Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
The derivative is a function, not a number, and the definition is how that function is built: f′(x)=limh→0hf(x+h)−f(x). Read it as a machine with three stages. Step off the base point by h, form the quotient, then let h shrink. What survives is a new function whose input is the same x and whose output is the slope at that input. The exam uses the definition in three ways. It asks you to compute a derivative from it for a simple function. It asks you to recognize a limit that has the shape of a quotient and name the function and point it belongs to. And it expects you to know that the three notations f′(x), dxdy, and dxdf(x) all name the same object. The common failure is treating the quotient as the answer, which leaves an expression in h where a function of x was required. Shrink the step last, and the answer arrives already correct.
Figure 1 — Three stages, one output. The definition takes an input, steps away from it, forms a quotient, and then collapses the step. What leaves the machine is a function, which is why the same rule can be evaluated at any input afterwards.
For f(x)=x2:
hf(x+h)−f(x)=h(x+h)2−x2=h2xh+h2=2x+h
Every term that contains h after the cancellation disappears when h→0, so f′(x)=2x. The cancellation is the whole trick. If an h survives in a denominator, the algebra stopped too early.
For f(x)=x1 the same three stages produce x(x+h)−1, which tends to −x21. For f(x)=x the quotient needs a conjugate, and the result is 2x1. The pattern is identical in all three cases: the algebra exists to remove the h from the denominator, and the limit does the rest.
| Notation | Reads as | Where it is used |
|---|---|---|
| f′(x) | the derivative of f at x | Function notation, composition, evaluation |
| dxdy | the derivative of y with respect to x | Contextual rates, implicit work, differential equations |
| dxdf(x) | the operator applied to f(x) | Statements of rules and quick computations |
The three are interchangeable in value. They differ in what they make easy to say: f′(a) names a number, dxdy names a rate against a stated input, and the operator form names the act of differentiating.
A limit such as limh→0hcos(π+h)+1 is the definition in costume. Identify the function as cosx and the base point as x=π, and the limit is f′(π)=−sinπ=0. The identification is the skill. Once the function and point are named, the value comes from any rule you trust.
Borrowed idea: Every definition computation runs the same three stages. Step. Quotient. Shrink. The algebra in the middle exists for one purpose only. It removes h from the denominator so that substitution stops producing zero over zero. That is the entire job. A student who expands carefully but takes the limit before cancelling writes an expression that is undefined at the point of interest and then reports a number pulled from nowhere, which is the single most common way this topic is lost in a free-response part that explicitly demanded the definition. Cancel first. Then set h to zero. Then say what the surviving expression is.
| Stage | What you write | What goes wrong if skipped |
|---|---|---|
| Step | f(x+h) expanded | The numerator stays unresolved |
| Subtract and divide | hf(x+h)−f(x) | The quotient is never formed |
| Cancel | Divide out the common h | Substitution gives 0/0 |
| Limit | limh→0 of the reduced form | The answer still contains h |
① Compute f′(x) from the definition for x2, then for x3, then for x4. The pattern in the coefficients is the power rule arriving on its own.
② Compute it once for x1 and once for x. Those two force the two algebraic moves the exam likes: common denominators and conjugates.
③ Take any derivative you already know and reverse-engineer the quotient. Naming the function and the point is the same skill the MCQ disguise tests.
Core Archetype (Simplest Form): Use the definition to find f′(x) for f(x)=3x2−4x.
Standard Execution:
- Step. f(x+h)=3(x+h)2−4(x+h)=3x2+6xh+3h2−4x−4h.
- Subtract. f(x+h)−f(x)=6xh+3h2−4h.
- Divide. h6xh+3h2−4h=6x+3h−4.
- Shrink. As h→0 the expression tends to 6x−4.
- Check. The power rule gives the same answer, which is the point of having both.
Exam Variation (Context / Parameter Shift): Evaluate limx→3x−3f(x)−f(3) where f(x)=x3.
Solution: The expression is the alternate form of the definition at a=3, so the limit is f′(3)=27. Recognizing the form answers the question without any algebra, and the algebra would confirm it: x−3x3−27=x2+3x+9, which tends to 27.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
limh→0h(2+h)3−8 equals which of the following?
MCQ 2
For f(x)=x, the difference quotient simplifies to x+h+x1. What is f′(x)?
MCQ 3
A stem asks for the rate at which a population grows with respect to time. Which notation states the requested quantity?
FRQ 4
Let f be the function defined by f(x)=x+12 for x>−1.
(a) Use the definition of the derivative to find f′(x). Show the algebraic steps that lead to your answer.
(b) Find an equation for the line tangent to the graph of f at x=1.
(c) Determine whether f′(x) is increasing or decreasing for x>−1, and justify your answer.
(d) Evaluate limh→0hf(3+h)−f(3) and explain how it relates to part (a).
| Move | |
|---|---|
| "Using the definition" | Substitute, expand, cancel, then limit |
| A limit with h below | Look for a difference of one function at two inputs |
| hf(x+h)−f(x) | A function of x after the limit |
| hf(a+h)−f(a) | A number at a |
| Leftover h in the answer | The limit was never taken |
| 0/0 before substituting | The cancellation was skipped |
Strengthen this topic with AP-style practice, or continue to next topic.
Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
The derivative is a function, not a number, and the definition is how that function is built: f′(x)=limh→0hf(x+h)−f(x). Read it as a machine with three stages. Step off the base point by h, form the quotient, then let h shrink. What survives is a new function whose input is the same x and whose output is the slope at that input. The exam uses the definition in three ways. It asks you to compute a derivative from it for a simple function. It asks you to recognize a limit that has the shape of a quotient and name the function and point it belongs to. And it expects you to know that the three notations f′(x), dxdy, and dxdf(x) all name the same object. The common failure is treating the quotient as the answer, which leaves an expression in h where a function of x was required. Shrink the step last, and the answer arrives already correct.
Figure 1 — Three stages, one output. The definition takes an input, steps away from it, forms a quotient, and then collapses the step. What leaves the machine is a function, which is why the same rule can be evaluated at any input afterwards.
For f(x)=x2:
hf(x+h)−f(x)=h(x+h)2−x2=h2xh+h2=2x+h
Every term that contains h after the cancellation disappears when h→0, so f′(x)=2x. The cancellation is the whole trick. If an h survives in a denominator, the algebra stopped too early.
For f(x)=x1 the same three stages produce x(x+h)−1, which tends to −x21. For f(x)=x the quotient needs a conjugate, and the result is 2x1. The pattern is identical in all three cases: the algebra exists to remove the h from the denominator, and the limit does the rest.
| Notation | Reads as | Where it is used |
|---|---|---|
| f′(x) | the derivative of f at x | Function notation, composition, evaluation |
| dxdy | the derivative of y with respect to x | Contextual rates, implicit work, differential equations |
| dxdf(x) | the operator applied to f(x) | Statements of rules and quick computations |
The three are interchangeable in value. They differ in what they make easy to say: f′(a) names a number, dxdy names a rate against a stated input, and the operator form names the act of differentiating.
A limit such as limh→0hcos(π+h)+1 is the definition in costume. Identify the function as cosx and the base point as x=π, and the limit is f′(π)=−sinπ=0. The identification is the skill. Once the function and point are named, the value comes from any rule you trust.
Borrowed idea: Every definition computation runs the same three stages. Step. Quotient. Shrink. The algebra in the middle exists for one purpose only. It removes h from the denominator so that substitution stops producing zero over zero. That is the entire job. A student who expands carefully but takes the limit before cancelling writes an expression that is undefined at the point of interest and then reports a number pulled from nowhere, which is the single most common way this topic is lost in a free-response part that explicitly demanded the definition. Cancel first. Then set h to zero. Then say what the surviving expression is.
| Stage | What you write | What goes wrong if skipped |
|---|---|---|
| Step | f(x+h) expanded | The numerator stays unresolved |
| Subtract and divide | hf(x+h)−f(x) | The quotient is never formed |
| Cancel | Divide out the common h | Substitution gives 0/0 |
| Limit | limh→0 of the reduced form | The answer still contains h |
① Compute f′(x) from the definition for x2, then for x3, then for x4. The pattern in the coefficients is the power rule arriving on its own.
② Compute it once for x1 and once for x. Those two force the two algebraic moves the exam likes: common denominators and conjugates.
③ Take any derivative you already know and reverse-engineer the quotient. Naming the function and the point is the same skill the MCQ disguise tests.
Core Archetype (Simplest Form): Use the definition to find f′(x) for f(x)=3x2−4x.
Standard Execution:
- Step. f(x+h)=3(x+h)2−4(x+h)=3x2+6xh+3h2−4x−4h.
- Subtract. f(x+h)−f(x)=6xh+3h2−4h.
- Divide. h6xh+3h2−4h=6x+3h−4.
- Shrink. As h→0 the expression tends to 6x−4.
- Check. The power rule gives the same answer, which is the point of having both.
Exam Variation (Context / Parameter Shift): Evaluate limx→3x−3f(x)−f(3) where f(x)=x3.
Solution: The expression is the alternate form of the definition at a=3, so the limit is f′(3)=27. Recognizing the form answers the question without any algebra, and the algebra would confirm it: x−3x3−27=x2+3x+9, which tends to 27.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
limh→0h(2+h)3−8 equals which of the following?
MCQ 2
For f(x)=x, the difference quotient simplifies to x+h+x1. What is f′(x)?
MCQ 3
A stem asks for the rate at which a population grows with respect to time. Which notation states the requested quantity?
FRQ 4
Let f be the function defined by f(x)=x+12 for x>−1.
(a) Use the definition of the derivative to find f′(x). Show the algebraic steps that lead to your answer.
(b) Find an equation for the line tangent to the graph of f at x=1.
(c) Determine whether f′(x) is increasing or decreasing for x>−1, and justify your answer.
(d) Evaluate limh→0hf(3+h)−f(3) and explain how it relates to part (a).
| Move | |
|---|---|
| "Using the definition" | Substitute, expand, cancel, then limit |
| A limit with h below | Look for a difference of one function at two inputs |
| hf(x+h)−f(x) | A function of x after the limit |
| hf(a+h)−f(a) | A number at a |
| Leftover h in the answer | The limit was never taken |
| 0/0 before substituting | The cancellation was skipped |
Strengthen this topic with AP-style practice, or continue to next topic.