Unit 2: Differentiation: Definition and Fundamental Properties

Interpreting the Derivative: Rates of Change, Secant Slopes, and Tangent Slopes

When you first meet the derivative, it helps to think of it as answering one powerful question: how fast is something changing right now? In Algebra, you often focused on an average rate of change across an interval, but Calculus builds a precise way to talk about an instantaneous rate of change at a single input value.

Average rate of change (difference quotient)

For a function $f(x)$, the average rate of change from $x=a$ to $x=b$ is the slope of the secant line through $\big(a,f(a)\big)$ and $\big(b,f(b)\big)$:

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

You may also see this described informally as:

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

This is “change in output per change in input” over an interval, not at a single instant.

Instantaneous rate of change (the derivative)

The other way of describing rate of change is at a specific point (like a speedometer reading at one moment). For a curved graph, you can’t just do “rise over run” using two points on the curve to get the slope at exactly one point. Instead, you approximate with secant lines and then take a limit.

From secant lines to a tangent line

Pick a point at $x=a$ and a nearby point at $x=a+h$. The slope of the secant line is

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

which simplifies to

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

As the two points get closer together (as $h$ approaches 0), the secant slope becomes more accurate, and in the limit it becomes the slope of the tangent line—when that tangent line exists.

A common informal description is that the tangent line “touches” the curve at one point, but the deeper idea is local agreement in direction (slope). A tangent line can intersect the curve again elsewhere; what matters is that it matches the curve’s slope at the point of tangency.

Why this matters (applications)

This interpretation is why derivatives show up everywhere:

• In physics, the derivative becomes velocity and acceleration.
• In economics, it models marginal cost and marginal revenue.
• In biology, it can represent population growth rates.
• In geometry, it gives the slope of a curve at a point.

Example (conceptual): average vs instantaneous

Suppose $s(t)$ measures position (meters) of a moving object at time $t$ (seconds).

Average velocity from $t=2$ to $t=5$ is

$\frac{s(5)-s(2)}{5-2}$

Instantaneous velocity at $t=2$ (if it exists) is the derivative $s'(2)$, defined via a limit of average velocities over shrinking intervals.

Exam Focus

• Typical question patterns:
  • Interpret $f'(a)$ as slope of the tangent line or instantaneous rate of change in context (with units).
  • Compare average rate of change on an interval to instantaneous rate at an endpoint.
  • Relate “increasing/decreasing” language to the sign of the derivative.
• Common mistakes:
  • Treating average rate of change and derivative as the same thing.
  • Forgetting units: if $f$ is in dollars and $x$ is in years, then $f'$ is dollars per year.
  • Thinking the tangent line must “touch only once” (tangent means matching slope locally, not “touching once”).

The Limit Definition of the Derivative

The derivative is defined using a limit because “instantaneous” change cannot be captured by a finite difference. You approximate the rate of change with secant slopes, then take the limit as the interval shrinks to zero.

Two equivalent limit definitions

At a point $x=a$, the derivative of $f$ at $a$ can be written as

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

An equivalent form is

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

Both express the same idea: a nearby point approaches $a$.

What it means for the derivative to exist

The derivative $f'(a)$ exists if the defining limit exists and is finite. Intuitively, the derivative exists if, when you zoom in near $x=a$, the graph looks more and more like a straight line. Differentiability fails at corners, cusps, vertical tangents (infinite slope), and at any discontinuity.

One-sided derivatives

Sometimes you approach from one side:

Right-hand derivative at $a$:

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

Left-hand derivative at $a$:

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

For $f'(a)$ to exist, both one-sided derivatives must exist and be equal.

Worked example 1: derivative from the definition

Find the derivative of $f(x)=x^2$ using the limit definition.

Start with

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

Compute

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

Substitute:

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

Expand the numerator:

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

So

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

Factor out and cancel $h$:

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

Now take the limit:

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

Worked example 2: a derivative at a specific point

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

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

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

Multiply by the conjugate:

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

Then

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

Simplify:

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

Cancel $h$ (valid for $h\neq 0$, which is fine inside a limit process):

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

Substitute $h=0$:

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

Exam Focus

• Typical question patterns:
  • Compute $f'(a)$ using the limit definition (often requiring algebra like factoring or conjugates).
  • Identify which limit expression matches the derivative at a point.
  • Use left and right definitions to test differentiability at a “sharp” point.
• Common mistakes:
  • Plugging in $h=0$ before simplifying, then stopping at $0/0$.
  • Algebra slips: expanding $f(x+h)$ incorrectly.
  • Canceling terms incorrectly (you can cancel a factor of $h$, not a term that is added).

Derivative Notation and What the Symbols Mean

Calculus uses several notations because different contexts emphasize different ideas: sometimes you want to highlight a function name, sometimes a dependent variable, and sometimes the “operator” viewpoint.

Notation reference

A key conceptual point is that $f'(x)$ is a new function (it takes inputs and produces slopes/rates), while $f'(a)$ is a single number (the slope/rate at one specific input).

Average rate vs derivative language

Average rate of change on $[a,b]$:

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

Instantaneous rate of change at $a$:

$f'(a)$

On AP questions, the wording often shifts between “rate of change,” “slope,” “velocity,” “marginal,” or “derivative.” These are often pointing to the same underlying idea.

Interpreting the units

Units are one of the best ways to check your reasoning.

• If $x$ is hours and $f(x)$ is miles, then $f'(x)$ is miles per hour.
• If $t$ is seconds and $s(t)$ is meters, then $s'(t)$ is meters per second.

Example: interpreting notation in context

Let $C(q)$ be the cost in dollars to produce $q$ items.

• $C'(50)$ is the instantaneous rate at which cost changes when producing 50 items, with units dollars per item.
• If $C'(50)=3.2$, then near 50 items, producing one additional item increases cost by about $3.20$.

Exam Focus

• Typical question patterns:
  • Translate between $f'(a)$, $\frac{dy}{dx}$, and $\frac{d}{dx}f(x)$.
  • Interpret $f'(a)$ in a word problem (including units).
  • Distinguish between $f'(x)$ (function) and $f'(a)$ (value).
• Common mistakes:
  • Writing $f'(x)$ when the question asks for $f'(3)$ (leaving the answer as a function instead of a number).
  • Dropping units or mixing them up.
  • Confusing $\frac{dy}{dx}$ with $\frac{\Delta y}{\Delta x}$ (instantaneous vs average).

Estimating Derivatives from Graphs and Tables

Many AP problems give you data rather than a formula. In that setting, you estimate the derivative by estimating a tangent slope (from a graph) or a limiting secant slope (from a table).

Estimating from a graph

If you’re asked to estimate $f'(a)$ from the graph of $f$, you are being asked for the slope of the tangent line at $x=a$.

A reliable strategy is:

1. Locate the point $\big(a,f(a)\big)$.
2. Draw the tangent line that matches the curve’s local direction at that point.
3. Pick two convenient points on your drawn tangent line (they do not need to be points on the curve) and compute slope.

Slope is always

$\frac{\text{rise}}{\text{run}}$

Because this is an estimate, your answer is often written as “approximately” and may be a decimal.

Estimating from a table

If you have values at $a$ and $a+h$, a one-sided secant estimate is

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

If you have values on both sides, a common and often more accurate estimate is the symmetric difference quotient:

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

Deciding whether the derivative is positive, negative, or zero

Even without an exact slope value, you can often determine the sign.

• If $f$ is increasing at $a$, then $f'(a)$ is positive.
• If $f$ is decreasing at $a$, then $f'(a)$ is negative.
• If the tangent is horizontal, then $f'(a)=0$.

A horizontal tangent gives a critical point candidate, not automatically a maximum or minimum.

Worked example: table-based estimate

A table gives values of $f$ near $x=2$:

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

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

Substitute values:

$f'(2)\approx \frac{6.08-5.34}{0.2}$

Compute:

$f'(2)\approx \frac{0.74}{0.2}=3.7$

Interpretation: near $x=2$, the function increases by about 3.7 output units per 1 input unit.

Worked example: graph-based sign reasoning

If a graph shows $f$ decreasing steeply at $x=1$, then $f'(1)$ is negative with relatively large magnitude. If the curve is almost flat there, $f'(1)$ is negative but closer to zero.

Exam Focus

• Typical question patterns:
  • Estimate $f'(a)$ from a table using secant slopes (often symmetric).
  • Estimate $f'(a)$ from a graph by drawing a tangent and computing its slope.
  • Determine where $f'(x)$ is positive/negative/zero from the shape of $f$.
• Common mistakes:
  • Using points on the curve instead of points on the tangent line when estimating from a graph.
  • Mixing up $f(a)$ with $f'(a)$ (reporting the y-value instead of the slope).
  • Forgetting that the slope is “rise over run” and accidentally inverting it.

Differentiability and Continuity

It’s tempting to assume that if a graph is continuous (no breaks), then it must have a derivative. Calculus separates these ideas.

• Continuity is about whether the graph has holes/jumps at a point.
• Differentiability is about whether the graph has a well-defined (finite) tangent slope at a point.

Key relationship

If $f$ is differentiable at $x=a$, then $f$ is continuous at $x=a$.

The converse is not always true: a function can be continuous but not differentiable.

Where differentiability fails (even if continuous)

A function may be continuous at $a$ but not differentiable there if it has:

1. A corner (left-hand and right-hand slopes are finite but different).
2. A cusp (slopes become infinite in opposite directions).
3. A vertical tangent (slope becomes infinite).

Differentiability also fails at discontinuities (jumps, holes, vertical asymptotes), since differentiability implies continuity.

How to reason using one-sided derivatives (especially for piecewise functions)

A useful workflow at a point where a formula changes is:

1. Check continuity at the point.
2. Compute the left-hand derivative and right-hand derivative.
3. If both exist and are equal, the derivative exists.

Skipping step 1 can waste time: if the function is discontinuous, it is automatically not differentiable.

Worked example: corner in an absolute value function

Consider $f(x)=|x|$ at $x=0$. The graph has a sharp corner at the origin.

For $x>0$, $|x|=x$ so the slope is 1.

For $x<0$, $|x|=-x$ so the slope is -1.

Because the one-sided slopes do not match, $f'(0)$ does not exist, even though $f$ is continuous at 0.

Worked example: discontinuity implies not differentiable

If $f$ has a jump at $x=3$, then even if $f(3)$ is defined, the limit

$\lim_{x\to 3}\frac{f(x)-f(3)}{x-3}$

cannot behave well because $f(x)$ does not approach $f(3)$ from both sides. So there is no derivative at $x=3$.

Exam Focus

• Typical question patterns:
  • Decide whether $f$ is differentiable at a point from a graph (identify corners/cusps/vertical tangents).
  • For piecewise functions, find values of constants that make $f$ differentiable at a junction.
  • Explain why differentiable implies continuous (often in words).
• Common mistakes:
  • Claiming “continuous implies differentiable.”
  • Confusing a vertical tangent with a corner (both fail differentiability, but for different reasons).
  • Checking only that left and right function values match (continuity) but not checking derivative match.

Fundamental Derivative Rules (Constant, Power, and Linearity)

Using the limit definition directly is accurate but often tedious, so calculus develops rules that let you compute derivatives efficiently while still being grounded in the limit definition.

Constant rule

If $f(x)=k$ where $k$ is a constant, then

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

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

This matches the limit idea because a constant function has no change in output.

Constant multiple rule

If a constant multiplies a function, you can “pull the constant out”:

$\frac{d}{dx}\big(c f(x)\big)=c f'(x)$

Intuitively, multiplying outputs by $c$ multiplies vertical change by $c$, so slopes scale by $c$.

Sum and difference rules (linearity)

For differentiable functions $f$ and $g$:

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

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

These linearity rules let you differentiate term-by-term for expressions built from addition/subtraction and constant multiples.

Power rule (built from first principles)

For integer powers,

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

A helpful description is: “multiply down and decrease the power.”

Examples:

• $x^4$ becomes $4x^3$
• $2x^2$ becomes $4x$

Why it makes sense: when you compute

$\lim_{h\to 0}\frac{(x+h)^n-x^n}{h}$

the algebra produces a factor of $h$ that cancels, and what remains approaches $nx^{n-1}$.

Worked example: differentiating a polynomial

Differentiate

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

Apply linearity and the power rule 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

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

Worked example: negative powers

Differentiate

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

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

Equivalently,

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

Worked example: combining linearity with power rule

Differentiate

$h(x)=5\sqrt{x}-\frac{3}{x^2}+4$

Rewrite using exponents:

$h(x)=5x^{1/2}-3x^{-2}+4$

Differentiate term-by-term:

$h'(x)=\frac{5}{2}x^{-1/2}+6x^{-3}$

Rewrite with positive exponents:

$h'(x)=\frac{5}{2\sqrt{x}}+\frac{6}{x^3}$

Common misconception: distributing the derivative incorrectly

Linearity works for addition/subtraction and constant multiples, but not for products or quotients. In general, it is not true that

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

This is exactly why the product rule and quotient rule are necessary.

Exam Focus

• Typical question patterns:
  • Differentiate polynomials and simple power functions quickly and accurately.
  • Use the power rule with negative and fractional exponents.
  • Recognize when a function is constant (derivative zero).
  • Rewrite radicals and fractions using exponents, then differentiate.
  • Use derivative rules to find slopes of tangent lines (compute $f'(a)$).
• Common mistakes:
  • Forgetting to multiply by the coefficient when applying the power rule.
  • Mishandling negative exponents when rewriting answers (for example, incorrectly writing $-3x^4$).
  • Treating $\frac{d}{dx}(x^n)$ as $x^{n-1}$ (forgetting the factor $n$).
  • Differentiating constants as if they change (writing $\frac{d}{dx}(4)=4$).
  • Applying linearity to products or quotients.

The Product and Quotient Rules

Many important functions are built by multiplying or dividing simpler functions. Since linearity does not cover those operations, you need special rules.

Product rule

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

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

One way to remember the structure is: differentiate the first, keep the second, plus keep the first, differentiate the second.

Another common mnemonic is:

• “1d2 + 2d1” (first times derivative of second, plus second times derivative of first)

A typical situation: if you have two polynomials multiplied, like

$(2x+7)(9x+8)$

you could expand first and then use the power rule, but the product rule is usually faster and less error-prone.

Quotient rule

If $g(x)\neq 0$ and both $f$ and $g$ are differentiable, then

$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$

A common memory aid is:

• “low d-high minus high d-low over low squared”

(denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared).

Worked example 1: product rule

Differentiate

$p(x)=(x^2+1)(3x^4-2x)$

Let

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

$g(x)=3x^4-2x$

Compute derivatives:

$f'(x)=2x$

$g'(x)=12x^3-2$

Apply product rule:

$p'(x)=2x(3x^4-2x)+(x^2+1)(12x^3-2)$

Worked example 2: quotient rule

Differentiate

$q(x)=\frac{x^2+1}{x-3}$

Let

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

$g(x)=x-3$

Then

$f'(x)=2x$

$g'(x)=1$

Apply quotient rule:

$q'(x)=\frac{2x(x-3)-(x^2+1)(1)}{(x-3)^2}$

Simplify numerator if desired:

$q'(x)=\frac{x^2-6x-1}{(x-3)^2}$

Common pitfalls

• Product rule is not “multiply derivatives.” You must follow $f'g+fg'$.
• Quotient rule sign errors are common: it is $f'g-fg'$, not $f'g+fg'$.
• Parentheses matter: missing them can change the entire derivative.

Exam Focus

• Typical question patterns:
  • Differentiate a product of two nontrivial functions (often polynomials).
  • Differentiate a rational function using the quotient rule.
  • Compute a tangent line slope to a product/quotient function at a point.
• Common mistakes:
  • Writing $f'(x)g'(x)$ for the derivative of a product.
  • Dropping parentheses in quotient rule, especially on the $f(x)g'(x)$ term.
  • Forgetting to square the denominator in the quotient rule.

Derivatives You Should Memorize: Trigonometric, Exponential, and Logarithmic

Some derivatives are best treated as core facts (often called “memory derivatives”). Once you know them, you combine them with the constant multiple, sum/difference, product, and quotient rules.

Core trigonometric derivative formulas

The most fundamental trig derivatives in AP Calculus AB are

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

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

From these, additional common trig derivatives follow:

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

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

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

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

Keep domain restrictions in mind, such as $\tan x$ and $\sec x$ being undefined where $\cos x=0$.

Exponential and logarithmic “memory derivatives”

Two additional derivatives that are typically memorized are

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

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

Why this matters (and a note about the chain rule)

Once you have these base derivatives, you can differentiate many mixed expressions using the rules already developed. In Unit 2, the main focus is often on functions like $\sin x$ rather than composites like $\sin(3x)$, since the chain rule is usually introduced later, though simple composites may still appear in some classrooms.

Worked example 1: sum with trig

Differentiate

$f(x)=4\sin x-3\cos x+x^2$

Differentiate term-by-term:

$f'(x)=4\cos x-3(-\sin x)+2x$

So

$f'(x)=4\cos x+3\sin x+2x$

Worked example 2: product with trig

Differentiate

$g(x)=x^2\sin x$

Use the product rule:

$g'(x)=2x\sin x+x^2\cos x$

Worked example 3: quotient with trig

Differentiate

$r(x)=\frac{\sin x}{x}$

Use the quotient rule:

$r'(x)=\frac{x\cos x-\sin x}{x^2}$

Common misconceptions

• Forgetting the negative sign in $\frac{d}{dx}(\cos x)=-\sin x$.
• Treating trig derivatives like the power rule.
• Differentiating $\tan x$ as $\sec x$ instead of $\sec^2 x$.

Exam Focus

• Typical question patterns:
  • Differentiate expressions involving sums/products/quotients with $\sin x$ and $\cos x$.
  • Evaluate $f'(a)$ for trig-containing functions at specific values of $a$.
  • Combine trig derivatives with algebraic derivatives (polynomials and rational functions).
  • Use the memorized derivatives of $e^x$ and $\ln x$ correctly when they appear.
• Common mistakes:
  • Missing the negative sign on $\frac{d}{dx}(\cos x)$.
  • Incorrect product/quotient rule setup when trig is involved.
  • Ignoring where trig functions are undefined (relevant when evaluating at a point).
  • Treating $\frac{d}{dx}(\ln x)$ as $\ln'(x)=\ln x$ instead of $\frac{1}{x}$.

Applying the Derivative to Motion: Velocity and Acceleration

Motion is one of the clearest real-world meanings of the derivative, and it forces you to interpret derivatives as rates, not just compute them symbolically.

Position, velocity, and acceleration

Let $s(t)$ be position along a line (meters) at time $t$ (seconds).

Average velocity on $[t_1,t_2]$:

$\frac{s(t_2)-s(t_1)}{t_2-t_1}$

Instantaneous velocity:

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

Acceleration:

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

Units check:

• $s$ in meters
• $v$ in meters per second
• $a$ in meters per second squared

Speed vs velocity

Velocity can be negative, indicating direction. Speed is the magnitude of velocity:

$\text{speed}=|v(t)|$

Interpreting signs and graphs

• If $v(t)>0$, position is increasing (moving in the positive direction).
• If $v(t)<0$, position is decreasing.
• If $a(t)>0$, velocity is increasing.

Be careful: “velocity increasing” does not necessarily mean moving right. For example, going from $-10$ to $-3$ is an increase in velocity (less negative), even though motion is still leftward.

Worked example 1: compute velocity and acceleration

Suppose

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

Then

$v(t)=3t^2-12t+9$

and

$a(t)=6t-12$

Velocity at $t=2$:

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

Interpretation: at 2 seconds, the object is moving in the negative direction at 3 position-units per second.

Worked example 2: average vs instantaneous velocity

Using the same $s(t)$, find the average velocity from $t=2$ to $t=2.5$:

$\text{avg v} = \frac{s(2.5)-s(2)}{2.5-2}$

Compute:

$s(2)=8-24+18=2$

$s(2.5)=15.625-37.5+22.5=0.625$

So

$\text{avg v} = \frac{0.625-2}{0.5}=-2.75$

This is close to (but not exactly) $v(2)=-3$ because $[2,2.5]$ is a short interval but not infinitesimal.

Common interpretation mistakes

• Confusing $s(t)$ (position) with $v(t)$ (velocity). A high position does not imply high velocity.
• Forgetting the absolute value when asked for speed.
• Using the wrong denominator for average velocity (it must be the time difference).

Exam Focus

• Typical question patterns:
  • Given $s(t)$, find $v(t)$ and $a(t)$ and interpret their meaning.
  • Compute average velocity on an interval and compare it to instantaneous velocity.
  • Use the sign of $v(t)$ to describe direction of motion, and $|v(t)|$ for speed.
• Common mistakes:
  • Reporting $v(t)$ when asked for $v(a)$ at a specific time.
  • Mixing up when to use average rate (secant slope) vs instantaneous rate (derivative).
  • Forgetting units or interpreting negative velocity as “slowing down” automatically.