Definition of the Derivative Comprehensive Worksheet
Foundational–Intermediate⌛ Est. 3–4 Hours70 Problems
A complete, rigorous resource covering the limit definition of the derivative — from the intuition of instantaneous rate of change through first-principles computation, non-differentiability analysis, and AP-level exam strategy. Suitable for AP Calculus AB/BC, university Calculus I, and any student who wants to understand where the derivative actually comes from.
Learning Objectives
State and apply both forms of the limit definition of the derivative: the $h \to 0$ form and the $x \to a$ alternative form.
Compute derivatives from first principles for polynomial, radical, rational, trigonometric, and exponential functions.
Simplify the difference quotient algebraically — including conjugate multiplication and trigonometric identities — to evaluate the limit.
Identify points of non-differentiability and explain the geometric meaning of each type.
Recognize a given limit expression as the derivative of a specific function at a specific point.
Interpret the derivative in context: instantaneous rate of change, slope of the tangent line, and marginal analysis.
Introduction
Before there were shortcut rules — before the power rule, the product rule, or any of the elegant machinery that makes differentiation feel almost effortless — there was a question. It was not an abstract question, and it did not arise in a lecture hall. It arose from a practical and urgent need: how do you describe the motion of a planet, the path of a cannonball, the speed of a falling body at a precise instant in time? Not the average speed over a minute, or a second, or even a millisecond. The speed right now, at this point, in this moment.
Isaac Newton and Gottfried Wilhelm Leibniz, working independently in the latter half of the seventeenth century, both arrived at essentially the same answer. To find the rate of change of a quantity at a single instant, you compute the ratio of change over a small interval and then let that interval shrink toward zero. The limit of that ratio — if it exists — is the derivative. Leibniz gave us the notation $dy/dx$ that persists to this day. Newton wrote of "fluxions." The concept was the same: the instantaneous rate of change as the limiting case of the average rate of change.
This is not merely historical background. The limit definition is the reason every rule you will ever learn about differentiation is true. The power rule, the chain rule, implicit differentiation — each of these can be derived from the definition, and each of them is only meaningful because the definition is what it is. Students who learn shortcuts without understanding the definition are, in a real sense, using tools they have borrowed but do not own. The definition is what makes the tools yours.
Geometrically, the story is just as compelling. The slope of a tangent line to a curve at a point is not immediately obvious — a tangent line, after all, only touches the curve at one point, and you ordinarily need two points to define a slope. The definition resolves this by approaching the tangent line as the limit of secant lines: lines through the given point and a second point on the curve, as the second point is dragged closer and closer to the first. When the two points coincide — in the limit — the secant line becomes the tangent line, and the average rate of change becomes the instantaneous rate of change.
In real-world modeling, this idea is everywhere. The instantaneous velocity of a moving object is the derivative of its position function. The marginal cost in economics — the cost of producing one additional unit — is modeled as the derivative of the total cost function. The rate at which a drug concentration changes in the bloodstream, the power output of an electrical signal at a given moment, the sensitivity of an option price to changes in the underlying asset: all of these are derivatives, all of them grounded in the same limit definition you will practice in this worksheet.
This worksheet is for students who are encountering the derivative for the first time, as well as those who have learned the shortcut rules and now want to revisit the foundation. It is also for students preparing for AP Calculus AB or BC, where the definition appears in both multiple-choice and free-response contexts — often in disguised forms that only students who genuinely understand the limit will recognize. And it is for students in university Calculus I who face the expectation of not just computing derivatives, but proving them.
Mastery of the limit definition looks like this: you can set up the difference quotient correctly for any function, expand $f(x+h)$ without algebraic errors, identify the algebraic technique needed to cancel the $h$ in the denominator, evaluate the resulting limit, and interpret your answer in context. You can also run the process in reverse: look at a limit and identify it as the derivative of a particular function at a particular point. That bidirectional fluency — from function to derivative and from limit to derivative — is the goal.
Used when evaluating the derivative at a specific point $a$. The variable approaches the point rather than the increment approaching zero.
Both forms are equivalent. The substitution $x = a + h$ converts the alternative form into the standard form. On AP exams, the alternative form appears frequently in limit-recognition problems where the student must identify $f$ and $a$ from a given expression.
The Difference Quotient
The expression $\dfrac{f(x+h) - f(x)}{h}$ is called the difference quotient. It represents the slope of the secant line through the points $(x, f(x))$ and $(x+h, f(x+h))$ on the graph of $f$. As $h \to 0$, the second point slides along the curve toward the first, the secant line rotates toward the tangent line, and the difference quotient approaches the instantaneous slope — the derivative.
Professor's Note: The most consequential algebraic step in every first-principles problem is expanding $f(x+h)$. Do this carefully, term by term, and never combine it with $f(x)$ before the expansion is complete. Rushed expansions that lose a term or a sign in the middle of the expression are the primary source of errors in this topic.
The Standard Algebraic Toolkit
Function Type
Key Technique
Goal
Polynomial (degree $n$)
Binomial expansion of $(x+h)^n$
Cancel the $h$ in the denominator
Rational ($1/g(x)$)
Combine fractions over common denominator
Factor and cancel $h$
Radical ($\sqrt{f(x)}$)
Multiply numerator and denominator by conjugate
Eliminate the radical difference; cancel $h$
Sine / Cosine
Sum-angle identity + limits $\lim_{h\to0}\tfrac{\sin h}{h}=1$ and $\lim_{h\to0}\tfrac{\cos h -1}{h}=0$
Reduce to known trig limits
Exponential ($e^x$)
Factor; use $\lim_{h\to0}\tfrac{e^h-1}{h}=1$
Identify the fundamental exponential limit
Non-Differentiability: When the Definition Fails
The derivative at $x = a$ fails to exist whenever the limit $\lim_{h \to 0} \dfrac{f(a+h) - f(a)}{h}$ does not exist. This happens in four geometrically distinct ways:
Type
Example
What Happens to the Limit
Corner / kink
$f(x) = |x|$ at $x=0$
Left-hand and right-hand limits exist but differ
Cusp
$f(x) = x^{2/3}$ at $x=0$
Both one-sided limits are $\pm\infty$
Vertical tangent
$f(x) = x^{1/3}$ at $x=0$
Limit is $+\infty$ (or $-\infty$) from both sides
Discontinuity
Any jump or removable discontinuity
Limit does not exist because $f$ is not even continuous
Key Theorem: Differentiability implies continuity. If $f'(a)$ exists, then $f$ is continuous at $a$. The converse is false: $f(x) = |x|$ is continuous at $0$ but not differentiable there.
Recognizing Limits as Derivatives
A core AP Calculus skill is running the definition backwards. Given a limit, identify it as $f'(a)$ for some $f$ and $a$. The strategy: look for the structure $\dfrac{\text{(something involving a number)} - \text{(same function at a fixed value)}}{\text{(difference from that fixed value)}}$. The fixed value is $a$; the function is whatever is being evaluated. For example:
The numerator is $x^4 - 3^4 = f(x) - f(3)$, the denominator is $x - 3$, and $a = 3$. By the power rule, $f'(3) = 4(3)^3 = 108$.
Comparing the Definition to Shortcut Rules
Once you have verified a shortcut rule using the definition (the power rule proof is an excellent exercise — see Example 5 below), you are licensed to use the shortcut. Mixing the definition and shortcut rules in a single problem is not wrong, but on exam questions that specifically ask for the derivative "using the definition," the full limit process is required. Reading the problem instruction carefully is not optional.
Worked Examples
Example 1 — Linear Function (Foundational)
Find $f'(x)$ using the definition for $f(x) = 3x - 5$.
A linear function has constant slope. The derivative $f'(x) = 3$ confirms that the slope of the tangent line is $3$ at every point — which is exactly the slope of the line itself. This consistency is a useful sanity check for linear functions.
Answer: $f'(x) = 3$
Example 2 — Quadratic: Careful Expansion
Find $f'(x)$ using the definition for $f(x) = x^2 - 4x + 1$.
$$\frac{f(x+h)-f(x)}{h} = \frac{h(2x + h - 4)}{h} = 2x + h - 4$$
Step 4 — Evaluate the limit
$$f'(x) = \lim_{h\to 0}(2x + h - 4) = 2x - 4$$
Common Mistake Warning
Students frequently write $(x+h)^2 = x^2 + h^2$, omitting the crucial middle term $2xh$. This error propagates through the entire calculation and produces $f'(x) = -4$, which is wrong. Always expand $(x+h)^2$ as $x^2 + 2xh + h^2$.
Answer: $f'(x) = 2x - 4$
Example 3 — Rational Function: Common Denominator Technique
Find $f'(x)$ using the definition for $f(x) = \dfrac{1}{x}$, $x \neq 0$.
The difference of two square roots cannot be simplified algebraically without this step. The conjugate converts $\sqrt{x+h}-\sqrt{x}$ into a difference of squares $(x+h)-x = h$, which cancels with the $h$ in the denominator. This is the standard technique for all radical difference quotients.
Answer: $f'(x) = \dfrac{1}{2\sqrt{x}}$
Example 5 — Power Rule Proof (Mini-Derivation)
Prove that $\dfrac{d}{dx}[x^n] = nx^{n-1}$ for positive integer $n$ using the limit definition.
You have now proven that the power rule is not an assertion but a consequence of the definition. Every application of $\dfrac{d}{dx}[x^n] = nx^{n-1}$ rests on this argument.
Proved: $\dfrac{d}{dx}[x^n] = nx^{n-1}$ for all positive integers $n$.
Example 6 — Derivative of $\sin x$ Using Trig Limits
Find $f'(x)$ using the definition for $f(x) = \sin x$.
Step 1 — Write the difference quotient
$$\frac{\sin(x+h) - \sin x}{h}$$
Step 2 — Apply the sum-angle identity
$\sin(x+h) = \sin x \cos h + \cos x \sin h$, so:
$$= \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} = \frac{\sin x(\cos h - 1) + \cos x \sin h}{h}$$
Step 3 — Split and use standard limits
$$= \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}$$
Recall: $\displaystyle\lim_{h\to 0}\frac{\sin h}{h} = 1$ and $\displaystyle\lim_{h\to 0}\frac{\cos h - 1}{h} = 0$.
Step 4 — Evaluate
$$f'(x) = \sin x \cdot 0 + \cos x \cdot 1 = \cos x$$
Note on the Trig Limits
These two limits — $(\sin h)/h \to 1$ and $(\cos h - 1)/h \to 0$ — are themselves proven using geometric arguments (the squeeze theorem) or L'Hôpital's rule. For AP and university exams, you are expected to cite them as known results rather than re-derive them each time.
Answer: $\dfrac{d}{dx}[\sin x] = \cos x$
Example 7 — Conceptual Trap: Recognizing a Limit as a Derivative
Evaluate $\displaystyle\lim_{h \to 0} \dfrac{(2+h)^5 - 32}{h}$ without expanding $(2+h)^5$.
Step 1 — Recognize the structure
Compare to the definition: $f'(a) = \displaystyle\lim_{h\to 0} \dfrac{f(a+h)-f(a)}{h}$.
Here: $a + h = 2 + h$, so $a = 2$. Also, $f(a) = f(2) = 32 = 2^5$, so $f(x) = x^5$.
Step 2 — Identify the derivative
The limit equals $f'(2)$ where $f(x) = x^5$. By the power rule: $f'(x) = 5x^4$.
Step 3 — Evaluate
$$f'(2) = 5(2)^4 = 5 \cdot 16 = 80$$
The Trap
The temptation is to expand $(2+h)^5$ using the binomial theorem. That produces the same answer but wastes time. The AP exam uses limits like this specifically to test whether students recognize the definition — not whether they can expand fifth powers.
Alternate: Using the Alternative Form
The same limit can be written as $\displaystyle\lim_{x\to 2}\dfrac{x^5 - 32}{x-2} = f'(2) = 80$. Both forms confirm the same value.
Answer: $80$
Example 8 — Non-Differentiability at a Corner
Show that $f(x) = |x|$ is not differentiable at $x = 0$ using the limit definition.
As $h \to 0^+$ (approaching from the right): $|h| = h$, so $\dfrac{|h|}{h} = \dfrac{h}{h} = 1$.
As $h \to 0^-$ (approaching from the left): $|h| = -h$, so $\dfrac{|h|}{h} = \dfrac{-h}{h} = -1$.
Step 3 — Conclude
The right-hand limit is $1$ and the left-hand limit is $-1$. Since they disagree, $\displaystyle\lim_{h\to 0}\dfrac{|h|}{h}$ does not exist, and therefore $f'(0)$ does not exist.
Geometric Meaning
At $x = 0$, the graph of $|x|$ has a sharp corner. The slope of the left branch ($-1$) and the slope of the right branch ($+1$) are different, so there is no single well-defined tangent line at the corner point.
Key Takeaway
Continuity does not guarantee differentiability. $f(x) = |x|$ is continuous everywhere — including at $0$ — but fails to be differentiable at $0$. This is the most important example of this distinction in introductory calculus.
$f(x) = |x|$ is not differentiable at $x = 0$: left- and right-hand limits of the difference quotient are $-1$ and $1$, respectively.
Practice Problems
A. Core Skill Development
Foundational
Use the limit definition $f'(x) = \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$ to find the derivative of each function. Do not use shortcut rules.
$f(x) = 7$
$f(x) = 4x + 9$
$f(x) = -2x + 3$
$f(x) = x^2$
$f(x) = 3x^2 - x$
$f(x) = x^3$
$f(x) = 2x^3 + x^2$
$f(x) = x^2 + 5x - 2$
$f(x) = \dfrac{1}{2x}$
$f(x) = \dfrac{3}{x+1}$, $x \neq -1$
B. Structural Recognition
Intermediate
Each problem below requires a specific algebraic technique to simplify the difference quotient. Before computing, identify the technique (conjugate, common denominator, binomial expansion, trig identity). Then solve using the definition.
$f(x) = \sqrt{2x+1}$, $x > -\tfrac{1}{2}$
$f(x) = \sqrt{x^2 + 1}$
$f(x) = \dfrac{1}{x^2}$, $x \neq 0$
$f(x) = \dfrac{x}{x-2}$, $x \neq 2$
$f(x) = \dfrac{1}{\sqrt{x}}$, $x > 0$ (use conjugate after combining)
$f(x) = x^4$ (use binomial theorem; expand fully)
$f(x) = \cos x$ (use sum-angle identity for cosine)
$f(x) = \tan x$ (write as $\sin x / \cos x$; use definition and known trig limits)
$f(x) = e^x$ (use $\lim_{h\to 0}\dfrac{e^h-1}{h} = 1$ as a known result)
C. Derivative at a Specific Point (Alternative Form)
Intermediate
Use the alternative form $f'(a) = \lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}$ to evaluate the derivative at the given point. Factor the numerator algebraically.
$f(x) = x^2 - 3x$, find $f'(2)$
$f(x) = x^3 + 1$, find $f'(-1)$
$f(x) = \sqrt{x+3}$, find $f'(1)$
$f(x) = \dfrac{1}{x-4}$, find $f'(5)$
$f(x) = x^4 - x$, find $f'(1)$
$f(x) = \dfrac{x^2-9}{x-3}$ — wait: is $f$ differentiable at $x = 3$? Discuss before computing.
$f(x) = |x-2|$, find $f'(2)$ — or explain why it does not exist.
$f(x) = x^3 - 6x^2 + 9x$, find $f'(3)$
D. Limit Recognition
Intermediate–Advanced
Each expression below is a limit that equals the derivative of some function at some point. Identify $f(x)$ and $a$, then evaluate using derivative rules — without expanding or computing the limit directly.
For each function and specified point, use one-sided limits of the difference quotient to determine whether the derivative exists. If it does not, classify the type of non-differentiability and describe the geometric feature.
$f(x) = |x+3|$ at $x = -3$
$f(x) = x^{2/3}$ at $x = 0$
$f(x) = \begin{cases} x^2 & x < 1 \\ 2x - 1 & x \geq 1 \end{cases}$ at $x = 1$
$f(x) = \begin{cases} x^2 & x \leq 1 \\ 3 - x & x > 1 \end{cases}$ at $x = 1$
$f(x) = x^{1/3}$ at $x = 0$
F. Exam-Level Questions
AP / University Exam Style
These problems combine multiple skills or require explanation alongside computation.
A function $f$ satisfies $f(2) = 3$ and $\displaystyle\lim_{x\to 2}\dfrac{f(x)-3}{x-2} = 7$. What is $f'(2)$? Explain your reasoning, citing the definition.
Use the limit definition to find the equation of the tangent line to $f(x) = x^2 + 2x$ at $x = 1$. Show all steps.
Suppose $f(x) = \sqrt{x}$ and $g(x) = x^2 + 1$. Using the definition, find the derivative of $f(g(x))$ at $x = 0$ — without using the chain rule. Then verify with the chain rule.
(Free-Response style) Let $f(x) = \dfrac{1}{x^2+1}$. (a) Use the definition to find $f'(0)$. (b) Interpret $f'(0)$ as the slope of a tangent line. (c) Write the equation of the tangent line at $x = 0$.
The graph of a function $f$ has a vertical tangent line at $x = 4$. Using the definition, explain why $f$ is continuous but not differentiable at $x = 4$, and describe what happens to the difference quotient as $h \to 0$.
A student writes: "If $f'(a)$ exists, then $f$ must be differentiable on an open interval around $a$." Is this true or false? Provide a precise explanation with an example.
G. Challenge Set
Challenge
For students seeking deeper engagement with the theoretical structure of the definition.
Use the limit definition to find $f'(x)$ for $f(x) = x^{1/3}$ for $x \neq 0$, using the algebraic identity $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ with $a = (x+h)^{1/3}$ and $b = x^{1/3}$.
Prove that if $f$ is differentiable at $a$, then $\displaystyle\lim_{h\to 0}\dfrac{f(a+h)-f(a-h)}{2h} = f'(a)$. (This is the symmetric difference quotient — commonly used in numerical analysis.)
Find a function $f$ such that the symmetric difference quotient $\dfrac{f(a+h)-f(a-h)}{2h}$ exists as $h\to 0$, but $f$ is not differentiable at $a$. What does this tell you about the symmetric quotient?
Error Diagnosis Problems
Each problem shows incorrect student work. Identify the error, explain why it is wrong, and produce the correct solution.
Student's work: For $f(x) = x^2$, the student writes $f(x+h) = x^2 + h^2$ and obtains $f'(x) = \displaystyle\lim_{h\to 0}\dfrac{h^2}{h} = \lim_{h\to 0} h = 0$. Identify the error and correct it.
Student's work: For $f(x) = \sqrt{x}$, the student writes $\dfrac{\sqrt{x+h}-\sqrt{x}}{h}$ and then "cancels" to get $\dfrac{1}{\sqrt{h}}$, concluding the limit does not exist. Identify the error and correct it.
Student's work: Evaluating $\displaystyle\lim_{h\to 0}\dfrac{(1+h)^3 - 1}{h}$, the student says "this is the derivative of $x^3$ at $x=1$, so the answer is $3(1)^2 = 3$." Is this reasoning correct? If so, why? If not, what is wrong?
Conceptual Reasoning Questions
These questions require written explanation. Computation alone is insufficient.
The definition requires that the same limit be obtained from both the left and the right. Explain, using the definition, why a function with a corner cannot be differentiable at the corner point — without referencing any specific example.
A student argues: "Since we take the limit as $h \to 0$, we can just set $h = 0$ in the difference quotient to evaluate the derivative." What is wrong with this reasoning? Why can't we simply substitute $h = 0$?
The quantity $\dfrac{f(x+h)-f(x)}{h}$ is the average rate of change over $[x, x+h]$. In what physical or practical sense does the derivative represent the "instantaneous" version of this? Is the word "instantaneous" literally meaningful, or is it shorthand for a mathematical process?
If $f'(a) = 0$, does the function necessarily have a local maximum or minimum at $a$? What does the definition tell you, and what additional analysis is required?
Explain why differentiability implies continuity using the limit definition. In your argument, show explicitly how the existence of $f'(a)$ forces $\displaystyle\lim_{x\to a} f(x) = f(a)$.
Real-World Modeling Problems
Physics: Instantaneous Velocity
A particle's position (in meters) at time $t$ seconds is given by $s(t) = 4t^2 - 3t + 1$.
Using the limit definition, find the velocity function $v(t) = s'(t)$.
Find the instantaneous velocity at $t = 2$ seconds. Include units and interpret the sign of your answer.
At what time is the particle momentarily at rest? What does this mean geometrically on the position-time graph?
Economics: Marginal Cost
A manufacturer's total cost (in dollars) of producing $q$ units is $C(q) = 0.02q^2 + 5q + 400$.
Use the limit definition to find $C'(q)$, the marginal cost function.
Evaluate $C'(50)$. Interpret this value: what does it mean in context, and what are the units?
The actual cost of producing the 51st unit is $C(51) - C(50)$. Compute this and compare it to $C'(50)$. What does the comparison reveal about the relationship between the derivative and the difference quotient?
Biology: Population Growth Rate
A bacterial population (in thousands) at time $t$ hours satisfies $P(t) = \sqrt{t+4}$.
Use the limit definition to find $P'(t)$.
Find the instantaneous growth rate at $t = 0$ and $t = 12$. What do these values tell you about how quickly the population is growing over time?
As $t \to \infty$, what happens to $P'(t)$? What biological phenomenon does this model?
Engineering: Signal Slope
A voltage signal is modeled by $V(t) = \sin(t)$ (in volts, $t$ in milliseconds).
Using the definition and the trig limits, show that $V'(t) = \cos t$.
At what time values is the rate of change of the voltage signal zero? What does this mean for the signal?
At $t = \pi/4$ ms, find the instantaneous rate of change of the voltage. Give units and interpret.
Optimization: Tangent Line as Linear Approximation
The function $f(x) = \sqrt{x}$ is used to approximate square roots near $x = 9$.
Use the definition to find $f'(9)$.
Write the equation of the tangent line to $f$ at $x = 9$. This tangent line is the linear approximation $L(x)$ of $\sqrt{x}$ near $x = 9$.
Use $L(x)$ to approximate $\sqrt{9.1}$. Then find the exact value and compute the error. What does the small error tell you about the quality of the linear approximation?
Exam Traps and Strategic Advice
Trap 1 — Expanding $(x+h)^2$ as $x^2 + h^2$
Dropping the middle term $2xh$ from $(x+h)^2$ is the single most common algebraic error in this topic. If you make this mistake, your difference quotient will not contain an $h$ that can be divided into the numerator, and the entire computation collapses. Before any other step, write $(x+h)^2 = x^2 + 2xh + h^2$ slowly and deliberately. This applies to $(x+h)^3$, $(x+h)^4$, and all higher powers.
Trap 2 — Substituting $h = 0$ Too Early
The entire reason we set up the limit is that substituting $h = 0$ directly gives $0/0$ — an indeterminate form. You must first simplify algebraically (cancel the $h$) before evaluating at $h = 0$. Students who substitute early get $0$ in the denominator, panic, and conclude the derivative does not exist. It almost always does — you just haven't finished the algebra.
Trap 3 — Forgetting to Subtract $f(x)$ Entirely
In the numerator $f(x+h) - f(x)$, the subtraction applies to the entire expression $f(x)$, not just the first term. For $f(x) = x^2 + 3x$, the difference is $(x^2+2xh+h^2+3x+3h) - (x^2+3x)$. Students frequently write $-x^2$ but forget to subtract $-3x$, leaving a residual term that corrupts the derivative. Use parentheses around $f(x)$ in the numerator as a reminder.
Trap 4 — Limit Recognition: Confusing $f$ and $a$
On AP multiple-choice, limits like $\lim_{h\to 0}\frac{(5+h)^3 - 125}{h}$ are designed to test definition recognition. The most common mistake is identifying $f(x) = (x+h)^3$ or missing that $125 = 5^3$. Systematically match the structure: the fixed number in both the "shifted" expression and the constant is $a$; the function being applied is $f$.
Trap 5 — Non-Differentiability at Piecewise Boundaries
For piecewise functions, many students check continuity at the boundary and conclude the function is differentiable. Continuity is necessary but not sufficient. You must also verify that the one-sided derivatives (limits of the difference quotient from left and right) agree. Even if the pieces join continuously, they may produce different slopes at the join, creating a corner — and a non-differentiable point.
AP Strategy: Show the Definition Setup
On free-response questions that ask for the derivative "using the definition," full credit requires writing out $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ explicitly, expanding $f(x+h)$, simplifying, and taking the limit. Jumping from setup to answer loses scoring points even if the answer is correct. Graders are looking for the process, not just the result.
Time-Saver: Use Shortcut + Verify
If an exam problem asks you to "use the definition" but you are short on time: compute the answer with a shortcut rule first. Then structure your definition work backward — knowing the answer helps you catch algebraic mistakes before they propagate. This is not mathematically rigorous advice, but it is strategically sound under pressure.
University Level: Cite the Trig Limits by Name
At the university level, citing $\lim_{h\to 0}\frac{\sin h}{h} = 1$ without proof is acceptable only if the instructor has established it previously. If asked to derive $\frac{d}{dx}[\sin x]$ rigorously, you may need to prove this limit using the squeeze theorem. Know whether your course requires the proof or accepts the limit as given.
Mixed Rule Detection
Each limit below either is a derivative in disguise, requires a specific algebraic technique, or appears to need the definition but is better evaluated another way. For each: (i) identify what you are looking at, (ii) justify your strategy, (iii) evaluate.
$\displaystyle\lim_{h\to 0}\dfrac{(x+h)^3 - x^3}{h}$ — Evaluate as a derivative; identify $f$ and compute $f'(x)$ directly.
$\displaystyle\lim_{h\to 0}\dfrac{\cos(x+h) - \cos x}{h}$ — Use trig identity; identify the two standard trig limits required.
$\displaystyle\lim_{h\to 0}\dfrac{\ln(e+h) - 1}{h}$ — Identify $f$ and $a$; evaluate without expanding the logarithm.
$\displaystyle\lim_{h\to 0}\dfrac{(x+h)^2 - x^2}{h^2}$ — Is this the derivative of $x^2$? Analyze carefully; the denominator is $h^2$, not $h$.
$\displaystyle\lim_{x\to 4}\dfrac{\sqrt{x} - 2}{x - 4}$ — Use the alternative derivative form; identify $f$ and $a$, then evaluate.
$\displaystyle\lim_{h\to 0}\dfrac{f(2+3h) - f(2)}{h}$ for differentiable $f$ — Express in terms of $f'(2)$. (Hint: rewrite the denominator.)
$\displaystyle\lim_{h\to 0}\dfrac{f(a+h) - f(a-h)}{2h}$ for differentiable $f$ — Prove this equals $f'(a)$ by writing it as the average of two standard-form quotients.
$\displaystyle\lim_{h\to 0}\dfrac{e^{x+h} - e^x}{h}$ — Factor $e^x$ from the numerator; apply the fundamental exponential limit.
Error: $(x+h)^2 \neq x^2 + h^2$. The correct expansion is $x^2+2xh+h^2$. With this, $f(x+h)-f(x) = 2xh+h^2$, the quotient is $2x+h$, and the limit is $2x$, not $0$.
ED52 — Radical cancellation error
Error: $\sqrt{x+h}$ and $\sqrt{x}$ cannot be "cancelled" by subtracting under the radical or dividing directly. The correct step is to multiply numerator and denominator by the conjugate $\sqrt{x+h}+\sqrt{x}$, obtaining $\dfrac{h}{h(\sqrt{x+h}+\sqrt{x})} = \dfrac{1}{\sqrt{x+h}+\sqrt{x}}$, with limit $\dfrac{1}{2\sqrt{x}}$.
ED53 — Limit recognition
Correct. The limit $\lim_{h\to 0}\frac{(1+h)^3-1}{h}$ matches the definition with $f(x)=x^3$ and $a=1$: $f'(1) = 3(1)^2 = 3$. The student's reasoning is sound and the answer is correct.
$C(51)-C(50) = [0.02(2601)+255+400]-[0.02(2500)+250+400] = [52.02+655]-[650] = \$7.02$. Compare to $C'(50)=\$7.00$. Error: $\$0.02$. The derivative is an excellent approximation of the marginal cost of the next unit.
What exactly is the limit definition of the derivative?
It is the formal mathematical statement that $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$, provided the limit exists. It defines the derivative as the limiting value of the average rate of change over smaller and smaller intervals, and it is the foundation from which all differentiation rules are proven.
Why learn this if shortcut rules exist?
The shortcut rules are proven using the definition. Without understanding the definition, you cannot prove the power rule, the product rule, or the chain rule — you can only memorize them. AP and university exams also test the definition directly, both computationally and conceptually. Students who know only the shortcuts will miss these questions.
What is the difference between the two forms of the definition?
The standard form ($h \to 0$) is used when finding a general formula for $f'(x)$. The alternative form ($x \to a$) is used when evaluating the derivative at a specific point $a$. They are mathematically equivalent — the substitution $x = a+h$ converts one into the other — but each is more convenient in different situations.
What algebraic technique should I use for radical functions?
Multiply both numerator and denominator by the conjugate of the radical expression in the numerator. For $\sqrt{x+h} - \sqrt{x}$, the conjugate is $\sqrt{x+h}+\sqrt{x}$. This converts the difference of square roots into a difference of squares, which simplifies to $h$ — canceling the $h$ in the denominator. This is the standard technique without exception.
How is the definition tested on AP Calculus exams?
In several ways: (1) Computing a derivative from first principles for a simple function. (2) Recognizing a given limit as a derivative of a specific function at a specific point. (3) Identifying non-differentiable points on a graph. (4) Interpreting the difference quotient as an average rate of change. Expect 1–3 multiple-choice questions and potential free-response components on both AB and BC exams.
When is a function not differentiable?
At any point where the limit of the difference quotient does not exist. This includes corners (where left and right limits differ), cusps (where both limits are infinite with opposite signs), vertical tangents (where both limits are the same infinite value), and discontinuities (where the limit fails entirely because the function is not even continuous). Continuity is necessary but not sufficient for differentiability.
What are the key trig limits I need for the definition?
Two fundamental limits: $\lim_{h\to 0}\frac{\sin h}{h} = 1$ and $\lim_{h\to 0}\frac{\cos h - 1}{h} = 0$. Both are used in deriving $\frac{d}{dx}[\sin x] = \cos x$ and $\frac{d}{dx}[\cos x] = -\sin x$ from first principles. They are accepted as known results on AP exams but may require proof in university courses.
Can I use the definition to prove the power rule for all real exponents?
For positive integer exponents, yes — the binomial theorem handles the proof cleanly. For rational exponents like $x^{1/3}$, a more subtle algebraic identity is needed (see Challenge Problem 1 in this worksheet). For irrational exponents, the proof requires logarithmic differentiation or the exponential definition, and is beyond the scope of AP Calculus but encountered in university analysis courses.
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