AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/2/2-6)
Unit 2 · Topic 2.6
2.6 Derivative Rules: Constant, Sum, Difference, and Constant Multiple
A few simple rules let you differentiate any polynomial term by term. The derivative of a constant is 0, the derivative of a sum or difference is the sum or difference of the derivatives, and a constant multiplier just comes along for the ride.
Key terms
- constant rule
- sum rule
- difference rule
- constant multiple rule
- polynomial
The rules
Together with the power rule, these four rules handle any polynomial. Each one follows from the limit definition and the limit properties in 1.5.
- Constant rule: d/dx (k) = 0. A constant's graph is a horizontal line, which has slope 0.
- Constant multiple rule: d/dx [k·f(x)] = k·f′(x). Stretching a graph vertically by k multiplies every slope by k.
- Sum rule: d/dx [f(x) + g(x)] = f′(x) + g′(x).
- Difference rule: d/dx [f(x) − g(x)] = f′(x) − g′(x).
Differentiating a polynomial term by term
Go one term at a time. For f(x) = 4x³ − 7x² + 2x − 9: the derivative of 4x³ is 12x², of −7x² is −14x, of 2x is 2, and of −9 is 0. So f′(x) = 12x² − 14x + 2.
Notice that adding a constant to a function doesn't change its derivative. Shifting a graph up or down doesn't change any slopes.
With practice you can do this in your head, one term at a time: multiply the coefficient by the exponent, then lower the exponent by one. Just keep track of signs, since a negative coefficient stays negative.
Simplify before you differentiate
These rules don't cover products or quotients directly. But you can often rewrite a product or quotient as a sum first:
- Expand products: (x + 1)²·x = x³ + 2x² + x, so the derivative is 3x² + 4x + 1.
- Split a quotient over a single term: (2x³ − 5x + 4)/x = 2x² − 5 + 4x⁻¹, so the derivative is 4x − 4x⁻².
Using the derivative
Once you have f′(x), you can answer slope questions. The slope at a point is f′(a). The tangent line is horizontal where f′(x) = 0. The tangent line is parallel to a given line where f′(x) equals that line's slope.
Example of a parallel tangent: for f(x) = x² + x, where is the tangent line parallel to y = 5x + 2? Parallel lines have equal slopes, so solve f′(x) = 2x + 1 = 5. That gives x = 2, at the point (2, 6).
These rules also work on functions given only by values. If f′(3) = 4 and g′(3) = −1, then the derivative of 2f(x) − 5g(x) at x = 3 is 2(4) − 5(−1) = 13.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Rewrite a quotient as a sum
Find g′(x) for g(x) = (2x³ − 5x + 4) / x.
Show the solutionHide the solution
- Step 1: Divide each term by x: g(x) = 2x² − 5 + 4x⁻¹.
- Step 2: Differentiate term by term: 4x − 0 + 4(−1)x⁻².
- Step 3: Simplify: g′(x) = 4x − 4/x².
Answer: g′(x) = 4x − 4/x²
- Example 2
Where is the tangent line horizontal?
Find all points where the graph of f(x) = x³ − 6x² + 9x + 1 has a horizontal tangent line.
Show the solutionHide the solution
- Step 1: Horizontal tangent means slope 0, so solve f′(x) = 0.
- Step 2: f′(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3).
- Step 3: f′(x) = 0 at x = 1 and x = 3.
- Step 4: Find the y-values: f(1) = 1 − 6 + 9 + 1 = 5 and f(3) = 27 − 54 + 27 + 1 = 1.
Answer: At the points (1, 5) and (3, 1).
- Example 3
Trap: differentiating a product one factor at a time
A student finds the derivative of y = (x + 1)²·x by differentiating each factor: 2(x + 1)·1 = 2x + 2. What's the correct derivative?
Show the solutionHide the solution
- Step 1: The sum rule works for sums, not products. You can't differentiate each factor separately and multiply.
- Step 2: Expand first: (x + 1)²·x = (x² + 2x + 1)·x = x³ + 2x² + x.
- Step 3: Differentiate term by term: y′ = 3x² + 4x + 1.
- Step 4: Check at x = 1: the correct slope is 3 + 4 + 1 = 8, but the student's answer gives 4, so the shortcut is wrong.
Answer: y′ = 3x² + 4x + 1
Common mistakes
- Leaving a constant term in the derivative. The derivative of −9 is 0.
- Differentiating a product as the product of the derivatives. Expand it, or use the product rule (2.8).
- Splitting a quotient like (x² + 1)/(x + 3) into separate terms. You can only split over a single-term denominator.
On the exam
- Questions about horizontal tangents and slopes at a point are common in the no-calculator multiple-choice section.
- When functions are given by tables, these rules let you combine derivative values, like finding the derivative of 3f(x) + g(x) at x = 2.
Connected topics
Videos
Check yourself
4 questions on 2.6 Derivative Rules: Constant, Sum, Difference, and Constant Multiple. Pick an answer to see if you got it, and why.
What is an equation of the line tangent to the graph of f(x) = x³ − 4x² + 3 at x = 2?
Let g(x) = 5f(x) − 2x³ + π², where f is a differentiable function with f′(1) = 3. What is g′(1)?
The line tangent to the graph of f at x = 2 is y = 4x − 5. Let g(x) = 3f(x) − 2x + 7. Which of the following is an equation of the line tangent to the graph of g at x = 2 ?
The graph of y = x² + bx + c is tangent to the line y = 3x at the point (1, 3), where b and c are constants. What are b and c ?
0 of 4 answered