Three Rules in One Pass
Figure 1 — Each card states one rule and applies it immediately. The rules act on single blocks, which is why a polynomial can be differentiated left to right without regrouping anything.
Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
Three rules do most of the work in this unit, and none of them requires a limit once the power rule is in hand. The derivative of a constant is zero, because a horizontal line has no slope. A constant multiple passes through the derivative unchanged, because scaling the output scales every slope by the same factor. A sum or difference differentiates term by term, because the limit of a sum is the sum of the limits. Those three, plus the power rule, differentiate every polynomial in one line. What they do not do is equally important: there is no rule that lets you differentiate a product, a quotient, or a composition by handling the pieces separately, and those three cases arrive in the next sections.
Figure 1 — Each card states one rule and applies it immediately. The rules act on single blocks, which is why a polynomial can be differentiated left to right without regrouping anything.
Each rule is a limit statement, and each follows from the limit laws:
dxd[c]=limh→0hc−c=0,dxd[cf]=climh→0hf(x+h)−f(x)=cf′(x)
dxd[f±g]=limh→0h[f(x+h)−f(x)]±[g(x+h)−g(x)]=f′(x)±g′(x)
Nothing in these arguments is specific to polynomials. They hold for every differentiable function, which is why the rules keep working when the pieces become trigonometric or exponential.
| You might hope | The truth | Where it is handled |
|---|---|---|
| (fg)′=f′g′ | False in general | Product rule |
| (gf)′=g′f′ | False in general | Quotient rule |
| (f∘g)′=f′∘g′ | False in general | Chain rule |
A single counterexample settles each row. With f(x)=x and g(x)=x, the product is x2, whose derivative is 2x, while f′g′=1. The failure is not a technicality; it is the reason the next three sections exist.
Because coefficients pass through, you can often reconstruct the original from the derivative. If f′(x)=6x2−4, then f must contain 2x3−4x plus some constant, since differentiating 2x3 gives 6x2 and differentiating −4x gives −4. That reconstruction is antidifferentiation, and it works for exactly the same reason the rules do.
Borrowed idea: Work from left to right, and treat each block as though the others were absent. A coefficient stays exactly where it is while its block is differentiated. A constant standing alone disappears, since it adds nothing to any slope. A subtraction sign belongs to the block that follows it, and a dropped sign stays invisible until the substitution step. When the expression is long, cover the other terms and differentiate what you can see. Then count the terms in your answer. Only the constant should have gone missing, and any other gap means a block was skipped.
| Given | Derivative | What the rules did |
|---|---|---|
| 7 | 0 | Constant rule |
| 7x2 | 14x | Multiple plus power |
| 7x2−7 | 14x | Constant rule removed the tail |
| 4x3 | 43x2 | Multiple 41 passed through |
| 2x−x2 | 2−2x | Difference, sign attached to the second block |
① Differentiate five polynomials with a constant term, then count the terms in each answer. The count should drop by exactly one, and that check catches most slips faster than re-deriving.
② Write a function whose derivative is 4x3−6x+1, then check it by differentiating back. Reconstructing forces you to see the rules in reverse.
③ Take a product such as x⋅x and differentiate it two ways, once correctly and once by the false rule (fg)′=f′g′. The difference between 2x and 1 is the reason the next section exists.
Core Archetype (Simplest Form): Differentiate f(x)=3x4−2x2+5x−9.
Standard Execution:
- Term by term. 3x4→12x3, −2x2→−4x, 5x→5, −9→0.
- Assemble. f′(x)=12x3−4x+5.
- Count. Four terms in, three out, and the missing one is the constant.
- Check a value. At x=1, f′=13, and the graph of f rises steeply there, which matches.
Exam Variation (Context / Parameter Shift): A quantity is modeled by Q(t)=4t4−3t2+8 for t≥0. Find the rate of change at t=2, with units of Q per unit of t, and determine whether Q is increasing or decreasing there.
Solution: The coefficient 41 passes through: Q′(t)=t3−6t. At t=2 this gives 8−12=−4 units of Q per unit of t. The negative sign means Q is decreasing at that instant, and the size says it is falling four units per unit of time.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
If f(x)=4x3−6x+11, what is f′(x)?
MCQ 2
If g(x)=2x5, what is g′(x)?
MCQ 3
Which statement is true for differentiable functions f and g?
A quantity is modeled by Q(t)=4t4−3t2+8 for 0≤t≤4, where Q is measured in liters and t in minutes.
(a) Find Q′(t).
(b) Find the rate of change of Q at t=2 minutes, with units, and state whether Q is increasing or decreasing there.
(c) Find every time in 0<t<4 at which Q′(t)=0.
(d) Explain why the constant 8 does not appear in Q′(t), and state what the value of Q at t=0 contributes to the rate of change.
| Statement | Example | |
|---|---|---|
| Constant | dxd[c]=0 | dxd[9]=0 |
| Constant multiple | dxd[cf]=cf′ | dxd[5x3]=15x2 |
| Sum | dxd[f+g]=f′+g′ | dxd[x2+x]=2x+1 |
| Difference | dxd[f−g]=f′−g′ | dxd[x2−x]=2x−1 |
| False product | (fg)′=f′g′ | x⋅x gives 2x, not 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
Three rules do most of the work in this unit, and none of them requires a limit once the power rule is in hand. The derivative of a constant is zero, because a horizontal line has no slope. A constant multiple passes through the derivative unchanged, because scaling the output scales every slope by the same factor. A sum or difference differentiates term by term, because the limit of a sum is the sum of the limits. Those three, plus the power rule, differentiate every polynomial in one line. What they do not do is equally important: there is no rule that lets you differentiate a product, a quotient, or a composition by handling the pieces separately, and those three cases arrive in the next sections.
Figure 1 — Each card states one rule and applies it immediately. The rules act on single blocks, which is why a polynomial can be differentiated left to right without regrouping anything.
Each rule is a limit statement, and each follows from the limit laws:
dxd[c]=limh→0hc−c=0,dxd[cf]=climh→0hf(x+h)−f(x)=cf′(x)
dxd[f±g]=limh→0h[f(x+h)−f(x)]±[g(x+h)−g(x)]=f′(x)±g′(x)
Nothing in these arguments is specific to polynomials. They hold for every differentiable function, which is why the rules keep working when the pieces become trigonometric or exponential.
| You might hope | The truth | Where it is handled |
|---|---|---|
| (fg)′=f′g′ | False in general | Product rule |
| (gf)′=g′f′ | False in general | Quotient rule |
| (f∘g)′=f′∘g′ | False in general | Chain rule |
A single counterexample settles each row. With f(x)=x and g(x)=x, the product is x2, whose derivative is 2x, while f′g′=1. The failure is not a technicality; it is the reason the next three sections exist.
Because coefficients pass through, you can often reconstruct the original from the derivative. If f′(x)=6x2−4, then f must contain 2x3−4x plus some constant, since differentiating 2x3 gives 6x2 and differentiating −4x gives −4. That reconstruction is antidifferentiation, and it works for exactly the same reason the rules do.
Borrowed idea: Work from left to right, and treat each block as though the others were absent. A coefficient stays exactly where it is while its block is differentiated. A constant standing alone disappears, since it adds nothing to any slope. A subtraction sign belongs to the block that follows it, and a dropped sign stays invisible until the substitution step. When the expression is long, cover the other terms and differentiate what you can see. Then count the terms in your answer. Only the constant should have gone missing, and any other gap means a block was skipped.
| Given | Derivative | What the rules did |
|---|---|---|
| 7 | 0 | Constant rule |
| 7x2 | 14x | Multiple plus power |
| 7x2−7 | 14x | Constant rule removed the tail |
| 4x3 | 43x2 | Multiple 41 passed through |
| 2x−x2 | 2−2x | Difference, sign attached to the second block |
① Differentiate five polynomials with a constant term, then count the terms in each answer. The count should drop by exactly one, and that check catches most slips faster than re-deriving.
② Write a function whose derivative is 4x3−6x+1, then check it by differentiating back. Reconstructing forces you to see the rules in reverse.
③ Take a product such as x⋅x and differentiate it two ways, once correctly and once by the false rule (fg)′=f′g′. The difference between 2x and 1 is the reason the next section exists.
Core Archetype (Simplest Form): Differentiate f(x)=3x4−2x2+5x−9.
Standard Execution:
- Term by term. 3x4→12x3, −2x2→−4x, 5x→5, −9→0.
- Assemble. f′(x)=12x3−4x+5.
- Count. Four terms in, three out, and the missing one is the constant.
- Check a value. At x=1, f′=13, and the graph of f rises steeply there, which matches.
Exam Variation (Context / Parameter Shift): A quantity is modeled by Q(t)=4t4−3t2+8 for t≥0. Find the rate of change at t=2, with units of Q per unit of t, and determine whether Q is increasing or decreasing there.
Solution: The coefficient 41 passes through: Q′(t)=t3−6t. At t=2 this gives 8−12=−4 units of Q per unit of t. The negative sign means Q is decreasing at that instant, and the size says it is falling four units per unit of time.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
If f(x)=4x3−6x+11, what is f′(x)?
MCQ 2
If g(x)=2x5, what is g′(x)?
MCQ 3
Which statement is true for differentiable functions f and g?
A quantity is modeled by Q(t)=4t4−3t2+8 for 0≤t≤4, where Q is measured in liters and t in minutes.
(a) Find Q′(t).
(b) Find the rate of change of Q at t=2 minutes, with units, and state whether Q is increasing or decreasing there.
(c) Find every time in 0<t<4 at which Q′(t)=0.
(d) Explain why the constant 8 does not appear in Q′(t), and state what the value of Q at t=0 contributes to the rate of change.
| Statement | Example | |
|---|---|---|
| Constant | dxd[c]=0 | dxd[9]=0 |
| Constant multiple | dxd[cf]=cf′ | dxd[5x3]=15x2 |
| Sum | dxd[f+g]=f′+g′ | dxd[x2+x]=2x+1 |
| Difference | dxd[f−g]=f′−g′ | dxd[x2−x]=2x−1 |
| False product | (fg)′=f′g′ | x⋅x gives 2x, not 1 |
Strengthen this topic with AP-style practice, or continue to next topic.