Unit 2: Differentiation: Definition and Fundamental Properties

From Average Rate of Change to the Derivative

What “rate of change” really means

A major goal of calculus is to describe how quantities change. Before calculus, you probably worked with average rate of change: how much an output changes per unit change in input over an interval. If a function $f(x)$ represents position (in meters) at time $x$ (in seconds), then the average velocity from $x=a$ to $x=b$ is

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

This same expression is also the slope of the secant line through the points $\big(a,f(a)\big)$ and $\big(b,f(b)\big)$ on the graph. In coordinate form, the same idea is often written as “rise over run”:

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

This works perfectly for a straight (linear) line, but for a curved graph, the slope changes from point to point, so you approximate using a secant line over a small interval.

Why this matters: in real life, many questions are about an instant (speed at exactly 3 seconds, marginal cost at exactly 100 units, slope at a point on a curve). Average rate of change can’t answer “at an instant” by itself because it needs an interval.

The key idea: zooming in to get instantaneous change

To get an instantaneous rate of change at $x=a$, you shrink the interval. Let the second point be $a+h$, where $h$ is a small number (positive or negative). The secant slope becomes the difference quotient

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

The closer the two points are, the more accurate the secant slope is as an approximation of the “true” slope at $x=a$. As $h$ gets closer to 0, the secant line “turns into” the tangent line—provided the graph behaves nicely at that point.

Slopes on curves: secant lines vs tangent lines

For a linear function, slope is constant, so “rise over run” gives the slope everywhere. For a non-linear curve, there isn’t one constant slope, so you approximate the slope at a point by drawing a secant line through two nearby points. The tangent line is the limiting position of these secant lines as the points merge.

Definition of the derivative at a point (instantaneous rate of change)

The derivative of $f$ at $x=a$ is the limit of the difference quotient (if the limit exists):

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

You will also see an equivalent form that approaches $a$ directly:

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

These are the same idea: measure the slope of a secant line and then take the limit as the two points merge. This is the formal version of instantaneous rate of change.

Why this matters: this limit definition is the foundation that justifies all differentiation rules. Even when you later use shortcuts like the power rule, the “truth underneath” is this limit.

Worked example: using the definition to find a derivative value

Find $f'(2)$ for $f(x)=x^2$.

Start with the definition:

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

Compute each piece:

$f(2+h)=(2+h)^2=4+4h+h^2$

$f(2)=4$

Substitute:

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

Simplify the numerator:

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

Factor out $h$ and cancel (this is the critical algebra step):

$f'(2)=\lim_{h\to 0}(4+h)$

Now take the limit:

$f'(2)=4$

Interpretation: the tangent line slope to $y=x^2$ at $x=2$ is 4.

What can go wrong

A very common error is trying to plug in $h=0$ too early. In the definition,

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

is undefined at $h=0$. You must simplify first (usually by factoring and canceling), and only then take the limit.

Exam Focus

• Typical question patterns:
  • “Use the definition of the derivative to find $f'(a)$ for a given function.”
  • “Write the expression for the derivative at $x=a$ as a limit.”
  • “Interpret the derivative as the slope of a tangent line or instantaneous rate of change.”
• Common mistakes:
  • Plugging in $h=0$ before algebraic simplification.
  • Cancelling incorrectly (you may cancel a factor of $h$, but never cancel terms across addition).
  • Mixing up the two equivalent forms and substituting the wrong point (for example, using $f(x)-f(h)$ instead of $f(a+h)-f(a)$).

The Derivative as a Function (Not Just a Number)

From “the slope at one point” to “a new function that gives slopes”

When you compute $f'(2)$, you get one number. But you can repeat the process at every input value and produce a brand-new function, the **derivative function** $f'(x)$, which gives the slope of the tangent line at each $x$.

Using the limit definition at a general $x$:

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

Why this matters: thinking of $f'(x)$ as a function lets you analyze how slope changes across an interval, which leads directly to graph analysis and optimization later.

Worked example: deriving $f'(x)$ from first principles

Let $f(x)=x^2$ again, but now find $f'(x)$.

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

Expand:

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

Simplify:

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

Factor and cancel:

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

Take the limit:

$f'(x)=2x$

Now you have a rule: at any input $x$, the tangent slope is $2x$.

Notation you must be comfortable with

AP Calculus uses multiple equivalent notations. They look different, but they represent the same derivative concept.

You should also recognize second derivative notation (derivative of the derivative):

A key conceptual point: $\frac{dy}{dx}$ is read “derivative of $y$ with respect to $x$.” On AP problems, this notation often appears in contexts involving units or rates.

Connecting derivative function to a graph

If $f'(x) > 0$ on an interval, then $f(x)$ is increasing there. If $f'(x) < 0$ on an interval, then $f(x)$ is decreasing there. If $f'(x) = 0$ at a point, the tangent is horizontal (but that alone does not guarantee a max or min).

Exam Focus

• Typical question patterns:
  • “Find $f'(x)$ using the limit definition.”
  • “Given a graph of $f$, sketch a graph of $f'$ (conceptually).”
  • “Evaluate $f'(a)$ from a table or from a derivative expression.”
• Common mistakes:
  • Treating $f'(x)$ as a single number rather than a function.
  • Confusing $f'(a)$ with $f(a)$.
  • Forgetting that a horizontal tangent (derivative zero) does not automatically mean a local extremum.

Interpreting Derivatives: Slope, Velocity, and Units

The derivative is a meaning-maker

AP problems often test whether you can interpret derivatives, not just compute them. The derivative is best understood as:

• the slope of the tangent line to $y=f(x)$ at a point
• the instantaneous rate of change of $f$ with respect to $x$

These are the same idea in two languages: geometry (slope) and real-world change (rate).

Units: one of the most tested interpretation skills

If $f(x)$ has units and $x$ has units, then $f'(x)$ has units “units of $f$ per unit of $x$.”

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

Motion interpretation (a classic application)

If $s(t)$ is position, then velocity and acceleration are derivatives:

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

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

This reinforces that derivatives can be taken repeatedly, and each derivative has a distinct meaning.

Tangent line approximation (linearization idea, informal)

Near $x=a$, the graph of $f$ looks almost like its tangent line if $f$ is smooth there. The tangent line at $x=a$ is

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

You may see questions asking you to approximate values using tangent lines.

Worked example: interpreting a derivative in context

Suppose $P(t)$ is the population of a town (people) at time $t$ (years). If $P'(5)=120$, then at $t=5$ years the population is increasing at a rate of 120 people per year.

Common misunderstanding: saying “the population is 120.” That would be $P(5)=120$, not $P'(5)=120$.

Worked example: writing a tangent line

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

First compute $f(4)$:

$f(4)=2$

Compute the derivative (later you’ll justify this with rules):

$f'(x)=\frac{1}{2\sqrt{x}}$

Evaluate at 4:

$f'(4)=\frac{1}{2\cdot 2}=\frac{1}{4}$

Use point-slope form:

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

Simplify if desired:

$y=\frac{1}{4}x+1$

Interpretation: near $x=4$, $\sqrt{x}$ is close to $\frac{1}{4}x+1$.

Exam Focus

• Typical question patterns:
  • “Interpret $f'(a)$ in context, including units.”
  • “Find the equation of the tangent line to $f$ at $x=a$.”
  • “Use a tangent line to approximate a nearby function value.”
• Common mistakes:
  • Confusing function value with derivative value (quantity vs rate).
  • Dropping or mis-stating units.
  • Using the wrong point in the tangent line formula (using $f'(a)$ as a point, or forgetting $\big(a,f(a)\big)$).

When Does a Derivative Exist? Differentiability and Continuity

Differentiability is a stronger condition than continuity

A function is differentiable at $x=a$ if the derivative $f'(a)$ exists (the limit defining it exists and is finite).

A function is continuous at $x=a$ if

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

Key relationship:

• If $f$ is differentiable at $a$, then $f$ is continuous at $a$.
• The converse is not always true: a function can be continuous but not differentiable.

Geometric situations where differentiability fails

From the tangent-line viewpoint, differentiability can fail when there is no single well-defined tangent slope.

1. Corner: left-hand slope and right-hand slope are finite but different.
2. Cusp: slopes approach infinity but in opposite directions.
3. Vertical tangent: slope becomes infinite (derivative does not exist as a finite number).
4. Discontinuity: if the function jumps, has a hole, or blows up, it cannot be differentiable there.

A crisp way to detect non-differentiability at $x=a$ is to compare one-sided limits of the difference quotient:

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

and

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

If these one-sided limits are not equal (or not finite), then $f'(a)$ does not exist.

Worked example: a corner with an absolute value

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

Compute the difference quotient:

$\frac{|h|-|0|}{h}=\frac{|h|}{h}$

If $h>0$, then $|h|=h$, so the quotient is 1. If $h<0$, then $|h|=-h$, so the quotient is $-1$. The left-hand limit is $-1$ and the right-hand limit is 1, so there is no single limit as $h\to 0$. Therefore $f'(0)$ does not exist.

But $|x|$ is continuous at 0, which is the classic example showing continuity does not guarantee differentiability.

Piecewise functions: continuity and differentiability at the join

When a function is defined piecewise, AP questions often ask you to find values of constants that make the function continuous or differentiable at a boundary point.

For continuity at $x=a$, match function values:

$\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$

For differentiability at $x=a$, also match derivatives (and continuity is required as well):

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

Exam Focus

• Typical question patterns:
  • “Is $f$ differentiable at $x=a$? Justify using continuity/corners/one-sided behavior.”
  • “Find constants that make a piecewise function continuous and differentiable at a point.”
  • “Given a graph, identify where $f'$ does not exist.”
• Common mistakes:
  • Claiming a function is differentiable just because it is continuous.
  • Forgetting to check both one-sided derivatives at a corner.
  • Saying “vertical tangent means derivative is 0” (it actually means the slope is infinite, so the derivative does not exist as a finite number).

Linear Properties of Derivatives and the Power Rule

Why rules exist: derivatives are limits with structure

The limit definition is the foundation, but calculating every derivative from scratch would be slow. Derivatives obey predictable algebraic rules because limits interact nicely with addition and constant multiples.

Constant rule and constant multiple rule

If $k$ is a constant and $f$ is differentiable, then:

$\frac{d}{dx}(k)=0$

$\frac{d}{dx}(kf(x))=kf'(x)$

Example (constant rule): if $f(x)=10$ then

$f'(x)=0$

Intuition: constants do not change, so their rate of change is 0. Multiplying by $k$ scales all outputs, so it scales the rate of change by $k$.

Sum and difference rule

If $f$ and $g$ are differentiable, then:

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

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

The power rule (for integer powers)

For powers of $x$, the most-used rule is the power rule:

$\frac{d}{dx}(x^n)=nx^{n-1}$

for integers $n$ (positive, zero, or negative), on domains where the expression is defined.

A common way to describe the power rule is: “multiply down and decrease the power.” For example, $x^4$ becomes $4x^3$, and $2x^2$ becomes $4x$.

Why this matters: combined with the sum rule, this gives derivatives of all polynomials quickly.

Worked example: polynomial differentiation

Let

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

Differentiate term-by-term:

$f'(x)=15x^4-14x+4$

Notice the constant term $-9$ disappears because its derivative is 0.

Negative powers and rational expressions

If

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

then

$f'(x)=-3x^{-4}$

You can rewrite to avoid negative exponents:

$f'(x)=-\frac{3}{x^4}$

A common mistake is to change the sign incorrectly or to reduce the exponent the wrong way (the exponent always goes down by 1).

A quick note on extending the power rule

You will eventually apply the power rule to fractional exponents as well, using the idea that roots are rational powers:

$\sqrt{x}=x^{1/2}$

Then

$\frac{d}{dx}(x^{1/2})=\frac{1}{2}x^{-1/2}$

This is consistent with the idea that derivatives of smooth power functions follow the same pattern.

Exam Focus

• Typical question patterns:
  • “Differentiate a polynomial or a sum of power functions.”
  • “Find $f'(a)$ for a given function expression.”
  • “Compute the derivative and interpret it (slope or rate) at a point.”
• Common mistakes:
  • Forgetting to multiply by the exponent (writing $x^{n-1}$ instead of $nx^{n-1}$).
  • Incorrectly claiming $\frac{d}{dx}(c)=c$ rather than 0.
  • Distributing the power rule over sums incorrectly (for example, thinking $\frac{d}{dx}((x+1)^2)=2(x+1)$ without the chain rule justification).

Product and Quotient Rules (When Functions Multiply or Divide)

Why “differentiate each part” is not enough

When you have a sum, you can differentiate term-by-term. But multiplication behaves differently: the rate of change of a product depends on how both factors change, not just one.

A common misconception is:

• Incorrect idea: $\frac{d}{dx}(f(x)g(x))=f'(x)g'(x)$

This is not true in general.

The product rule

If $u$ and $v$ are differentiable functions of $x$, then

$\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$

Many students remember this as “1d2 + 2d1”: first times derivative of second, plus second times derivative of first.

Worked example: using the product rule

Differentiate

$h(x)=(x^2+1)(3x-5)$

Identify parts:

$f(x)=x^2+1$

$g(x)=3x-5$

Derivatives:

$f'(x)=2x$

$g'(x)=3$

Apply product rule:

$h'(x)=(2x)(3x-5)+(x^2+1)(3)$

You may expand if desired:

$h'(x)=6x^2-10x+3x^2+3$

$h'(x)=9x^2-10x+3$

The quotient rule

If $u$ and $v$ are differentiable and $v\neq 0$, then

$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$

The order in the numerator matters. A helpful memory aid is “low d-high minus high d-low, over low squared.”

Worked example: using the quotient rule

Differentiate

$q(x)=\frac{x^2+1}{x-2}$

Let

$f(x)=x^2+1$

$g(x)=x-2$

Then

$f'(x)=2x$

$g'(x)=1$

Apply the quotient rule:

$q'(x)=\frac{(2x)(x-2)-(x^2+1)(1)}{(x-2)^2}$

Simplify the numerator:

$q'(x)=\frac{2x^2-4x-x^2-1}{(x-2)^2}$

$q'(x)=\frac{x^2-4x-1}{(x-2)^2}$

Choosing a method: simplify first or use a rule?

Sometimes it’s easier to expand before differentiating; sometimes not.

• For products of polynomials, expanding first can be quicker and less error-prone.
• For complicated products (involving trig, exponentials, radicals), the product rule is usually better.

For quotients, you can sometimes rewrite as a product using negative exponents, which may let you use the product rule and power rule instead.

Exam Focus

• Typical question patterns:
  • “Differentiate a function that is a product or quotient of two expressions.”
  • “Find and evaluate $f'(a)$ where $f$ involves multiplication/division.”
  • “Show setup clearly: identify $f$ and $g$ and apply the rule correctly.”
• Common mistakes:
  • Using $f'(x)g'(x)$ for a product.
  • Reversing terms in the quotient rule numerator (sign errors are very common).
  • Forgetting to square the denominator in the quotient rule.

Derivatives of Exponential Functions

Exponentials as “self-replicating change”

Exponential functions model processes where the rate of change is proportional to the current amount (population growth, continuously compounded interest, radioactive decay). In calculus, exponentials are powerful because their derivatives have especially clean forms.

The natural exponential function

The function $e^x$ is special because its rate of change matches its value:

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

This is one of the classic “memory derivatives” that is usually faster to memorize than to re-derive during an exam.

General exponential functions

For $a>0$ and $a\neq 1$,

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

Why this matters: it connects all exponentials to the natural log $\ln(a)$.

Worked example: differentiating an exponential expression

Differentiate

$f(x)=5e^x-2\cdot 3^x$

Use linearity and the exponential rules:

$f'(x)=5e^x-2\cdot 3^x\ln(3)$

Growth/decay interpretation

If a model is

$P(t)=P_0e^{kt}$

then

$P'(t)=kP_0e^{kt}$

So the instantaneous growth rate satisfies

$P'(t)=kP(t)$

Here $k$ has units of “per unit time.” If $k>0$ you have growth; if $k<0$ you have decay. A common misunderstanding is thinking $k$ is the amount added each time unit; it is not additive, it is multiplicative (a proportional rate).

Exam Focus

• Typical question patterns:
  • “Differentiate functions containing $e^x$ or $a^x$.”
  • “Interpret parameters in $P(t)=P_0e^{kt}$ using derivatives.”
  • “Evaluate an exponential derivative at a point (often in context).”
• Common mistakes:
  • Forgetting the $\ln(a)$ factor for $a^x$.
  • Confusing $e^x$ with $x^e$ (very different types of functions).
  • Treating exponential growth as linear (mixing up constant additive change with proportional change).

Derivatives of Logarithmic Functions

Logarithms undo exponentials

A logarithm answers: “What exponent produces this value?” The natural logarithm $\ln(x)$ is the inverse of $e^x$.

The derivative of the natural logarithm

For $x>0$,

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

This is another classic “memory derivative” that you should know automatically.

Other logarithms and change of base

For $a>0$ and $a\neq 1$,

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

In AP Calculus, you can often rewrite using change of base:

$\log_a(x)=\frac{\ln(x)}{\ln(a)}$

Then differentiate using constants and the derivative of $\ln(x)$.

Worked example: differentiating a log function

Differentiate

$g(x)=4\ln(x)-\log_2(x)$

Differentiate term-by-term:

$g'(x)=\frac{4}{x}-\frac{1}{x\ln(2)}$

You may combine terms if desired:

$g'(x)=\frac{4\ln(2)-1}{x\ln(2)}$

Logarithms and rates: interpreting $\frac{1}{x}$

The derivative

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

tells you that $\ln(x)$ changes quickly when $x$ is small and slowly when $x$ is large. This matches the idea that logarithms compress large scales.

Exam Focus

• Typical question patterns:
  • “Differentiate expressions involving $\ln(x)$ or $\log_a(x)$.”
  • “Rewrite a logarithm using change of base and differentiate.”
  • “Interpret derivative behavior for log functions (increasing, decreasing, rate).”
• Common mistakes:
  • Ignoring domain: $\ln(x)$ requires $x>0$ (unless using $\ln(|x|)$ in later contexts).
  • Differentiating $\log_a(x)$ as $\frac{1}{x}$ without the $\ln(a)$ factor.
  • Treating $\ln(x)$ as algebraic and trying to use the power rule on it.

Derivatives of Trigonometric Functions

Trig derivatives connect geometry and change

Trigonometric functions model periodic behavior (sound waves, seasonal temperatures, circular motion). Their derivatives follow patterns that reflect how sine and cosine are related.

Core trig derivative rules

These are fundamental derivatives you must know:

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

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

From these, you can also differentiate tangent:

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

These are also commonly treated as “memory derivatives” because they come up constantly.

Worked example: differentiating a trig combination

Differentiate

$f(x)=2\sin(x)-3\cos(x)+\tan(x)$

Differentiate term-by-term:

$f'(x)=2\cos(x)+3\sin(x)+\sec^2(x)$

Notice the sign change: derivative of $-3\cos(x)$ becomes $+3\sin(x)$ because $\frac{d}{dx}(\cos(x))=-\sin(x)$.

A key conceptual check: derivative values at special angles

Because $\sin$ and $\cos$ values at special angles are familiar, you can sanity-check derivatives.

For example, since $\sin(0)=0$ and near 0 the sine graph increases, you expect $\sin'(0)$ to be positive, and indeed

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

Similarly, $\cos(0)=1$ and near 0 cosine decreases, so the derivative should not be positive. In fact,

$\cos'(0)=-\sin(0)=0$

At exactly 0 the slope is 0, matching the flatness of cosine at its maximum.

Common misconception: degrees vs radians

All standard trig derivatives are true when the input is measured in radians. In AP Calculus, trig differentiation assumes radians unless explicitly stated otherwise.

Exam Focus

• Typical question patterns:
  • “Differentiate functions involving $\sin(x)$, $\cos(x)$, and $\tan(x)$.”
  • “Evaluate a trig derivative at a point (often at a special angle).”
  • “Use trig derivatives in tangent line problems.”
• Common mistakes:
  • Missing the negative sign in $\frac{d}{dx}(\cos(x))=-\sin(x)$.
  • Confusing $\sec^2(x)$ with $\sec(x)^2$ notation (they mean the same thing, but students sometimes write $\sec(x^2)$ by mistake).
  • Forgetting the radians requirement and making inconsistent interpretations.

Putting It Together: Derivative Skills You’re Expected to Use Fluently

How AP expects you to combine ideas

By the end of this unit, you’re expected to move between three modes smoothly:

1. Limit definition mode: set up and sometimes compute derivatives from first principles.
2. Rule mode: use derivative rules accurately and efficiently (constant, constant multiple, sum/difference, power, product, quotient, and key “memory derivatives” like $\sin(x)$, $\cos(x)$, $e^x$, and $\ln(x)$).
3. Interpretation mode: explain what a derivative means in context, including units.

Many exam questions are designed to test not just whether you can compute, but whether you know when a derivative exists and what it means.

Worked mixed example: derivative at a point, tangent line, and interpretation

Let

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

1) Find $f'(x)$.

Use the power rule:

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

2) Find the slope of the tangent at $x=-1$.

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

3) Find the tangent line at $x=-1$.

Compute the point:

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

Point-slope form with slope 1 through $(-1,1)$:

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

Simplify:

$y=x+2$

4) Interpret: at $x=-1$, the function is increasing at a rate of 1 output unit per input unit, and the best local linear approximation near $x=-1$ is $y=x+2$.

What goes wrong when students “pattern match” too fast

Errors often come from trying to apply a memorized rule to an expression without understanding its structure.

• If you see sums of terms, linearity is safe.
• If you see products or quotients, decide whether to simplify first or apply the product/quotient rule.
• If you are asked explicitly for a derivative “using the definition,” you must show the difference quotient limit setup, not just the final derivative rule.

Exam Focus

• Typical question patterns:
  • “Find $f'(x)$ and use it to compute a tangent line equation.”
  • “A free response problem that mixes computation and interpretation (including units).”
  • “Determine where a function is differentiable based on a graph or piecewise definition.”
• Common mistakes:
  • Ignoring the instruction “use the definition” and applying shortcuts without justification.
  • Algebra errors when simplifying difference quotients (especially factoring and canceling).
  • Writing a tangent line with the wrong slope (using $f(a)$ instead of $f'(a)$) or wrong point (forgetting $\big(a,f(a)\big)$).