Same Rule, Different Costumes
Figure 1 — Each row is one term in three states. The middle column is the only real work: once the term is written as a power, the derivative follows in a single step and the rule itself never changes.
Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
The power rule is stated for a constant exponent, and "constant" is broader than "whole number". dxdxn=nxn−1 holds for every real n, which means roots and reciprocals are already covered once they are rewritten as powers: x=x1/2 and x31=x−3. The derivation does not assume n is an integer either. Expand (x+h)n, and every term after the first carries a factor of h, so the limit leaves exactly nxn−1. Two consequences matter on the exam. The rewrite is the work, not the differentiation. And a fractional power can fail at the endpoint of its domain, where the slope becomes infinite rather than finite.
Figure 1 — Each row is one term in three states. The middle column is the only real work: once the term is written as a power, the derivative follows in a single step and the rule itself never changes.
| Written as | As a power | Derivative | Derivative, original form |
|---|---|---|---|
| x | x1/2 | 21x−1/2 | 2x1 |
| 3x | x1/3 | 31x−2/3 | 33x21 |
| x1 | x−1 | −x−2 | −x21 |
| x41 | x−4 | −4x−5 | −x54 |
| xx | x3/2 | 23x1/2 | 23x |
The right-hand column is what the answer key usually prints, and the two middle columns are what you actually compute. Translating between them is a skill in itself, and it is the one that decides whether an otherwise correct derivative matches an option.
Expand (x+h)n with the binomial theorem and subtract xn:
(x+h)n−xn=nxn−1h+2n(n−1)xn−2h2+⋯
Divide by h and every term after the first still carries a factor of h:
limh→0h(x+h)n−xn=nxn−1+0+⋯=nxn−1
The exponent drops because exactly one of the n factors is replaced by the increment, and there are n ways to choose it. Nothing in that counting argument requires n to be a whole number, which is why the fractional case inherits the rule rather than needing a separate proof.
For f(x)=x the derivative 2x1 is undefined at x=0, and the graph agrees: the tangent there is vertical, so no finite slope exists. This is the vertical-tangent failure from the previous topic, appearing inside a formula instead of a picture. The lesson is that the rule reports a slope wherever the rewritten power is differentiable, and the endpoint of a root's domain has to be checked separately.
Borrowed idea: Rewrite before you differentiate. Every root and every reciprocal is a power wearing a costume. Turn the root into a fractional exponent, turn the reciprocal into a negative one, and the rule runs unchanged. Does the exponent have to be a whole number? No, the same algebra works for any constant, which is why the rewrite is worth the ten seconds. What about the endpoint where a root starts? There the slope may be infinite, and the derivative simply does not exist. So check where the expression lives before you report a number.
| You see | Rewrite | Derivative | Watch for |
|---|---|---|---|
| x | x1/2 | 21x−1/2 | slope infinite at 0 |
| x1 | x−1/2 | −21x−3/2 | sign and domain |
| x25 | 5x−2 | −10x−3 | coefficient survives |
| x2x | x5/2 | 25x3/2 | add the exponents first |
| xx3 | x2 | 2x | simplify before differentiating |
① Write ten expressions in mixed costumes, convert each to a single power, and differentiate. Then convert every answer back to the original costume and check that the two forms agree at a test value.
② Differentiate x three ways: by the rule, by the definition with a conjugate, and by reading the graph at x=4. All three should give 41.
③ Find where f(x)=x2/3 has a finite derivative and where it does not. The formula gives 32x−1/3, which is undefined at 0, and that single point is the cusp from the previous topic.
Core Archetype (Simplest Form): Differentiate f(x)=3x−x2.
Standard Execution:
- Rewrite. f(x)=3x1/2−2x−1.
- Differentiate. f′(x)=23x−1/2+2x−2.
- Translate back. f′(x)=2x3+x22.
- Check. At x=1 both forms give 3.5, and the original function is decreasing in its reciprocal term, which matches the positive second term.
Exam Variation (Context / Parameter Shift): Water drains from a tank so that the depth at time t minutes is D(t)=129−t centimeters for 0≤t<9. Find the rate at which the depth is changing at t=5, with units, and explain what happens as t→9−.
Solution: Rewrite as D(t)=12(9−t)1/2, so D′(t)=6(9−t)−1/2⋅(−1)=−9−t6. At t=5 this gives −3 centimeters per minute. As t→9− the denominator tends to zero, so the rate becomes unbounded: the tank empties with a vertical tangent, which is the endpoint failure again.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
If f(x)=x3/2, what is f′(x)?
MCQ 2
If g(x)=3x1, what is g′(x)?
MCQ 3
Why does the power rule apply to x1/2 even though the exponent is not a whole number?
Water drains from a tank so that the depth of water at time t minutes is modeled by D(t)=129−t centimeters for 0≤t<9.
(a) Write D(t) as a power of (9−t) and find D′(t).
(b) Find the rate at which the depth is changing at t=5 minutes. Include units.
(c) Explain why D′(9) does not exist, using the form of D′(t) rather than the graph.
(d) A second tank drains so that its depth is E(t)=12(9−t)1/3. Compare E′(5) with D′(5) and explain which tank is draining faster at that instant.
| Rewrite | Derivative | |
|---|---|---|
| xn | none | nxn−1 |
| x | x1/2 | 21x−1/2 |
| nx | x1/n | n1x1/n−1 |
| xn1 | x−n | −nx−n−1 |
| xaxb | xa+b | (a+b)xa+b−1 |
| xbxa | xa−b | (a−b)xa−b−1 |
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 power rule is stated for a constant exponent, and "constant" is broader than "whole number". dxdxn=nxn−1 holds for every real n, which means roots and reciprocals are already covered once they are rewritten as powers: x=x1/2 and x31=x−3. The derivation does not assume n is an integer either. Expand (x+h)n, and every term after the first carries a factor of h, so the limit leaves exactly nxn−1. Two consequences matter on the exam. The rewrite is the work, not the differentiation. And a fractional power can fail at the endpoint of its domain, where the slope becomes infinite rather than finite.
Figure 1 — Each row is one term in three states. The middle column is the only real work: once the term is written as a power, the derivative follows in a single step and the rule itself never changes.
| Written as | As a power | Derivative | Derivative, original form |
|---|---|---|---|
| x | x1/2 | 21x−1/2 | 2x1 |
| 3x | x1/3 | 31x−2/3 | 33x21 |
| x1 | x−1 | −x−2 | −x21 |
| x41 | x−4 | −4x−5 | −x54 |
| xx | x3/2 | 23x1/2 | 23x |
The right-hand column is what the answer key usually prints, and the two middle columns are what you actually compute. Translating between them is a skill in itself, and it is the one that decides whether an otherwise correct derivative matches an option.
Expand (x+h)n with the binomial theorem and subtract xn:
(x+h)n−xn=nxn−1h+2n(n−1)xn−2h2+⋯
Divide by h and every term after the first still carries a factor of h:
limh→0h(x+h)n−xn=nxn−1+0+⋯=nxn−1
The exponent drops because exactly one of the n factors is replaced by the increment, and there are n ways to choose it. Nothing in that counting argument requires n to be a whole number, which is why the fractional case inherits the rule rather than needing a separate proof.
For f(x)=x the derivative 2x1 is undefined at x=0, and the graph agrees: the tangent there is vertical, so no finite slope exists. This is the vertical-tangent failure from the previous topic, appearing inside a formula instead of a picture. The lesson is that the rule reports a slope wherever the rewritten power is differentiable, and the endpoint of a root's domain has to be checked separately.
Borrowed idea: Rewrite before you differentiate. Every root and every reciprocal is a power wearing a costume. Turn the root into a fractional exponent, turn the reciprocal into a negative one, and the rule runs unchanged. Does the exponent have to be a whole number? No, the same algebra works for any constant, which is why the rewrite is worth the ten seconds. What about the endpoint where a root starts? There the slope may be infinite, and the derivative simply does not exist. So check where the expression lives before you report a number.
| You see | Rewrite | Derivative | Watch for |
|---|---|---|---|
| x | x1/2 | 21x−1/2 | slope infinite at 0 |
| x1 | x−1/2 | −21x−3/2 | sign and domain |
| x25 | 5x−2 | −10x−3 | coefficient survives |
| x2x | x5/2 | 25x3/2 | add the exponents first |
| xx3 | x2 | 2x | simplify before differentiating |
① Write ten expressions in mixed costumes, convert each to a single power, and differentiate. Then convert every answer back to the original costume and check that the two forms agree at a test value.
② Differentiate x three ways: by the rule, by the definition with a conjugate, and by reading the graph at x=4. All three should give 41.
③ Find where f(x)=x2/3 has a finite derivative and where it does not. The formula gives 32x−1/3, which is undefined at 0, and that single point is the cusp from the previous topic.
Core Archetype (Simplest Form): Differentiate f(x)=3x−x2.
Standard Execution:
- Rewrite. f(x)=3x1/2−2x−1.
- Differentiate. f′(x)=23x−1/2+2x−2.
- Translate back. f′(x)=2x3+x22.
- Check. At x=1 both forms give 3.5, and the original function is decreasing in its reciprocal term, which matches the positive second term.
Exam Variation (Context / Parameter Shift): Water drains from a tank so that the depth at time t minutes is D(t)=129−t centimeters for 0≤t<9. Find the rate at which the depth is changing at t=5, with units, and explain what happens as t→9−.
Solution: Rewrite as D(t)=12(9−t)1/2, so D′(t)=6(9−t)−1/2⋅(−1)=−9−t6. At t=5 this gives −3 centimeters per minute. As t→9− the denominator tends to zero, so the rate becomes unbounded: the tank empties with a vertical tangent, which is the endpoint failure again.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
If f(x)=x3/2, what is f′(x)?
MCQ 2
If g(x)=3x1, what is g′(x)?
MCQ 3
Why does the power rule apply to x1/2 even though the exponent is not a whole number?
Water drains from a tank so that the depth of water at time t minutes is modeled by D(t)=129−t centimeters for 0≤t<9.
(a) Write D(t) as a power of (9−t) and find D′(t).
(b) Find the rate at which the depth is changing at t=5 minutes. Include units.
(c) Explain why D′(9) does not exist, using the form of D′(t) rather than the graph.
(d) A second tank drains so that its depth is E(t)=12(9−t)1/3. Compare E′(5) with D′(5) and explain which tank is draining faster at that instant.
| Rewrite | Derivative | |
|---|---|---|
| xn | none | nxn−1 |
| x | x1/2 | 21x−1/2 |
| nx | x1/n | n1x1/n−1 |
| xn1 | x−n | −nx−n−1 |
| xaxb | xa+b | (a+b)xa+b−1 |
| xbxa | xa−b | (a−b)xa−b−1 |
Strengthen this topic with AP-style practice, or continue to next topic.