The big idea: To differentiate a single power term, multiply by the power and then reduce the power by one.
That is all there is to it.
- the coefficient (the number in front)
- the power (the exponent)
- the original term
- the derivative — multiply by n, reduce power by 1
Two steps every time: bring the power down (multiply the coefficient by it), then subtract 1 from the power.
Three examples — watch the pattern
Differentiate (a) x⁵ (b) 3x² (c) 4x⁷
Step by step
- (a) Multiply coefficient 1 by the power 5, reduce power from 5 to 4.
- (b) Multiply coefficient 3 by the power 2 = 6, reduce power from 2 to 1.
- (c) Multiply coefficient 4 by the power 7 = 28, reduce power from 7 to 6.
Final answer
5x⁴, 6x, 28x⁶
Most common error: Students often forget to multiply by the power, and just subtract 1 from it.
For example, writing d/dx[3x²] = 3x (wrong) instead of 6x (correct).
Always multiply first, then reduce.
Interactive graph: power rule and tangent slope
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The big idea: A polynomial is a sum of power terms.
To differentiate a polynomial, apply the power rule to each term separately.
Add and subtract the results.
Worked example 1
Find f′(x) if f(x) = 4x³ + 2x² − 5x + 7.
Step by step
- Differentiate each term separately.
- Apply the power rule to each term.
- Write the final answer.
Final answer
f′(x) = 12x² + 4x − 5
Worked example 2
Find dy/dx if y = 6x⁴ − 3x + 8.
Step by step
- Apply the power rule term by term.
Final answer
dy/dx = 24x³ − 3
Two special cases to memorise: Constant term: The derivative of any number on its own is 0. d/dx[7] = 0. Linear term (ax): The derivative of ax is just a. d/dx[−5x] = −5.
You can verify: −5x = −5x¹, so d/dx = (1)(−5)x⁰ = −5 × 1 = −5.
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The big idea: Once you have f′(x), you can find the gradient at any specific point by substituting the x-value into f′(x).
The result is a single number — the gradient of the curve there.
Worked example
Find the gradient of f(x) = 3x² − 4x + 1 at the point where x = 2.
Step by step
- Step 1: Differentiate f(x) to get f′(x).
- Step 2: Substitute x = 2 into f′(x).
Final answer
The gradient at x = 2 is 8.
The most common exam error: Substituting x = 2 into f(x) instead of f′(x).
f(2) = 3(4) − 4(2) + 1 = 5 — this is the y-value of the curve at x = 2, NOT the gradient.
Always differentiate first, THEN substitute.
Extension: find x given the gradient
For g(x) = x³ − 3x, find the x-values where the gradient equals 9.
Step by step
- Step 1: Find g′(x).
- Step 2: Set g′(x) = 9.
- Step 3: Solve for x.
Final answer
x = 2 and x = −2.
IB marks the process, not just the answer: Write f′(x) = ... on its own line before substituting.
IB gives a method mark just for showing the correct derivative, even if you then make an arithmetic error when substituting.
The big idea: A few cases need special attention: constant terms, linear terms, and expressions that look unusual but still obey the power rule once you expand them.
| Expression | Derivative | Why |
|---|---|---|
| d/dx[5] | 0 | A constant has no x to change — its rate of change is 0 |
| d/dx[7x] | 7 | 7x = 7x¹ → bring down the 1, reduce to x⁰ = 1, so 7 × 1 = 7 |
| d/dx[x] | 1 | x = 1·x¹ → same logic → 1 |
| d/dx[x³] | 3x² | Standard power rule |
| d/dx[−x⁴] | −4x³ | Coefficient is −1. Multiply: (4)(−1) = −4 |
Expanding brackets first
Find dy/dx if y = x(3x − 2).
Step by step
- Expand the brackets first — you cannot differentiate a product directly (at this level).
- Now apply the power rule term by term.
Final answer
dy/dx = 6x − 2
Expand before differentiating: If you see brackets like x(3x − 2) or (x + 2)², always expand first, then differentiate.
Trying to differentiate without expanding leads to errors at this level.
Negative powers — for completeness
Differentiate f(x) = 3x⁻² using the power rule.
Step by step
- The power rule works for negative powers too. Bring down the power, reduce by 1.
Final answer
f′(x) = −6x⁻³