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🎯⭐ INTERACTIVE LESSON

Definition of the Derivative

Learn step-by-step with interactive practice!

Definition of the Derivative - Complete Interactive Lesson

Part 1: From Average to Instantaneous Rate of Change

∫ The Derivative as a Limit

Part 1 of 7 — From Average to Instantaneous Rate of Change


Topics in This Part

Section
📖 Average Rate of Change (Secant Lines)
Instantaneous Rate of Change (Tangent Lines)
📌 The Limit Definition of the Derivative
Computing Derivatives from the Definition
Alternate Form of the Definition

🔑 Key Concept: The derivative f′(a)f'(a) is the instantaneous rate of change of ff at x=ax = a, defined as the limit of average rates of change as the interval shrinks to zero.

📖 Average Rate of Change

The average rate of change of ff on [a,b][a,b] is the slope of the secant line through (a,f(a))(a, f(a)) and (b,f(b))(b, f(b)):

AROC=f(b)−f(a)b−a\boxed{\text{AROC} = \frac{f(b) - f(a)}{b - a}}

Physical interpretation: If f(t)f(t) = position at time tt, then AROC is the average velocity on [a,b][a,b].


Example: Average Velocity

A car's position is s(t)=t2s(t) = t^2 meters at time tt seconds.

Average velocity from t=1t=1 to t=3t=3:

s(3)−s(1)3−1=9−12=4 m/s\frac{s(3)-s(1)}{3-1} = \frac{9-1}{2} = 4 \text{ m/s}

But what is the velocity at exactly t=1t = 1? We need to let the interval shrink...

🔑 Key Idea: Average rate of change → secant line slope. Make the interval infinitely small → tangent line slope.

📌 The Limit Definition of the Derivative

As h→0h \to 0, the secant line becomes the tangent line:

f′(a)=lim⁡h→0f(a+h)−f(a)h\boxed{f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}}

This is the most fundamental formula in calculus.


Computing f′(x)f'(x) from the Definition

Example: Find f′(x)f'(x) for f(x)=x2f(x) = x^2.

StepComputation
Write the limitf′(x)=lim⁡h→0(x+h)2−x2hf'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}
Expand=lim⁡h→0x2+2xh+h2−x2h= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}
Cancel=lim⁡h→02xh+h2h= \lim_{h \to 0} \frac{2xh + h^2}{h}
Factor out hh=lim⁡h→0(2x+h)= \lim_{h \to 0} (2x + h)
Evaluate=2x= 2x

f(x)=x2  ⟹  f′(x)=2x\boxed{f(x) = x^2 \implies f'(x) = 2x}

AP Tip: On the AP exam, you MUST show the limit process — you cannot just write down the answer using shortcut rules when asked to use the definition.

Check Your Understanding 🎯

Alternate Form of the Derivative

f′(a)=lim⁡x→af(x)−f(a)x−a\boxed{f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}}

Standard FormAlternate Form
lim⁡h→0f(a+h)−f(a)h\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}lim⁡x→af(x)−f(a)x−a\lim_{x \to a} \frac{f(x)-f(a)}{x-a}
Uses increment hhUses the point xx directly
Best for: finding f′(x)f'(x) as a functionBest for: evaluating f′(a)f'(a) at a specific point

Recognizing Derivatives in Disguise

On the AP exam, you may be given a limit and asked to identify it as a derivative:

Example: lim⁡h→0sin⁡(π/6+h)−1/2h\lim_{h \to 0} \frac{\sin(\pi/6 + h) - 1/2}{h}

This is f′(π/6)f'(\pi/6) where f(x)=sin⁡xf(x) = \sin x, since sin⁡(π/6)=1/2\sin(\pi/6) = 1/2.

Answer: f′(π/6)=cos⁡(π/6)=32f'(\pi/6) = \cos(\pi/6) = \frac{\sqrt{3}}{2}

AP Tip: If you see a limit that looks like f(a+h)−f(a)h\frac{f(a+h)-f(a)}{h} or f(x)−f(a)x−a\frac{f(x)-f(a)}{x-a}, identify ff and aa first — then use derivative rules instead of computing the limit directly.

Recognizing Derivatives 🎯

Identify the Derivative 🔍

Match each limit to the derivative it represents.

Compute from the Definition ✍️

Part 2: When Derivatives Exist (and When They Don't)

∫ Differentiability

Part 2 of 7 — When Derivatives Exist (and When They Don't)


Topics in This Part

Section
📖 Differentiability Implies Continuity
Four Ways Derivatives Fail to Exist
📌 Piecewise Differentiability Check
Local Linearity

🔑 Key Concept: Differentiability is STRONGER than continuity. Every differentiable function is continuous, but not every continuous function is differentiable.

📖 Differentiability ⟹ Continuity

f differentiable at c  ⟹  f continuous at c\boxed{f \text{ differentiable at } c \implies f \text{ continuous at } c}

Contrapositive: If ff is NOT continuous at cc, then ff is NOT differentiable at cc.

Warning: The converse is FALSE!

  • f(x)=∣x∣f(x) = |x| is continuous at x=0x = 0 but NOT differentiable.

The Hierarchy

Differentiable  ⟹  Continuous  ⟹  Limit Exists\text{Differentiable} \implies \text{Continuous} \implies \text{Limit Exists}

None of these arrows reverse! Each arrow is a one-way implication.

AP Tip: "Differentiable ⟹ Continuous" appears on nearly every AP exam. Know it cold, and remember the converse is false.

Four Ways Derivatives Fail to Exist

TypeWhat HappensExampleAt
CornerLeft and right slopes differ$f(x) =x
CuspSlopes → ±∞\pm\infty from opposite sidesf(x)=x2/3f(x) = x^{2/3}x=0x = 0
Vertical tangentSlope → ±∞\pm\infty from same sidef(x)=x1/3f(x) = x^{1/3}x=0x = 0
DiscontinuityFunction jumps or is undefinedf(x)=⌊x⌋f(x) = \lfloor x \rfloorx=nx = n

Corner: f(x)=∣x∣f(x) = |x| at x=0x = 0

f′(0−)=lim⁡h→0−∣0+h∣−∣0∣h=lim⁡h→0−−hh=−1f'(0^-) = \lim_{h \to 0^-} \frac{|0+h|-|0|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1

f′(0+)=lim⁡h→0+∣0+h∣−∣0∣h=lim⁡h→0+hh=1f'(0^+) = \lim_{h \to 0^+} \frac{|0+h|-|0|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1

Since −1≠1-1 \neq 1, f′(0)f'(0) does not exist.

🔑 Key Fact: At a corner, the function is continuous but the left and right derivatives are different finite numbers.

Check Your Understanding 🎯

📌 Checking Differentiability for Piecewise Functions

Two-Step Process

Step 1: Check continuity (necessary condition)

  • Evaluate left and right limits at the breakpoint

Step 2: Check that derivatives match (sufficient condition)

  • Compute derivatives of each piece and evaluate at the breakpoint

Example: f(x)={x2x≤12x−1x>1f(x) = \begin{cases} x^2 & x \leq 1 \\ 2x - 1 & x > 1 \end{cases}

CheckLeft PieceRight PieceMatch?
Continuitylim⁡x→1−x2=1\lim_{x \to 1^-} x^2 = 1lim⁡x→1+(2x−1)=1\lim_{x \to 1^+} (2x-1) = 1✓
Derivativef′(x)=2x→2f'(x) = 2x \to 2f′(x)=2→2f'(x) = 2 \to 2✓

Both pass → ff IS differentiable at x=1x = 1.


Example: g(x)={x2x≤13x−2x>1g(x) = \begin{cases} x^2 & x \leq 1 \\ 3x - 2 & x > 1 \end{cases}

CheckLeft PieceRight PieceMatch?
Continuitylim⁡=1\lim = 1lim⁡=1\lim = 1✓
Derivative2x→22x \to 233✗

Continuous but not differentiable at x=1x = 1 (corner).

AP Tip: For piecewise functions, ALWAYS check continuity FIRST. If it fails, stop — the function is not differentiable.

Piecewise Differentiability 🎯

Differentiability Check 🔍

Find the Value ✍️

Part 3: Reading Derivatives from Graphs

∫ Graphical Interpretation of Derivatives

Part 3 of 7 — Reading Derivatives from Graphs


Topics in This Part

Section
📖 Derivative = Slope of Tangent Line
From Graph of ff to Graph of f′f'
📌 Estimating Derivatives from Tables
Reading ff from f′f' (Reverse Direction)
The First Derivative Test

🔑 Key Concept: The derivative gives the slope of the tangent line. Positive derivative means increasing; negative means decreasing; zero means horizontal tangent.

📖 From Graph of ff to Graph of f′f'

This is one of the most important skills on the AP exam:

Feature of ffCorresponding Feature of f′f'
ff increasingf′>0f' > 0 (above xx-axis)
ff decreasingf′<0f' < 0 (below xx-axis)
Local max of fff′=0f' = 0 and changes ++ to −-
Local min of fff′=0f' = 0 and changes −- to ++
Inflection point of ffLocal max or min of f′f'
ff concave upf′f' is increasing
ff concave downf′f' is decreasing
Steep slopeLarge $
Gentle slopeSmall $

Common Trap

f′(c)=0 does NOT always mean f has an extremum at c\boxed{f'(c) = 0 \text{ does NOT always mean } f \text{ has an extremum at } c}

Example: f(x)=x3f(x) = x^3 at x=0x = 0: f′(0)=0f'(0) = 0 but f′f' doesn't change sign → inflection point, NOT an extremum.

AP Tip: On graph-matching problems, always check: (1) where ff has horizontal tangents → f′=0f' = 0, (2) where ff is steepest → f′f' peaks, (3) inflection points of ff → extrema of f′f'.

Graph Reading 🎯

📌 Estimating Derivatives from Data Tables

When given a table of values (common on AP FRQ):

MethodFormulaAccuracy
Forward differencef′(a)≈f(a+h)−f(a)hf'(a) \approx \frac{f(a+h)-f(a)}{h}Good
Backward differencef′(a)≈f(a)−f(a−h)hf'(a) \approx \frac{f(a)-f(a-h)}{h}Good
Symmetric (central)f′(a)≈f(a+h)−f(a−h)2hf'(a) \approx \frac{f(a+h)-f(a-h)}{2h}Best

Example with a Table

xx01234
f(x)f(x)58132029

Estimate f′(2)f'(2):

f′(2)≈f(3)−f(1)3−1=20−82=6f'(2) \approx \frac{f(3)-f(1)}{3-1} = \frac{20-8}{2} = 6

AP Tip: When estimating derivatives from tables, use the symmetric difference quotient whenever possible. The AP exam scoring guidelines explicitly prefer this method.

The First Derivative Test

Sign change of f′ at c  ⟹  local extremum of f at c\boxed{\text{Sign change of } f' \text{ at } c \implies \text{local extremum of } f \text{ at } c}

f′f' Before ccf′f' After ccConclusion
++−-Local maximum at cc
−-++Local minimum at cc
++++No extremum (increasing through cc)
−-−-No extremum (decreasing through cc)

🔑 Key Fact: The First Derivative Test is the primary tool for classifying critical points on the AP exam.

First Derivative Test 🎯

From ff to f′f' 🔍

Estimate from Data ✍️

Part 4: The Language of Derivatives

∫ Derivative Notation & Units

Part 4 of 7 — The Language of Derivatives


Topics in This Part

Section
📖 The Four Common Notations
Leibniz Notation — Why It's Special
📌 Higher-Order Derivatives
Units of Derivatives
Interpreting Derivatives in Context

🔑 Key Concept: Different notations emphasize different aspects of the derivative. Leibniz notation (dy/dxdy/dx) is especially powerful because it suggests the chain rule and carries units naturally.

📖 The Four Common Notations

NotationNameRead AsBest For
f′(x)f'(x)Lagrange"ff prime of xx"General rules, abstract functions
dydx\frac{dy}{dx}Leibniz"dee-why dee-ex"Chain rule, related rates, implicit diff
ddx[f(x)]\frac{d}{dx}[f(x)]Operator"dee-dee-ex of f(x)f(x)""Take the derivative of ___"
y˙\dot{y}Newton"yy dot"Time derivatives in physics

Leibniz Notation: More Than a Symbol

dydx\frac{dy}{dx} is NOT a fraction, but it behaves like one in key situations:

dydx=dydu⋅dudx(Chain Rule — "cancels" like fractions!)\boxed{\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \quad \text{(Chain Rule — "cancels" like fractions!)}}

Evaluated at a point: dydx∣x=3\left.\frac{dy}{dx}\right|_{x=3} means f′(3)f'(3)

AP Tip: The AP exam uses all notations interchangeably. Be comfortable reading f′(x)f'(x), dydx\frac{dy}{dx}, and ddx[f(x)]\frac{d}{dx}[f(x)] — they all mean the same thing.

📌 Higher-Order Derivatives

OrderLagrangeLeibnizMeaning
Firstf′(x)f'(x)dydx\frac{dy}{dx}Slope / rate of change
Secondf′′(x)f''(x)d2ydx2\frac{d^2y}{dx^2}Concavity / acceleration
Thirdf′′′(x)f'''(x)d3ydx3\frac{d^3y}{dx^3}Jerk (rate of change of acceleration)
nn-thf(n)(x)f^{(n)}(x)dnydxn\frac{d^ny}{dx^n}—

Physical Interpretation Chain

Position s(t)→d/dtVelocity v(t)→d/dtAcceleration a(t)\text{Position } s(t) \xrightarrow{d/dt} \text{Velocity } v(t) \xrightarrow{d/dt} \text{Acceleration } a(t)

s′(t)=v(t),s′′(t)=v′(t)=a(t)s'(t) = v(t), \qquad s''(t) = v'(t) = a(t)

🔑 Key Fact: The second derivative tells you about concavity: f′′>0f'' > 0 → concave up, f′′<0f'' < 0 → concave down.

Notation & Higher Derivatives 🎯

Units of Derivatives

Units of dydx=units of yunits of x\boxed{\text{Units of } \frac{dy}{dx} = \frac{\text{units of } y}{\text{units of } x}}

Contextyy Unitsxx Unitsdy/dxdy/dx UnitsMeaning
Position vs timemeterssecondsm/sVelocity
Water volume vs timegallonsminutesgal/minFlow rate
Cost vs quantitydollarsitems$/itemMarginal cost
Population vs timepeopleyearspeople/yrGrowth rate
Temperature vs position°Ccm°C/cmTemp gradient

AP Tip: On AP FRQ, you MUST include units when interpreting a derivative in context. "At t=5t = 5 hours, the temperature is changing at a rate of −3-3 degrees per hour."

Interpreting Derivatives 🎯

Match the Notation 🔍

Interpret in Context ✍️

Part 5: The Tangent Line Equation

∫ Tangent Lines and Linear Approximation

Part 5 of 7 — The Tangent Line Equation


Topics in This Part

Section
📖 Equation of the Tangent Line
Normal Lines (Perpendicular)
📌 Linear Approximation (Linearization)
Over- vs. Under-Estimates

🔑 Key Concept: The tangent line at x=ax = a is the best linear approximation to ff near aa. This idea is the foundation of differential calculus.

📖 Equation of the Tangent Line

y−f(a)=f′(a)(x−a)\boxed{y - f(a) = f'(a)(x - a)}

Three ingredients:

  1. The point: (a,f(a))(a, f(a))
  2. The slope: m=f′(a)m = f'(a)
  3. Plug into point-slope form

Worked Example

Find the tangent line to f(x)=x3f(x) = x^3 at x=2x = 2.

StepComputation
Pointf(2)=8f(2) = 8 → (2,8)(2, 8)
Slopef′(x)=3x2f'(x) = 3x^2 → f′(2)=12f'(2) = 12
Equationy−8=12(x−2)y - 8 = 12(x - 2)
Simplifiedy=12x−16y = 12x - 16

Normal Line

The normal line is perpendicular to the tangent. If tangent slope is mm:

Normal slope=−1m\boxed{\text{Normal slope} = -\frac{1}{m}}

For the example above: normal slope =−112= -\frac{1}{12}, so y−8=−112(x−2)y - 8 = -\frac{1}{12}(x-2).

AP Tip: Normal lines appear less frequently than tangent lines, but they do show up! Remember: perpendicular slopes are negative reciprocals.

Tangent Lines 🎯

📌 Linear Approximation (Linearization)

Near x=ax = a, the tangent line approximates the function:

f(x)≈L(x)=f(a)+f′(a)(x−a)\boxed{f(x) \approx L(x) = f(a) + f'(a)(x - a)}

L(x)L(x) is called the linearization of ff at x=ax = a.


Example: Approximate 4.1\sqrt{4.1}

Using f(x)=xf(x) = \sqrt{x} at a=4a = 4:

ComponentValue
f(a)=f(4)f(a) = f(4)22
f′(x)=12xf'(x) = \frac{1}{2\sqrt{x}}—
f′(4)f'(4)14\frac{1}{4}
L(4.1)L(4.1)2+14(0.1)=2.0252 + \frac{1}{4}(0.1) = 2.025
Actual 4.1\sqrt{4.1}2.02485...2.02485...
Error0.000150.00015

The approximation is excellent for small Δx=x−a\Delta x = x - a.

🔑 Key Fact: Linear approximation works best when xx is close to aa. The farther away, the worse the approximation.

Over- vs. Under-Estimates

Concave up  ⟹  tangent below curve  ⟹  underestimate\boxed{\text{Concave up} \implies \text{tangent below curve} \implies \text{underestimate}} Concave down  ⟹  tangent above curve  ⟹  overestimate\boxed{\text{Concave down} \implies \text{tangent above curve} \implies \text{overestimate}}

Concavityf′′f'' SignTangent Relative to CurveApproximation Is
Concave upf′′>0f'' > 0BelowUnderestimate
Concave downf′′<0f'' < 0AboveOverestimate

Example

For 4.1\sqrt{4.1}: f′′(x)=−14x3/2<0f''(x) = -\frac{1}{4x^{3/2}} < 0 → concave down → tangent is above → overestimate.

Indeed: 2.025>2.02485...2.025 > 2.02485... ✓

AP Tip: "Is this an overestimate or underestimate? Justify your answer." is a classic AP FRQ follow-up. Always cite concavity (f′′f'' sign).

Linear Approximation 🎯

Tangent Line Practice 🔍

Linear Approximation ✍️

Part 6: Derivative Definition Practice

∫ Problem-Solving Workshop

Part 6 of 7 — Derivative Definition Practice


Strategy Guide

Problem TypeMethod
"Find f′(x)f'(x) using the definition"Write the limit, expand, simplify, cancel hh, evaluate
"Evaluate this limit" (looks like a derivative)Recognize as f′(a)f'(a), use rules instead
"Is ff differentiable at x=cx = c?"Check continuity, then check left and right derivatives

🔑 Key Principle: Many limit problems are derivatives in disguise. Recognizing this saves massive computation.

📖 Worked Example: f(x)=1xf(x) = \frac{1}{x} from the Definition

f′(x)=lim⁡h→01x+h−1xhf'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}

StepWork
Common denominator=lim⁡h→0x−(x+h)x(x+h)h= \lim_{h \to 0} \frac{\frac{x - (x+h)}{x(x+h)}}{h}
Simplify numerator=lim⁡h→0−hhx(x+h)= \lim_{h \to 0} \frac{-h}{hx(x+h)}
Cancel hh=lim⁡h→0−1x(x+h)= \lim_{h \to 0} \frac{-1}{x(x+h)}
Evaluate=−1x2= \frac{-1}{x^2}

f(x)=1x  ⟹  f′(x)=−1x2\boxed{f(x) = \frac{1}{x} \implies f'(x) = -\frac{1}{x^2}}


Worked Example: f(x)=xf(x) = \sqrt{x} from the Definition

f′(x)=lim⁡h→0x+h−xh⋅x+h+xx+h+xf'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}

=lim⁡h→0(x+h)−xh(x+h+x)=lim⁡h→01x+h+x=12x= \lim_{h \to 0} \frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})} = \lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}} = \frac{1}{2\sqrt{x}}

AP Tip: For radicals, always conjugate-multiply. For fractions, find common denominators. These are the two key algebraic moves.

Definition Practice 🎯

📌 Recognizing Derivatives in Disguise

If a limit looks like f(a+h)−f(a)h or f(x)−f(a)x−a, identify f and a.\boxed{\text{If a limit looks like } \frac{f(a+h)-f(a)}{h} \text{ or } \frac{f(x)-f(a)}{x-a}, \text{ identify } f \text{ and } a.}

Limit Expressionf(x)f(x)aaf′(a)f'(a)Answer
lim⁡h→0e2+h−e2h\lim_{h \to 0} \frac{e^{2+h}-e^2}{h}exe^x22e2e^2e2e^2
lim⁡x→3x2−9x−3\lim_{x \to 3} \frac{x^2-9}{x-3}x2x^2332(3)2(3)66
lim⁡h→0ln⁡(1+h)h\lim_{h \to 0} \frac{\ln(1+h)}{h}ln⁡(1+x)\ln(1+x)0011+0\frac{1}{1+0}11
lim⁡h→0cos⁡(π+h)+1h\lim_{h \to 0} \frac{\cos(\pi+h)+1}{h}cos⁡x\cos xπ\pi−sin⁡π-\sin\pi00

🔑 Key Fact: This technique turns hard limit computations into easy derivative evaluations. Look for this pattern on EVERY limit problem!

Recognize & Evaluate 🎯

Mixed Practice 🔍

Recognize the Derivative ✍️

Part 7: Comprehensive Review

∫ Review & AP Exam Applications

Part 7 of 7 — Comprehensive Review


The Derivative: Three Perspectives

PerspectiveInterpretation
GeometricSlope of the tangent line to ff at x=ax = a
PhysicalInstantaneous rate of change of ff at aa
Algebraiclim⁡h→0f(a+h)−f(a)h\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}

🔑 Key Principle: Mastering the definition of the derivative means understanding ALL three perspectives and knowing when each is most useful.

📖 Complete Formulas Reference

f′(x)=lim⁡h→0f(x+h)−f(x)h=lim⁡x→af(x)−f(a)x−a\boxed{f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{x \to a} \frac{f(x)-f(a)}{x-a}}

Tangent: y−f(a)=f′(a)(x−a)\boxed{\text{Tangent: } y - f(a) = f'(a)(x-a)}

Linear approx: f(x)≈f(a)+f′(a)(x−a)\boxed{\text{Linear approx: } f(x) \approx f(a) + f'(a)(x-a)}


The Differentiability Hierarchy

Differentiable  ⟹  Continuous  ⟹  Limit Exists\text{Differentiable} \implies \text{Continuous} \implies \text{Limit Exists}

None reverse! ∣x∣|x| is continuous but not differentiable. ⌊x⌋\lfloor x \rfloor has limits from one side but isn't continuous.


Motion Connections

s(t)→d/dtv(t)=s′(t)→d/dta(t)=s′′(t)s(t) \xrightarrow{d/dt} v(t) = s'(t) \xrightarrow{d/dt} a(t) = s''(t)

ConceptMeaning
v(t)=0v(t) = 0Particle at rest
v(t)>0v(t) > 0Moving right/up
v(t)<0v(t) < 0Moving left/down
a(t)a(t) and v(t)v(t) same signSpeeding up
a(t)a(t) and v(t)v(t) opposite signSlowing down

Comprehensive Review 🎯

📌 Common AP Exam Question Types

Question TypeWhat to Do
"Find f′(a)f'(a) using the definition"Write the limit, expand, simplify, cancel hh, evaluate
"f′(3)=−2f'(3) = -2. Interpret in context.""At x=3x=3, ff is decreasing at 2 [units] per [unit]"
"Is ff differentiable at cc?"Check continuity AND left/right derivatives
"Find the tangent line"y−f(a)=f′(a)(x−a)y - f(a) = f'(a)(x-a)
"Approximate f(x)f(x)"Linear approx: f(a)+f′(a)(x−a)f(a) + f'(a)(x-a)
"Over or underestimate?"Check f′′f'' sign (concavity)
"Evaluate this limit"Is it a derivative in disguise? Identify ff and aa

AP FRQ Interpretation Template

"At time t=[value]t = [\text{value}] [units], the [quantity] is [increasing/decreasing] at a rate of ∣f′(a)∣|f'(a)| [units of yy] per [units of xx]."

Example: If T(t)T(t) = temperature (°C) at time tt (hours) and T′(3)=−1.5T'(3) = -1.5:

"At t=3t = 3 hours, the temperature is decreasing at a rate of 1.5 degrees Celsius per hour."

AP Tip: You MUST include units, state increasing/decreasing, and use "rate of" language. This is worth 1–2 points on every contextual interpretation question.

AP Exam Practice 🎯

Final Review 🔍

Tangent Line Application ✍️