Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 5 Calculus · 5.3 Differentiating powers and polynomials
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Differentiate axⁿ for any integer n, positive, zero or negative | SL, HL | "Find f′(x)" (1 to 2 marks) |
| Differentiate a sum or difference of such terms, term by term | SL, HL | "Given f(x) = 3x⁴ − 5x² + 2x − 8, find f′(x)" (2 marks) |
| Rewrite a function as a sum of powers before differentiating | SL, HL | Expand a bracket, split a fraction, write 1/x² as x⁻² (2 to 4 marks) |
| Find the gradient of a curve at a given point | SL, HL | "Find the gradient of the curve at x = 2" (2 marks) |
| Find the point or points where the gradient has a given value | SL, HL | "Find the coordinates of the points where the gradient is −9" (4 to 5 marks) |
| Use gradient information to find unknown constants | SL, HL | "The curve passes through (1, 5) with gradient 11 there; find a and b" (4 to 5 marks) |
| Interpret a derivative found this way as a rate of change | SL, HL | Paper 2 context: marginal cost, rate of change of volume (2 to 3 marks) |
Before you start
You need 5.1: the derivative f′(x) is the gradient function, and dy/dx is the rate of change of y with respect to x. You also need the laws of exponents from 1.5 and 1.7, because the whole technique rests on writing things as powers of x: 1/x³ = x⁻³, x⁰ = 1, and xᵃ × xᵇ = xᵃ⁺ᵇ. The formula booklet gives the rule itself, "f(x) = xⁿ ⇒ f′(x) = nxⁿ⁻¹", so the marks are for using it cleanly, not for remembering it. Fractional powers such as √x follow in 5.6 with exactly the same rule.
1The idea in one paragraph
The gradient of x² is 2x, of x³ is 3x², of x⁴ is 4x³. The pattern is always the same: bring the power down in front, then reduce the power by one. That is the power rule, and it works for every integer power, negative ones included. Two more facts make it a complete method for polynomials. A number multiplying a term just multiplies its derivative, and a sum is differentiated one term at a time. A constant on its own differentiates to zero, because it only moves a graph up or down and does not change any gradient. The one skill that is not automatic is getting the function into the right shape first: brackets expanded, fractions split, every term written as a number times a power of x.
2The power rule, and where it comes from
In 5.1 the chord method showed that the derivative of x² is 2x. Do the same for x³. Take a point (x, x³) and a nearby point with x-coordinate x + h:
So the derivative of x³ is 3x². The same expansion for x⁴ leaves 4x³, and for x⁵ leaves 5x⁴. The pattern is the rule.
If f(x) = xⁿ, then f′(x) = nxⁿ⁻¹. Bring the power down, reduce it by one. (Formula booklet.)
| f(x) | x² | x³ | x⁴ | x⁷ | x | 1 = x⁰ | x⁻¹ | x⁻² |
|---|---|---|---|---|---|---|---|---|
| f′(x) | 2x | 3x² | 4x³ | 7x⁶ | 1 | 0 | −x⁻² | −2x⁻³ |
Three entries in that table deserve a second look.
- x = x¹ gives 1 × x⁰ = 1. The line y = x has gradient 1 everywhere.
- 1 = x⁰ gives 0 × x⁻¹ = 0. A horizontal line has gradient 0.
- x⁻² gives −2x⁻³. For a negative power, reducing the power by one makes it more negative: −2 − 1 = −3, not −1. This is the single most common slip with this rule.
The guide states the rule for integer n, and that is all this subtopic needs; 5.6 extends it, unchanged, to any rational power.
3Constants and multiples
A constant. If f(x) = c, a number, then f′(x) = 0. Figure 1 shows why that matters inside a longer function: y = x² + 3 is y = x² moved up 3, and moving a graph up does not tilt it. At every x the two curves have exactly the same gradient, so the +3 contributes nothing to the derivative.
A multiple. If f(x) = axⁿ, then f′(x) = anxⁿ⁻¹. The number in front just comes along for the ride and multiplies the result.
Why it works: multiplying a function by 3 stretches its graph vertically by a factor of 3, so every rise is 3 times as big for the same run, and every gradient is multiplied by 3. Figure 2 shows y = x² and y = 3x² at x = 1: gradient 2 and gradient 6.
A term in x alone. If f(x) = ax, then f′(x) = a. The straight line y = ax has gradient a everywhere. So d/dx(7x) = 7 and d/dx(−x) = −1.
4Sums and differences: term by term
The derivative of a sum is the sum of the derivatives. So a polynomial is differentiated one term at a time, keeping the signs.
Write the answer with the same order and signs as the question and check each term separately. On Paper 1 this is often worth two marks, one for "at least two terms right" and one for "all right", so a single wrong term costs a mark but not both.
The rule for sums does not extend to products or quotients. The derivative of (2x − 1)(x + 3) is not 2 × 1, the product of the derivatives of the brackets. That is why the next two sections exist.
5Negative powers
A term such as 6/x² is a power of x in disguise. Write it with a negative index first, then apply the rule.
Watch the signs: a negative multiple times a negative power gives a positive term, as −2 × (−1) = +2 did above.
Figure 3 shows what the rule says about y = 1/x. Its derivative is −x⁻² = −1/x², which is negative for every x ≠ 0 because x² is always positive. Every tangent to y = 1/x slopes down, on both branches, exactly as the graph shows.
A common trap is the term 3/(2x). The 2 is in the denominator with the x, so 3/(2x) = (3/2)x⁻¹, and its derivative is −(3/2)x⁻². It is not 6x⁻¹: only the x moves up with a negative index, and the 2 stays on the bottom as a halving.
6Rewrite first: brackets and fractions
The rule only applies to a sum of terms of the form axⁿ. Anything else must be rewritten into that shape before you differentiate. Figure 4 summarises the routine; here are the two rewrites the exam uses.
Expand brackets.
Split a fraction over a single-term denominator. Divide every term on top by the bottom.
This only works because the denominator is a single term. You cannot split (x² + 1)/(x + 3) this way; a fraction like that needs the quotient rule from 5.6.
Both at once. A harder one of the kind Paper 1 sets as a "show that":
7Using the derivative
A derivative is only half the question. The other half is using it, and there are three standard uses.
The gradient at a point. Substitute the x-coordinate into f′(x), not into f(x).
For y = (x³ − 4x + 2)/x at x = 2: dy/dx = 2x − 2x⁻², so the gradient is 2(2) − 2 ÷ 4 = 4 − 0.5 = 3.5. If the question gives the y-coordinate too, as in "at the point (2, 1)", you do not need it for the gradient; you need it only when you go on to write the tangent's equation in 5.4.
The points where the gradient has a given value. Set f′(x) equal to that value and solve for x. Then find y from the original function. Figure 5 shows the case f(x) = x³ − 3x with gradient 9.
Two points, because a cubic's gradient function is a quadratic and a quadratic equation can have two solutions. Keep both unless the question restricts x. The two tangents in Figure 5 are parallel, since they share a gradient.
A special case is a gradient of zero: solving f′(x) = 0 finds the stationary points, as 5.2 did.
Finding unknown constants. Each piece of information gives an equation. The curve y = ax³ + bx passes through (1, 5) and has gradient 11 there.
"Passes through" is a statement about y; "gradient" is a statement about dy/dx. Use each on the right function.
Rates of change in context. In 5.1 the chord method estimated dV/dr ≈ 113 for a sphere of radius 3 cm. The rule gives it exactly:
In economics the same move gives marginal cost, the rate at which total cost rises with output. If a workshop's weekly cost of making q chairs is C(q) = 0.02q³ − 1.5q² + 60q + 400 dollars, then C′(q) = 0.06q² − 3q + 60, and C′(30) = 54 − 90 + 60 = 24. When 30 chairs a week are being made, cost is rising at about $24 per extra chair. Interpret it with the units and the value of q, as 5.1 set out.
8Checking with technology
On Paper 2 your GDC can check any derivative you find by hand. Pick an awkward x, say 1.7, and compare the calculator's numerical derivative at 1.7 with your f′(1.7). If they disagree beyond rounding, one term is wrong; check the negative powers first. The GDC cannot give you f′(x) as an expression, so when a question says "find f′(x)" it wants the algebra, on either paper. When it says "find the gradient at x = 1.7" on Paper 2, the GDC value on its own is acceptable: write "using GDC" and the answer.
9Where marks are lost
Reducing a negative power the wrong way. x⁻² differentiates to −2x⁻³, not −2x⁻¹. Reducing by one always makes the power smaller, so −2 goes to −3.
Leaving the constant in. The derivative of 4x² + 7 is 8x, not 8x + 7. Constants vanish.
Differentiating a product bracket by bracket. (2x − 1)(x + 3) does not differentiate to 2 × 1 = 2. Expand first, or use the product rule from 5.6.
Dividing only the first term of a fraction. (x³ − 4x + 2)/x is x² − 4 + 2/x. Writing x² − 4x + 2 divides the first term only.
Substituting into f instead of f′ for a gradient. The gradient at x = 2 is f′(2). f(2) is a height.
Finding y from f′ instead of f. After solving f′(x) = 9 for x, the point on the curve has y = f(x). Putting x into f′ just gives 9 back.
Dropping a solution. f′(x) = 9 for a cubic is a quadratic: two values of x, two points, unless the domain rules one out.
Writing 1/x as x⁻¹ and then its derivative as 1/x⁻². Keep the working in index form until the end, then convert back once, carefully.
10Work it right
- Rewrite first: expand brackets, split single-term denominators, change every 1/xⁿ into x⁻ⁿ. Write that line down; it often earns a method mark.
- Differentiate term by term: multiply by the power, then subtract one from the power.
- Check each term, especially the sign of every negative-power term, and that constants have gone.
- For a gradient at a point, substitute into f′(x), and show the substitution.
- For "where is the gradient k?", set f′(x) = k, solve, keep every solution, then get y from f(x).
- For unknown constants, one equation from the point (using f), one from the gradient (using f′), then solve simultaneously.
- In context, give the rate with units and the value it applies at.
11Try it
Marks in brackets. Q1 to Q4 are Paper 1 style, no calculator. Q5 is Paper 2 style, with a GDC.
Q1. Let f(x) = 3x⁴ − 5x² + 2x − 8.
(a) Find f′(x). 2 marks
(b) Find the gradient of the graph of f at x = −1. 2 marks
Q2. Consider y = (2x³ − 5x + 4) ÷ x, for x ≠ 0.
(a) Find dy/dx. 3 marks
(b) Find the gradient of the curve at x = 2. 1 mark
Q3. The curve C has equation y = x³ − 6x² + 5. Find the coordinates of the points on C where the gradient is −9. 5 marks
Q4. Let f(x) = ax² + b/x, where a and b are constants. The graph of f passes through the point (1, 5), and the gradient of the graph at this point is 4. Find the value of a and the value of b. 5 marks
Q5. A workshop's weekly cost, in euros, of producing q items is C(q) = 0.01q³ − 0.6q² + 15q + 200.
(a) Find C′(q). 2 marks
(b) Find C′(40), and interpret your answer in context. 2 marks
(c) Find the level of production at which the cost is increasing at €24 per item. 2 marks
12In one breath
To differentiate xⁿ, bring the power down and reduce it by one, nxⁿ⁻¹, for every integer n: x⁴ gives 4x³, x gives 1, a constant gives 0, and x⁻² gives −2x⁻³ because one less than −2 is −3. A number in front multiplies the answer, and a sum is done term by term, so 4x⁵ − 3x² + 7x − 9 gives 20x⁴ − 6x + 7. The rule only works on sums of powers, so rewrite first: expand brackets, split a fraction over a single term, and turn every 1/xⁿ into x⁻ⁿ. Then use it: substitute into f′ for a gradient at a point, set f′(x) = k and solve to find where the gradient is k (then get y from f), and use one equation from a point and one from a gradient to find unknown constants. In context the derivative is a rate, such as marginal cost or dV/dr, with units of the top per unit of the bottom.
Answers
Q1. (a) f′(x) = 12x³ − 10x + 2. A1 for two terms correct, A1 for all three and no constant.
(b) f′(−1) = 12(−1) − 10(−1) + 2 = −12 + 10 + 2 = 0. M1 for substituting x = −1 into their f′(x), A1 for 0. Substituting into f(x) scores M0.
Q2. (a) y = 2x² − 5 + 4x⁻¹, so dy/dx = 4x − 4x⁻² (or 4x − 4/x²). M1 for splitting the fraction into terms, A1 for 4x, A1 for −4x⁻². Differentiating top and bottom separately scores 0.
(b) At x = 2: 8 − 4/4 = 7. A1. Follow-through from their (a).
Q3. dy/dx = 3x² − 12x. Setting 3x² − 12x = −9 gives 3x² − 12x + 9 = 0, so x² − 4x + 3 = 0 and (x − 1)(x − 3) = 0, giving x = 1 or x = 3. Then y = 1 − 6 + 5 = 0 and y = 27 − 54 + 5 = −22. The points are (1, 0) and (3, −22). A1 for dy/dx, M1 for setting their dy/dx = −9, A1 for x = 1 and x = 3, M1 for substituting their x-values into the equation of C, A1 for both points.
Q4. From the point: a(1)² + b/1 = 5, so a + b = 5. f′(x) = 2ax − bx⁻², so f′(1) = 2a − b = 4. Adding: 3a = 9, so a = 3 and b = 2. A1 for a + b = 5, M1 for differentiating b/x as −bx⁻², A1 for 2a − b = 4, M1 for solving simultaneously, A1 for both values.
Q5. (a) C′(q) = 0.03q² − 1.2q + 15. A1 for two terms, A1 for all three with the constant 200 gone.
(b) C′(40) = 0.03(1600) − 48 + 15 = 15. When 40 items are being produced each week, the cost is increasing at €15 per extra item. A1 for 15, A1 for an interpretation with the units, "per item" and q = 40.
(c) 0.03q² − 1.2q + 15 = 24, solved on the GDC: q = 46.5 (3 s.f.), rejecting the negative root q = −6.46 because production cannot be negative. M1 for setting C′(q) = 24, A1 for 46.5 only. Accept q ≈ 46 as a whole number of items if justified.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 5.3 Differentiating powers and polynomials. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.