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 $h<0$, then $|h|=-h$, so

$\frac{f(0+h)-f(0)}{h}=\frac{|h|-0}{h}=\frac{-h}{h}=-1$

Thus

$f'_-(0)=\lim_{h\to 0^-}(-1)=-1$

Right side: if $h>0$, then $|h|=h$, so

$\frac{|h|}{h}=1$

Thus

$f'_+(0)=\lim_{h\to 0^+}(1)=1$

Because $-1 \neq 1$, the derivative at $0$ does not exist. The graph is continuous there, but it has a sharp corner.

Worked example 2: identifying non-differentiability from a graph

If a graph shows a sharp point at $x=2$, you should immediately think “corner/cusp” and conclude $f'(2)$ does not exist, even if the function value exists and the graph is unbroken.

A common misconception is to assume “if you can draw it without lifting your pencil, it’s differentiable.” That’s false: you can draw $|x|$ without lifting your pencil, but it is not differentiable at the corner.

Exam Focus

Typical question patterns:

• “Given a graph, list the $x$-values where $f$ is not differentiable and explain why (corner, cusp, discontinuity, vertical tangent).”
• “True/false with justification: differentiable implies continuous; continuous implies differentiable.”
• “Use one-sided derivatives (conceptually or numerically) to determine differentiability at a point.”

Common mistakes:

• Claiming continuity automatically guarantees differentiability.
• Saying the derivative exists at a vertical tangent because “the line is still a tangent.”
• Confusing a corner (finite but different one-sided slopes) with a discontinuity (break in the graph).

Fundamental Differentiation Rules (Linearity and the Power Rule)

Computing every derivative from the limit definition is tedious, so we use rules that generalize many of those limit computations.

The constant rule

If $f(x)=k$ where $k$ is a constant, then the slope is always $0$:

$\frac{d}{dx}[k]=0$

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

Constant multiple rule

If you have a constant multiplied by a function, you can pull the constant out:

$\frac{d}{dx}[cf(x)]=c\frac{d}{dx}[f(x)]$

Sum and difference rules

Derivatives distribute across addition and subtraction:

$\frac{d}{dx}[f(x)+g(x)]=f'(x)+g'(x)$

$\frac{d}{dx}[f(x)-g(x)]=f'(x)-g'(x)$

These are often grouped as the linearity properties of differentiation.

The power rule

If $f(x)=x^n$ then

$\frac{d}{dx}[x^n]=nx^{n-1}$

A good way to describe it is: “multiply down and decrease the power.” For example, $x^4$ becomes $4x^3$ and $2x^2$ becomes $4x$. The power rule works for polynomials, and (in standard calculus) it also works for negative integer exponents.

What goes wrong: common rule misuse

It’s easy to overgeneralize the sum rule into false statements like

$\frac{d}{dx}[f(x)g(x)]=f'(x)g'(x)$

That is not true in general. When you see multiplication or division of functions, you need different rules (product/quotient), discussed below.

Worked example 1: differentiating a polynomial

Differentiate $f(x)=3x^5-2x^3+7x-4$.

$f'(x)=3\cdot 5x^4-2\cdot 3x^2+7-0$

So

$f'(x)=15x^4-6x^2+7$

Worked example 2: using rules to find a tangent line

Find the tangent line to $f(x)=x^3-5x$ at $x=2$.

Differentiate:

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

Slope at $x=2$:

$f'(2)=3(2^2)-5=7$

Point:

$f(2)=2^3-5(2)=-2$

Tangent line:

$y-(-2)=7(x-2)$

So

$y=7x-16$

Worked example 3: careful with negative exponents

If $f(x)=x^{-2}$, apply the power rule:

$f'(x)=-2x^{-3}$

Rewrite with positive exponents:

$f'(x)=-\frac{2}{x^3}$

Exam Focus

Typical question patterns:

• “Differentiate a polynomial function and evaluate the derivative at a point.”
• “Find an equation of a tangent line using $f(a)$ and $f'(a)$.”
• “Given a derivative expression, reason about slopes (sign/size) at certain inputs.”

Common mistakes:

• Dropping constants or mishandling negative constants.
• Applying the power rule incorrectly (writing $nx^{n-1}$ as $nx^n$).
• Treating products like sums (trying to differentiate “term-by-term” across multiplication).

Product Rule and Quotient Rule

Once multiplication or division shows up between functions (especially polynomials), linearity rules are not enough.

Product rule

If $f(x)=uv$, then

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

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

Why it’s useful: if you have two polynomials multiplied by each other like $(2x+7)(9x+8)$, you could multiply it out and then use the power rule, but that takes time. The product rule differentiates the product directly.

Quotient rule

If $f(x)=\frac{u}{v}$, then

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

A common mnemonic is “low d high - high d low over low squared” (denominator times derivative of numerator minus numerator times derivative of denominator, all over the denominator squared).

Exam Focus

Typical question patterns:

• “Differentiate a product like $(2x+7)(9x+8)$ without expanding first.”
• “Differentiate a rational function using the quotient rule and simplify.”

Common mistakes:

• Forgetting the subtraction order in the numerator of the quotient rule.
• Squaring the wrong part (only the denominator becomes $v^2$).
• Trying to use linearity rules on products/quotients.

Derivatives of Trigonometric, Exponential, and Logarithmic Functions (Memory Derivatives)

Some derivatives are usually memorized because they’re easier to recall than to re-derive during a test. In this unit, trig derivatives are a major milestone, and it’s also common to know the basic exponential and logarithmic ones.

Trigonometric derivatives (radians required)

The standard results

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

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

are true when $x$ is in radians. On the AP exam, unless explicitly stated otherwise, trig function inputs are in radians.

Interpretation:

• The rate of change of $\sin(x)$ at an angle equals $\cos(x)$.
• The rate of change of $\cos(x)$ at an angle equals $-\sin(x)$.

Exponential and logarithmic “memory derivatives”

These are also commonly memorized:

$\frac{d}{dx}[e^x]=e^x$

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

For $\ln(x)$, the real-valued domain is typically $x>0$.

Worked example 1: basic trig differentiation

Differentiate $f(x)=4\sin(x)-3\cos(x)$.

$f'(x)=4\cos(x)-3(-\sin(x))$

So

$f'(x)=4\cos(x)+3\sin(x)$

Worked example 2: tangent line to a trig function

Find the equation of the tangent line to $f(x)=\sin(x)$ at $x=0$.

$f'(x)=\cos(x)$

So

$f'(0)=\cos(0)=1$

And

$f(0)=\sin(0)=0$

Tangent line:

$y-0=1(x-0)$

So

$y=x$

This is a famous local fact: near $0$ (in radians), $\sin(x)$ behaves very much like $x$.

What goes wrong: common trig derivative errors

• Forgetting the negative sign: $\frac{d}{dx}[\cos(x)]$ is not $\sin(x)$.
• Differentiating $\sin$ into $\sin$ (confusing with integrals later).
• Assuming the formulas hold in degrees without adjustment.

Exam Focus

Typical question patterns:

• “Differentiate expressions involving $\sin(x)$ and $\cos(x)$ using linearity rules.”
• “Evaluate $f'(a)$ for trig functions at common angles like $0$, $\frac{\pi}{2}$, $\pi$.”
• “Use basic memory derivatives like $\frac{d}{dx}[e^x]$ and $\frac{d}{dx}[\ln(x)]$ when they appear.”

Common mistakes:

• Missing the negative in $-\sin(x)$.
• Using degree-mode thinking for special values.
• Mixing up which function stays the same under differentiation (for $e^x$).

Connecting Multiple Representations: Graphs, Behavior, and Meaning

Unit 2 isn’t just about taking derivatives; it’s about understanding what derivatives mean and how they appear in equations, tables, graphs, and words.

From the graph of $f$ to information about $f'$

Even before you learn how to graph derivatives formally, you can reason about them:

• If $f$ is increasing on an interval, then tangent slopes are positive, so $f'(x)>0$ there.
• If $f$ is decreasing, then $f'(x)<0$.
• If $f$ has a horizontal tangent at $x=a$, then $f'(a)=0$.

From the sign of $f'$ to behavior of $f$

This same relationship works in reverse: the derivative tells you where the original function rises or falls.

Velocity as a derivative (a core application)

If $s(t)$ is position along a line, then instantaneous velocity is

$v(t)=s'(t)$

Velocity gives both speed (magnitude) and direction (sign). If $v(t)>0$, position increases; if $v(t)<0$, position decreases.

Worked example 1: interpreting a derivative graphically

Suppose a function $f$ rises steeply near $x=1$, flattens around $x=2$, and then decreases after $x=3$.

Then:

• $f'(1)$ is positive and relatively large.
• $f'(2)$ is near $0$.
• $f'(x)$ becomes negative after $x=3$.

Worked example 2: motion interpretation

Let $s(t)=t^3-6t^2$ represent position (meters). Find the velocity at $t=2$.

$s'(t)=3t^2-12t$

$s'(2)=3(4)-24=-12$

Interpretation: at $t=2$ seconds, the velocity is $-12$ meters per second (moving in the negative direction). A negative velocity does not automatically mean “slowing down”; it means direction. The speed here is $12$.

Exam Focus

Typical question patterns:

• “Given a graph of $f$, identify intervals where $f'(x)>0$, $f'(x)<0$, or $f'(x)=0$.”
• “Interpret the meaning of $f'(a)$ in a real-world context (especially motion).”
• “Compare derivative values using graphical steepness or tabular differences.”

Common mistakes:

• Confusing the value of the function with the slope (a high $y$-value does not imply a high derivative).
• Saying “negative derivative means negative output.” It means decreasing.
• Interpreting negative velocity as “deceleration” rather than direction.

Using the Definition Strategically (Algebra Tools You Must Control)

Even though most derivatives will be computed using rules, you must be fluent with the limit definition and the algebra that makes it work. The AP exam will sometimes force the definition for unfamiliar forms, proofs, or conceptual checks.

The main obstacle: indeterminate forms

When you plug $h=0$ into

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

you often get

$\frac{0}{0}$

That does not mean the limit is zero; it means you must simplify first.

Two recurring algebra strategies

1) Factor and cancel (common with polynomials)

2) Multiply by the conjugate (common with radicals)

These are methods for rewriting the expression so the factor causing $\frac{0}{0}$ cancels.

Worked example 1: derivative from definition with factoring

Find $f'(x)$ for $f(x)=x^2+3x$ using the limit definition.

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

Compute:

$f(x+h)=(x+h)^2+3(x+h)=x^2+2xh+h^2+3x+3h$

Subtract $f(x)=x^2+3x$:

$f(x+h)-f(x)=2xh+h^2+3h$

Divide by $h$:

$\frac{2xh+h^2+3h}{h}=2x+h+3$

Take the limit:

$f'(x)=\lim_{h\to 0}(2x+h+3)=2x+3$

Worked example 2: derivative from definition with a radical

Find $f'(x)$ for $f(x)=\sqrt{x}$.

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

Multiply by the conjugate:

$\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}$

Simplify:

$\frac{h}{h(\sqrt{x+h}+\sqrt{x})}=\frac{1}{\sqrt{x+h}+\sqrt{x}}$

Now take the limit:

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

What goes wrong: “canceling” incorrectly

You can only cancel common factors, not terms. For example, from

$\frac{h^2+h}{h}$

you can factor and cancel:

$\frac{h(h+1)}{h}=h+1$

But from

$\frac{h^2+1}{h}$

you cannot cancel the $h$ because it is not a factor of the entire numerator.

Exam Focus

Typical question patterns:

• “Use the definition to find $f'(x)$ for a function with a radical or polynomial.”
• “Show work using algebraic simplification before evaluating a limit.”
• “Evaluate a derivative at a point using the definition (often with a specific numeric $a$).”

Common mistakes:

• Evaluating the limit before removing the indeterminate form.
• Conjugate sign mistakes.
• Canceling terms instead of factors.

Practice-Style AP Free-Response Skills Within Unit 2

Many Unit 2 free-response questions blend skills: definition of derivative, tangent lines, differentiability, and interpretation. The goal is usually not long algebra; it’s showing you understand what the derivative means and how to use it.

Skill 1: Explaining reasoning clearly

AP scoring rewards correct statements with clear linkage. For example, if asked whether $f$ is differentiable at $x=a$ based on a graph, a strong justification is:

“Not differentiable because the graph has a corner at $x=a$, so the left-hand and right-hand tangent slopes are not equal.”

Skill 2: Building a tangent line model

If you’re given $f(a)$ and $f'(a)$, you can always write the tangent line:

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

This shows up constantly, sometimes in disguise as “local linear behavior.”

Skill 3: Knowing when you must use the definition

If you only know values from a table, you can’t differentiate by rules; you estimate with difference quotients. If you are explicitly told “use the definition,” then rules (even if you know them) are not the point.

Worked example: mixed skills (table + tangent line)

You are told $f(2)=5$ and that near $x=2$ the table gives $f(2.1)=5.4$ and $f(1.9)=4.7$.

1) Estimate $f'(2)$ using a symmetric difference quotient:

$f'(2)\approx\frac{f(2.1)-f(1.9)}{2(0.1)}=\frac{5.4-4.7}{0.2}=3.5$

2) Write the tangent line at $x=2$ using your estimate:

$y-5=3.5(x-2)$

Exam Focus

Typical question patterns:

• “Justify differentiability (or lack of it) using a graph and correct vocabulary.”
• “Estimate $f'(a)$ from tabular data and use it to write a tangent line.”
• “Interpret derivatives in context and include correct units.”

Common mistakes:

• Writing a tangent line equation with slope $f(a)$ instead of $f'(a)$.
• Estimating $f'(a)$ using points that are not close to $a$ (poor local estimate).
• Giving a conclusion (like “not differentiable”) without a reason tied to corners/cusps/vertical tangents/discontinuities.