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Differentiation of polynomials and tangent lines (QCE Math Methods Unit 2)

Syllabus dot point

“Apply the power rule, the sum rule, and the constant-multiple rule to differentiate polynomial functions, and use the derivative to find tangent and normal line equations”

7 min read

Quick answer

Differentiate any polynomial term by term using the power rule ddxxn=nxn−1\dfrac{d}{dx} x^n = n x^{n-1} together with the sum and constant-multiple rules; the derivative at x=ax = a is the gradient of the tangent line, and the normal at the same point has gradient −1/f′(a)-1/f'(a).

Jump to a section
  1. What this dot point is asking
  2. The three rules
  3. Why the rules let you differentiate any polynomial
  4. Standard manipulations
  5. Higher derivatives
  6. Tangent and normal lines
  7. Rewriting before differentiating
  8. Exam-style questions

What this dot point is asking

QCAA wants you to differentiate polynomial functions using the three core rules (power, sum, constant multiple), and to use the derivative as the gradient function for tangent and normal line problems.

The three rules

Power rule. For any rational nn:

ddxxn=nxn−1\frac{d}{dx} x^n = n x^{n-1}

This was proved from first principles for positive integer nn in the previous dot point. The same form holds for nn rational or negative, with the standard domain restrictions.

Sum (and difference) rule. The derivative of a sum is the sum of the derivatives:

ddx[f(x)+g(x)]=f′(x)+g′(x)\frac{d}{dx} [f(x) + g(x)] = f'(x) + g'(x)

Constant multiple rule. A constant pulls outside the derivative:

ddx[cf(x)]=cf′(x)\frac{d}{dx} [c f(x)] = c f'(x)

Constant function. The derivative of a constant is zero:

ddxc=0\frac{d}{dx} c = 0

Combined, these let you differentiate any polynomial term by term.

Why the rules let you differentiate any polynomial

A polynomial is a sum of constant multiples of powers of xx, so the constant-multiple rule handles each coefficient, the power rule handles each power, and the sum rule combines them. That is why differentiation of a polynomial is purely mechanical: apply the power rule to every term, multiply by its coefficient, and add. The same three rules underpin the calculus of more complicated functions in Year 12, where they combine with the product, quotient and chain rules.

Standard manipulations

Coefficient times power
ddx(5x3)=5⋅3x2=15x2\dfrac{d}{dx} (5 x^3) = 5 \cdot 3 x^2 = 15 x^2.
Negative power
ddx(x−2)=−2x−3=−2/x3\dfrac{d}{dx} (x^{-2}) = -2 x^{-3} = -2/x^3.
Rational power
ddx(x1/2)=12x−1/2=1/(2x)\dfrac{d}{dx} (x^{1/2}) = \tfrac{1}{2} x^{-1/2} = 1/(2 \sqrt x).
Combine terms before differentiating
Expand brackets and split fractions if needed. For x3−2xx\dfrac{x^3 - 2x}{x}, first simplify to x2−2x^2 - 2, then differentiate to 2x2x.

Higher derivatives

Differentiating a polynomial gives another polynomial, which can itself be differentiated. The second derivative f′′(x)f''(x) measures the rate of change of the gradient and describes concavity: f′′(x)>0f''(x) > 0 means the curve is concave up (holding water) and f′′(x)<0f''(x) < 0 means concave down. For f(x)=2x3−5x2+4x−1f(x) = 2x^3 - 5x^2 + 4x - 1, the first derivative is 6x2−10x+46x^2 - 10x + 4 and the second is 12x−1012x - 10. Second derivatives become important when classifying stationary points and analysing motion, where they represent acceleration.

Tangent and normal lines

The tangent to y=f(x)y = f(x) at x=ax = a:

  • Point: (a,f(a))(a, f(a)).
  • Gradient: m=f′(a)m = f'(a).
  • Equation: y−f(a)=f′(a)(x−a)y - f(a) = f'(a) (x - a).

The normal at the same point:

  • Perpendicular to the tangent.
  • Gradient: −1/f′(a)-1/f'(a) (provided f′(a)≠0f'(a) \neq 0).
  • Equation: y−f(a)=−1f′(a)(x−a)y - f(a) = -\dfrac{1}{f'(a)} (x - a).

If f′(a)=0f'(a) = 0 the tangent is horizontal and the normal is vertical (x=ax = a).

Rewriting before differentiating

The power rule applies cleanly only to terms in the form cxncx^n, so an expression must first be put into that shape. Expand any brackets, split fractions over a common denominator, and rewrite roots and reciprocals as powers: x\sqrt x becomes x1/2x^{1/2}, and 1x2\dfrac{1}{x^2} becomes x−2x^{-2}. Skipping this step is the most common reason a derivative comes out wrong, because the power rule cannot be applied to a product or quotient directly at this stage of the course.

Tangent to a cubic

Find the point

Find the tangent to y=x3−4xy = x^3 - 4x at x=1x = 1. The point is (1, 1−4)=(1,−3)(1, \ 1 - 4) = (1, -3).

Find the gradient

f′(x)=3x2−4f'(x) = 3x^2 - 4, so f′(1)=3−4=−1f'(1) = 3 - 4 = -1.

Write the tangent equation

y−(−3)=−1(x−1)  ⇒  y=−x−2.y - (-3) = -1(x - 1) \;\Rightarrow\; y = -x - 2.

Markers reward the point, the derivative gradient, and the tangent in point-slope form.

Common traps
Forgetting to drop the power by one
ddxx4=4x3\dfrac{d}{dx} x^4 = 4 x^3, not 4x44x^4.
Misapplying the rule to a constant times xx
ddx(3x)=3\dfrac{d}{dx} (3x) = 3, not 3x3x or 3x03 x^0 stated awkwardly.
Confusing tangent with secant
A secant joins two points on the curve; a tangent touches at one. The derivative gives the tangent gradient.
Forgetting the negative-reciprocal rule for normals
mnormal=−1/mtangentm_{\text{normal}} = -1/m_{\text{tangent}}. A common slip is using −mtangent-m_{\text{tangent}} or 1/mtangent1/m_{\text{tangent}}.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice5 marks
Paper 2 (complex familiar). For f(x)=2x3−5x2+4x−1f(x) = 2x^3 - 5x^2 + 4x - 1, determine (a) f′(x)f'(x), (b) the gradient at x=2x = 2, (c) the equation of the tangent at (2,f(2))(2, f(2)).
Show worked answer →

(a) Power rule term by term: f′(x)=6x2−10x+4f'(x) = 6x^2 - 10x + 4.

(b) f′(2)=6(4)−10(2)+4=8f'(2) = 6(4) - 10(2) + 4 = 8.

(c) f(2)=16−20+8−1=3f(2) = 16 - 20 + 8 - 1 = 3, so the point is (2,3)(2, 3) with gradient 88. Tangent: y−3=8(x−2)y - 3 = 8(x - 2), so y=8x−13y = 8x - 13.

Markers reward the term-by-term derivative, the gradient at the point, and the tangent in point-slope form.

Practice4 marks
Paper 1 (technique). For y=2x3−xxy = \dfrac{2x^3 - x}{x} (where x≠0x \neq 0), (a) determine dydx\dfrac{dy}{dx} by first simplifying, and (b) determine the gradient of the normal at x=1x = 1.
Show worked answer →

(a) Simplify first: y=2x3−xx=2x2−1y = \dfrac{2x^3 - x}{x} = 2x^2 - 1, so dydx=4x\dfrac{dy}{dx} = 4x.

(b) At x=1x = 1 the tangent gradient is 44, so the normal gradient is −14-\dfrac{1}{4}.

Markers reward simplifying before differentiating, the derivative, and the negative-reciprocal normal gradient.

Practise this

Sources & how we know this

ExamExplained