Unit 2: Differentiation: Definition and Fundamental Properties

The derivative as a limit: from average change to instantaneous change

Average rate of change (the idea you already know)

When a function describes how one quantity depends on another, you often want to measure how fast the output is changing as the input changes. Over an interval, that “how fast” is the average rate of change, which is the slope of the secant line through two points.

If the function is f(x) and you look from input a to input b, the average rate of change is

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

In coordinate language, this is the familiar

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

This is a slope (“rise over run”) and is the rate of change over an interval of time (or an interval of any input). In context, it might mean average velocity (change in position over change in time), average cost per item, average temperature increase per hour, and so on.

Instantaneous rate of change and the tangent line

Many real questions are about a precise moment: your speed right now, or the exact steepness of a curve at a particular point. That is the instantaneous rate of change, which is what the average rate of change becomes when the interval shrinks to a single point.

For a straight (linear) line, slope is always “rise over run.” For a curved graph, you can’t pick two points on the curve and call that “the slope at a point” because different pairs give different secant slopes. Instead:

• A secant line intersects the graph at two points and approximates slope.
• A tangent line touches the curve at exactly one point and has the same direction the curve is heading at that point.

You approximate the slope at a point by using secant lines with points closer and closer together (the closer the points are, the more accurate the approximation). The “perfect” limiting secant is the tangent line.

To turn “shrink the interval” into mathematics, start with the secant slope between x=a and x=a+h:

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

Simplify the denominator:

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

Now let h\to 0 to force the second point to approach the first.

Definition of the derivative at a point

The derivative of f at a is defined by the limit (when it exists):

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

This is called the definition of the derivative. Conceptually, the derivative is the rate of change at a specific point, so we use a limit to find the slope as the change in input becomes infinitesimally small.

What it is:

• A single number (for a fixed a) representing the slope of the tangent line to y=f(x) at x=a.
• The instantaneous rate of change of f with respect to x at x=a.

How it works conceptually:

• Compute slopes of secant lines for smaller and smaller h.
• If those slopes approach a single value, that value is the derivative.

Equivalent limit definition (approaching with x\to a)

You may also see

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

This is the same idea written with the second point’s x-value approaching a.

Worked example: derivative from the definition

Find the derivative of

f(x)=x^2

at a general point x=a using the limit definition.

Start with

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

Compute:

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

f(a)=a^2

Substitute:

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

Expand and simplify the numerator:

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

So

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

Factor out h:

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

Cancel h (this is exactly why you must simplify before substituting h=0):

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

Now substitute h=0:

f'(a)=2a

So the derivative function is

f'(x)=2x

Common misconception to avoid: plugging in h=0 too early gives \frac{0}{0}, which is undefined.

Tangent line equation using the derivative

Once you know the slope at x=a, you can write the tangent line at that point. The point on the curve is \bigl(a,f(a)\bigr) and the slope is f'(a).

Point-slope form:

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

Exam Focus

• Typical question patterns:
  • “Use the limit definition to find f'(x) for a given function.”
  • “Find the slope of the tangent line to y=f(x) at x=a.”
  • “Write the equation of the tangent line at a given point.”
• Common mistakes:
  • Substituting h=0 before simplifying and getting stuck at \frac{0}{0}.
  • Mixing up secant slope (average rate) with tangent slope (instantaneous rate).
  • Using the wrong point in point-slope form (for a tangent at x=a, the point must use f(a)).

Derivative notation, meaning, and units

Notation you must be fluent with

AP Calculus uses multiple equivalent notations for derivatives. They mean the same derivative but emphasize different interpretations.

A key distinction:

• f'(a) is a number.
• f'(x) is a new function that gives the slope at every input where it exists.

First and second derivative shorthand (common in notes)

You will also see derivative notation presented like this:

In typed or handwritten notes, the second derivative is sometimes written with “double-prime” marks (for example, it may look like f”); the standard calculus notation to interpret that is f''(x).

The derivative as a function (not just a slope)

When the limit exists for each x in an interval, you can define the derivative function:

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

It outputs the slope of f at each input. This matters because many problems ask about how fast a function changes across a whole region.

Units of the derivative

Tracking units is a strong way to check interpretation. If f has units “output units” and x has units “input units,” then f'(x) has units

\frac{\text{output units}}{\text{input units}}

Example: if s(t) is position in meters and t is time in seconds, then s'(t) is meters per second.

Differentiation at a point vs. over an interval

These are related but not the same:

1. Average rate of change on [a,b]:

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

2. Instantaneous rate of change at a:

f'(a)

3. The derivative function:

f'(x)

Worked example: interpreting derivative language

Suppose:

• C(q) is the cost in dollars to produce q items.

Then:

• C'(q) is marginal cost, measured in dollars per item.
• C'(100) is the approximate cost to produce the 101st item (more precisely: the rate at which cost is changing at q=100).

Exam Focus

• Typical question patterns:
  • “Interpret the meaning of f'(a) in context, including units.”
  • “Given a statement about a rate of change, identify what derivative it represents.”
  • “Distinguish between f'(a) and f'(x).”
• Common mistakes:
  • Treating f'(a) like a point or like a function output without specifying the input.
  • Ignoring units, leading to misinterpretations (especially in motion and economics contexts).
  • Confusing average rate of change with instantaneous rate of change.

Estimating derivatives from graphs, tables, and verbal descriptions

Estimating slope from a graph

Often you are not given a formula for f(x). If you are given a graph and asked to estimate f'(a), you are estimating the slope of the tangent line.

A reliable process:

1. Draw the tangent line as carefully as you can at x=a.
2. Choose two convenient points on your tangent line (not necessarily on the curve) to compute slope.
3. Compute slope as rise over run.

A common pitfall is using two points on the curve near a. That gives a secant slope, which may be close, but can be inaccurate if the curve bends strongly.

Estimating from a table (difference quotients)

If you have a table of values, you approximate the derivative at x=a using nearby points.

Forward difference:

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

Backward difference:

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

Symmetric (central) difference (often best when data are smooth):

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

Why symmetric is often better: it balances error from looking only to one side. AP problems sometimes provide data on both sides specifically to encourage a central estimate.

One-sided derivatives and corners

Even if a function is continuous, the derivative at a point might fail to exist. A key diagnostic is whether left-hand and right-hand slopes match.

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}

Then f'(a) exists only if both one-sided derivatives exist and are equal.

Graphically:

• A corner or cusp occurs when the left-hand slope and right-hand slope are different (or infinite in different ways).
• A vertical tangent corresponds to an “infinite” slope; in many AP contexts you treat this as not differentiable at that point.

Example: estimating from a table

Suppose you have values near x=2:

Estimate f'(2) using a symmetric difference with h=0.1:

f'(2)\approx \frac{f(2.1)-f(1.9)}{2(0.1)}

f'(2)\approx \frac{6.31-5.72}{0.2}

f'(2)\approx \frac{0.59}{0.2}=2.95

Interpretation: near x=2, the function is increasing at about 2.95 output units per input unit.

Exam Focus

• Typical question patterns:
  • “Estimate f'(a) from a graph (slope of the tangent).”
  • “Estimate f'(a) from a table using a difference quotient.”
  • “Decide whether f'(a) exists based on graph features (corner, cusp, discontinuity).”
• Common mistakes:
  • Using points on the curve instead of points on the tangent line when estimating graph slopes.
  • Forgetting the denominator in a symmetric difference is 2h, not h.
  • Assuming continuity automatically implies differentiability.

Differentiability, continuity, and when derivatives fail to exist

What it means to be differentiable

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

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

If that limit exists and is finite, the graph has a well-defined tangent slope at a.

Differentiability implies continuity

A crucial theorem in AP Calculus:

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

In symbols, if f'(a) exists, then

\lim_{x\to a}f(x)=f(a)

Important warning: the converse is false.

• Continuous does not necessarily mean differentiable.

Common reasons a function is not differentiable

If you see any of these at x=a, the derivative typically does not exist.

1. Discontinuity (jumps, holes, infinite discontinuities)
2. Corner (left and right slopes differ)
3. Cusp (slopes become infinite in opposite directions)
4. Vertical tangent (slope is not finite)

Example: a classic corner

Consider

f(x)=|x|

At x=0, the function is continuous, but the slope from the left is -1 and the slope from the right is 1, so the derivative does not exist at 0.

Differentiability on an interval

You may be asked about differentiability on an interval like (-2,3) or [0,5].

Two common subtleties:

• Endpoints: differentiability at an endpoint of a closed interval is usually discussed via one-sided derivatives.
• Piecewise functions: you must check differentiability at the “break points,” even if each piece is differentiable on its own interior.

Exam Focus

• Typical question patterns:
  • “Is f differentiable at x=a? Justify using a graph or one-sided derivatives.”
  • “Explain why differentiability implies continuity (or use the implication to conclude continuity).”
  • “Identify points where f'(x) does not exist from a graph.”
• Common mistakes:
  • Stating “continuous so differentiable” (not true).
  • Missing non-differentiable points in piecewise graphs at corners or cusps.
  • Treating a vertical tangent as having derivative 0 (it does not; it is not a finite slope).

Fundamental derivative rules (built from the limit definition)

Using the limit definition every time is tedious, so calculus develops rules that make taking derivatives fast while still being grounded in the definition.

Constant rule

If

f(x)=c

where c is a constant, then

f'(x)=0

Example: if f(x)=10, then f'(x)=0.

Power rule (for positive integer powers)

For

f(x)=x^n

where n is a positive integer, the derivative is

f'(x)=nx^{n-1}

A helpful description: multiply down and decrease the power.

Examples:

• x^4 becomes 4x^3
• 2x^2 becomes 4x

The power rule works for polynomials.

Constant multiple rule

If

f(x)=c\,g(x)

then

f'(x)=c\,g'(x)

Meaning: you can “pull the constant out,” scaling the slope by the same constant.

Sum and difference rules

If

f(x)=g(x)+h(x)

then

f'(x)=g'(x)+h'(x)

If

f(x)=g(x)-h(x)

then

f'(x)=g'(x)-h'(x)

Worked example: differentiating a polynomial

Differentiate

f(x)=3x^4-5x^2+7x-9

Differentiate term-by-term:

\frac{d}{dx}(3x^4)=12x^3

\frac{d}{dx}(-5x^2)=-10x

\frac{d}{dx}(7x)=7

\frac{d}{dx}(-9)=0

So

f'(x)=12x^3-10x+7

Common mistake to avoid: subtracting 1 from the coefficient instead of the exponent, or rewriting 7x as 7x^0 and then forgetting x^0=1 so the derivative is just 7.

Worked example: tangent line using basic rules

Find the tangent line to

f(x)=x^3-2x

at x=1.

Derivative:

f'(x)=3x^2-2

Slope at x=1:

f'(1)=3(1)^2-2=1

Point on the curve:

f(1)=(1)^3-2(1)=-1

Point-slope form:

y-(-1)=1(x-1)

Simplify:

y+1=x-1

y=x-2

Exam Focus

• Typical question patterns:
  • “Find f'(x) for a polynomial or a simple combination using rules.”
  • “Find the slope of the tangent line at x=a and write the tangent line equation.”
  • “Given that f'(a) represents a rate, compute it from a formula and interpret.”
• Common mistakes:
  • Forgetting the constant multiple rule (for example, differentiating 3x^4 as 4x^3).
  • Dropping signs when applying the sum/difference rule.
  • Confusing f'(a) (a number) with f(a) (a function value).

Product and quotient rules

Some functions are built by multiplying or dividing simpler functions. You can sometimes expand and simplify first (for instance, multiplying two polynomials and then using the power rule), but that can take time and increase algebra errors. The product and quotient rules give a direct path.

Product rule

If

f(x)=u\,v

then

f'(x)=u\frac{dv}{dx}+v\frac{du}{dx}

A common memory cue is “1d2 + 2d1” (first times derivative of second, plus second times derivative of first).

Example situation: for something like

f(x)=(2x+7)(9x+8)

you could multiply it out and then use the power rule, but the product rule avoids that extra expansion.

Quotient rule

If

f(x)=\frac{u}{v}

then

f'(x)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}

A common memory cue is “low d high - high d low / low^2” (denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared).

Exam Focus

• Typical question patterns:
  • “Differentiate a product of two functions (especially polynomials or trig/exponential pieces).”
  • “Differentiate a quotient and simplify; evaluate at a point.”
  • “Use a memory cue correctly to avoid sign/order errors.”
• Common mistakes:
  • Forgetting that the product rule is not “multiply the derivatives.”
  • Dropping parentheses in the quotient rule, leading to sign mistakes.
  • Squaring only part of the denominator instead of all of v^2.

Derivatives of trigonometric functions

These are often treated as “memory derivatives” because they are easier to memorize than to derive on the spot.

Trigonometric functions are central in calculus because they model periodic motion (waves, rotations, oscillations). Their derivatives are also periodic.

Core trig derivatives

You should know these as fundamental facts:

\frac{d}{dx}(\sin x)=\cos x

\frac{d}{dx}(\cos x)=-\sin x

From these, additional standard trig derivatives include:

\frac{d}{dx}(\tan x)=\sec^2 x

\frac{d}{dx}(\sec x)=\sec x\tan x

\frac{d}{dx}(\csc x)=-\csc x\cot x

\frac{d}{dx}(\cot x)=-\csc^2 x

Radian measure (important subtlety)

These derivative formulas assume the input is measured in radians. On the AP exam, trig inputs are understood to be radians unless stated otherwise.

Example: slope of a trig function at a point

Let

f(x)=\sin x+2\cos x

Differentiate term-by-term:

f'(x)=\cos x-2\sin x

Then the slope at x=0 is

f'(0)=\cos 0-2\sin 0=1-0=1

Example: interpreting a trig derivative in context

Suppose a Ferris wheel height above ground is modeled as

h(t)=20+10\sin t

where h is in meters and t is in seconds.

Instantaneous vertical velocity:

h'(t)=10\cos t

At t=0,

h'(0)=10\cos 0=10

Interpretation: at that moment, height is increasing at 10 meters per second.

Common mistake to avoid: differentiating \cos x as \sin x without the negative sign.

Exam Focus

• Typical question patterns:
  • “Differentiate an expression involving \sin x and \cos x using basic rules.”
  • “Compute f'(a) for a trig function and interpret the result.”
  • “Use a derivative to find a tangent line slope on a trig graph.”
• Common mistakes:
  • Missing the negative sign in \frac{d}{dx}(\cos x)=-\sin x.
  • Mixing up which trig derivative goes with which function (especially \tan x and \sec x).
  • Forgetting that these standard derivatives assume radian input.

Derivatives of exponential and logarithmic functions

These are also commonly treated as “memory derivatives.”

Exponential and logarithmic functions model growth and decay and “undoing growth.” Their derivatives are powerful because they preserve recognizable forms.

Natural exponential function

For

f(x)=e^x

the derivative is

\frac{d}{dx}(e^x)=e^x

Meaning: the rate of change of e^x at any point equals its current value.

Example:

f(x)=5e^x-x^3

Then

f'(x)=5e^x-3x^2

Natural logarithm

For the natural logarithm,

\frac{d}{dx}(\ln x)=\frac{1}{x}

This is defined for x>0.

Example:

g(x)=\ln x+4x^2

Then

g'(x)=\frac{1}{x}+8x

Common mistake to avoid: the derivative of \ln x is not found by changing the inside (it is exactly \frac{1}{x}).

Exponential functions with other bases

For a>0 and a\neq 1,

\frac{d}{dx}(a^x)=a^x\ln a

Example:

f(x)=2^x

Then

f'(x)=2^x\ln 2

Logarithms with other bases

Use change of base:

\log_a x=\frac{\ln x}{\ln a}

Then

\frac{d}{dx}(\log_a x)=\frac{1}{x\ln a}

Exam Focus

• Typical question patterns:
  • “Differentiate expressions involving e^x and \ln x using basic rules.”
  • “Interpret f'(a) for an exponential model (growth/decay) in context.”
  • “Differentiate a^x or \log_a x (often with bases like 2 or 10).”
• Common mistakes:
  • Treating \frac{d}{dx}(e^x) as xe^{x-1} (confusing with the power rule).
  • Forgetting domain restrictions: \ln x requires x>0.
  • Forgetting the factor \ln a in \frac{d}{dx}(a^x)=a^x\ln a.

Connecting derivatives to motion and other real-world rates

A major theme in AP Calculus is that derivatives are not just abstract slopes; they measure real rates.

Position, velocity, and acceleration

If s(t) is position as a function of time, then:

• Velocity is the derivative of position:

v(t)=s'(t)

• Acceleration is the derivative of velocity (second derivative of position):

a(t)=v'(t)=s''(t)

Units:

• If s is meters and t is seconds, then v is meters per second and a is meters per second squared.

Example: computing velocity from position

Let

s(t)=t^3-6t^2+9t

Then

v(t)=s'(t)=3t^2-12t+9

And

a(t)=v'(t)=6t-12

Interpretation:

v(2)=3(4)-12(2)+9=12-24+9=-3

So at t=2 seconds, velocity is -3 (position units per time unit), meaning the object is moving in the negative direction.

Common mistake to avoid: confusing negative velocity with “slowing down.” Negative velocity is direction; speeding up/slowing down depends on velocity and acceleration together.

Marginal analysis (economics)

If C(x) is total cost to produce x units, then C'(x) is marginal cost.

If R(x) is revenue, then R'(x) is marginal revenue.

If P(x) is profit, then

P(x)=R(x)-C(x)

and

P'(x)=R'(x)-C'(x)

Local linearity (the “zoomed-in line” idea)

If f is differentiable at a, then near a the graph looks almost like a line. That tangent line gives a local approximation:

f(x)\approx f(a)+f'(a)(x-a)

Exam Focus

• Typical question patterns:
  • “Given s(t), find v(t) and interpret v(a).”
  • “Given a contextual function, state what f'(a) means in words and units.”
  • “Use the tangent line idea to approximate behavior near a point (often informally).”
• Common mistakes:
  • Mixing up position s(t) with velocity s'(t).
  • Ignoring units when interpreting derivatives in context.
  • Treating a derivative value as an average rate rather than an instantaneous rate.

Higher-order derivatives and basic notation (first look)

What a second derivative represents

If f'(x) tells you the rate of change of f, then the second derivative tells you the rate of change of the rate of change.

Notation:

f''(x)

and if y=f(x):

\frac{d^2y}{dx^2}

In motion contexts: position, then velocity, then acceleration.

Computing a second derivative

You find a second derivative by differentiating again.

Example:

f(x)=x^4-3x^2

First derivative:

f'(x)=4x^3-6x

Second derivative:

f''(x)=12x^2-6

Exam Focus

• Typical question patterns:
  • “Find f'(x) and f''(x) for a polynomial.”
  • “Interpret f''(a) in a motion context if position is given.”
  • “Use notation correctly (distinguish f'(a), f''(a), and functions).”
• Common mistakes:
  • Forgetting that f''(x) is the derivative of f'(x), not \bigl(f(x)\bigr)^2.
  • Confusing the meaning of a negative second derivative (it does not automatically mean the function is decreasing).
  • Dropping constants or making sign errors on the second differentiation.