Practice Maths

Differentiating Polynomials

Key Terms

The power rule: d/dx(xn) = nxn−1 for any real n.
Constant rule
: d/dx(c) = 0. Constants vanish when differentiated.
Constant multiple rule
: d/dx[cf(x)] = c · f′(x).
Sum/difference rule
: d/dx[f(x) ± g(x)] = f′(x) ± g′(x).
To differentiate a polynomial: apply the power rule to each term.
Every polynomial can be differentiated term by term. The derivative of a degree-n polynomial is degree n−1.
Differentiation Rules for Polynomials
d/dx(xn) = nxn−1    (Power Rule) d/dx(c) = 0    (Constant Rule) d/dx[cf(x)] = c f′(x)    (Constant Multiple) d/dx[f(x)±g(x)] = f′(x)±g′(x)    (Sum/Difference)
f(x) f′(x) Rule used
70Constant
x55x4Power
4x³12x²Power × constant multiple
3x²−5x+26x−5Sum/difference
x4−2x³+7x4x³−6x²+7All rules

y = x³ (blue) and its derivative y′ = 3x² (orange) on the same axes.

x y 1 −1 y=x³ y′=3x²
Hot Tip Bring the power down and reduce it by 1. For example, d/dx(5x4): bring down 4, multiply: 4×5=20, reduce power: x3. Answer: 20x3. Always write each term in the form axn before applying the power rule. If the function is given as a product, expand it first.

Worked Example 1 — Differentiating a polynomial

Question: Differentiate y = 4x5 − 3x² + 7x − 2.

dy/dx = 4(5x4) − 3(2x) + 7(1) − 0

dy/dx = 20x4 − 6x + 7

Worked Example 2 — Expand then differentiate

Question: Differentiate y = x(x − 3)².

Expand: (x − 3)² = x² − 6x + 9, so y = x(x² − 6x + 9) = x³ − 6x² + 9x

dy/dx = 3x² − 12x + 9

Why the Power Rule Works

The power rule d/dx(xn) = nxn−1 is not just a formula to memorise — it emerges directly from first principles. For f(x) = xn, applying the limit definition and using the binomial theorem gives f(x+h) − f(x) = nxn−1h + lower powers of h. After dividing by h and taking the limit, all terms with h vanish, leaving nxn−1. The key insight is that the dominant term in the numerator after expansion always has exactly one factor of h, so it survives the division. Every other term has h² or higher and therefore vanishes. This is why the power rule “brings down the exponent and reduces it by one” — both actions are visible in the algebra.

Term-by-Term Differentiation and Why It Holds

The sum/difference rule states that you can differentiate a polynomial by differentiating each term separately. This works because limits distribute across addition and subtraction: lim[f(x+h) + g(x+h) − f(x) − g(x)] / h splits into the sum of two separate limits. In practice, this means that for y = 4x5 − 3x² + 7x − 2, you apply the power rule to each term independently: 4(5x4) − 3(2x) + 7(1) − 0 = 20x4 − 6x + 7. The constant −2 differentiates to zero because it contributes no change to the gradient — a horizontal shift doesn’t alter slope. The degree of the derivative is always one less than the degree of the original polynomial.

Exam Tip: Before differentiating, expand any products or factorisations. The power rule only applies directly to expressions in the form axn. For example, y = x(x − 3)² must be expanded to y = x³ − 6x² + 9x first, then differentiated term by term to get dy/dx = 3x² − 12x + 9. Students who try to differentiate without expanding make errors involving the product rule (which is a Year 12 technique).

The Second Derivative

If you differentiate f′(x), you obtain the second derivative f′′(x) (also written d²y/dx²). For f(x) = 2x³ − 6x² + 4, the first derivative is f′(x) = 6x² − 12x, and differentiating again gives f′′(x) = 12x − 12. The second derivative measures the rate of change of the gradient — in other words, how quickly the slope is changing. Where f′′(x) > 0, the curve is concave up (like a cup); where f′′(x) < 0, it is concave down. The point where f′′(x) = 0 and concavity changes is called an inflection point. For a cubic, the inflection point can be found by solving f′′(x) = 0.

Finding Tangent Lines and Parallel Tangents

A tangent to a curve y = f(x) at the point (a, f(a)) has gradient m = f′(a) and equation y − f(a) = f′(a)(x − a). To find where two tangents are parallel to a given line y = mx + c, set f′(x) = m and solve for x, then find the corresponding y-values on the curve. For example, to find tangents to y = x³ − 3x² − 9x + 2 that are parallel to y = −12x + 5 (gradient −12): set 3x² − 6x − 9 = −12, giving 3x² − 6x + 3 = 0, so x² − 2x + 1 = 0, thus (x−1)² = 0 and x = 1. Substitute back: y = 1 − 3 − 9 + 2 = −9. Tangent: y + 9 = −12(x − 1), giving y = −12x + 3.

Exam Tip: A horizontal tangent occurs where dy/dx = 0. To find where the gradient equals a given value m, set dy/dx = m and solve. These are two of the most common exam question types in calculus. Always substitute the x-value back into the original function (not the derivative) to find the y-coordinate of the point on the curve.

Interpreting the Derivative in Context

In real-world problems, the derivative gives the rate of change of one quantity with respect to another. If A = x² is the area of a square with side length x, then dA/dx = 2x is the rate at which area changes as side length increases. At x = 3, dA/dx = 6, meaning the area increases at approximately 6 cm² per cm of additional side length. For volume V = x³, dV/dx = 3x². At x = 2, the rate of change of volume with respect to side length is 12 cm³ per cm. Interpreting derivatives in context requires specifying units and identifying what the independent and dependent variables represent.

Mastery Practice

See Answers ➔
  1. Fluency

    Differentiate each function using the power rule.

    1. (a) f(x) = x6
    2. (b) f(x) = x4
    3. (c) f(x) = x10
    4. (d) f(x) = x
    5. (e) f(x) = 1 (constant)
  2. Fluency

    Differentiate each function.

    1. (a) f(x) = 5x³
    2. (b) f(x) = −2x4
    3. (c) f(x) = ½x²
    4. (d) f(x) = 8x
    5. (e) f(x) = −3x5
  3. Fluency

    Differentiate each polynomial.

    1. (a) y = 3x² + 5x − 4
    2. (b) y = x³ − 6x² + 2
    3. (c) y = 4x4 − 3x³ + 2x² − x + 7
    4. (d) y = −2x³ + 9x
  4. Fluency

    Find f′(x) for each function, then evaluate f′(x) at the given value.

    1. (a) f(x) = 3x² − 2x at x = 4
    2. (b) f(x) = x³ + x at x = 2
    3. (c) f(x) = −x² + 6x − 1 at x = 3
    4. (d) f(x) = 2x4 − 5x² + 1 at x = 1
  5. Understanding

    Expand then differentiate.

    Method: Expand the expression into individual terms first, then differentiate term by term.
    1. (a) y = (x + 2)(x − 5)
    2. (b) y = (2x − 1)²
    3. (c) y = x²(x − 3)
    4. (d) y = (x + 1)(x − 2)(x + 3)
  6. Understanding

    Finding where the gradient has a given value.

    1. (a) For y = x² − 4x + 1, find the x-value where the gradient is 6.
    2. (b) For y = x³ − 3x, find the x-values where the gradient is 0.
    3. (c) For y = 2x³ − 3x² − 12x, find the x-values where the gradient is −12.
  7. Understanding

    Gradient of a tangent at a specific point.

    1. (a) Find the gradient of the tangent to y = x³ − 2x at (2, 4).
    2. (b) Find the gradient of the tangent to y = 3x² − x + 5 at (−1, 9).
    3. (c) Determine whether the tangent to y = x4 − 4x at (1, −3) has a positive or negative gradient and find its equation.
  8. Understanding

    Working with the second derivative.

    1. (a) For y = 2x³ − 6x² + 4, find dy/dx and d²y/dx².
    2. (b) For f(x) = x4 − 2x², find f′(x) and f′′(x).
    3. (c) Find the value of x where f′′(x) = 0 for f(x) = x³ − 6x² + 9x.
  9. Problem Solving

    Applying differentiation to find tangents and parallels.

    Challenge. Consider y = x³ − 3x² − 9x + 2.
    1. (a) Find dy/dx.
    2. (b) Find the equations of any tangents to the curve that are parallel to the line y = −12x + 5.
    3. (c) Find the x-values where the curve has a horizontal tangent.
  10. Problem Solving

    Area and rate problems.

    Challenge. A square has side length x cm. Its area is A = x² and its volume (as a cube) is V = x³.
    1. (a) Find dA/dx. Interpret this in terms of how area changes with side length.
    2. (b) Find dV/dx. What does dV/dx represent at x = 3?
    3. (c) A cuboid has volume V(x) = x³ − 2x² + 3x. Find the rate at which its volume changes at x = 2.