important because it breaks the common mistaken belief that “continuous means differentiable.”

Worked example: differentiable implies continuous (why it’s true)

You do not usually have to prove this from scratch on AP, but understanding the logic helps you avoid errors.

Assume f is differentiable at a, so

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

exists. Consider rewriting the change in function value:

f(a+h)-f(a)=\left(\frac{f(a+h)-f(a)}{h}\right)h

If the derivative limit exists and is finite, then the quotient stays near a finite number for small h, and multiplying by h forces the whole product toward 0 as h\to 0. That means

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

which implies

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

That is exactly continuity at a.

Exam Focus

• Typical question patterns:
  • Decide whether a function is differentiable at a point based on a graph (identify corners, cusps, discontinuities, vertical tangents).
  • Given a piecewise function, find values of constants that make the function continuous and differentiable at a junction.
  • Explain (in words) why differentiability implies continuity or provide a counterexample to the converse.
• Common mistakes:
  • Claiming a function is differentiable just because it is continuous.
  • Missing a corner on a graph and assuming a tangent exists everywhere.
  • Treating a vertical tangent as having derivative 0 (it’s the opposite issue: slope is unbounded, not flat).