Differentiation of polynomials and tangent lines (QCE Math Methods Unit 2)
“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”
Differentiate any polynomial term by term using the power rule together with the sum and constant-multiple rules; the derivative at is the gradient of the tangent line, and the normal at the same point has gradient .
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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 :
This was proved from first principles for positive integer in the previous dot point. The same form holds for rational or negative, with the standard domain restrictions.
Sum (and difference) rule. The derivative of a sum is the sum of the derivatives:
Constant multiple rule. A constant pulls outside the derivative:
Constant function. The derivative of a constant is zero:
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 , 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
- .
- Negative power
- .
- Rational power
- .
- Combine terms before differentiating
- Expand brackets and split fractions if needed. For , first simplify to , then differentiate to .
Higher derivatives
Differentiating a polynomial gives another polynomial, which can itself be differentiated. The second derivative measures the rate of change of the gradient and describes concavity: means the curve is concave up (holding water) and means concave down. For , the first derivative is and the second is . Second derivatives become important when classifying stationary points and analysing motion, where they represent acceleration.
Tangent and normal lines
The tangent to at :
- Point: .
- Gradient: .
- Equation: .
The normal at the same point:
- Perpendicular to the tangent.
- Gradient: (provided ).
- Equation: .
If the tangent is horizontal and the normal is vertical ().
Rewriting before differentiating
The power rule applies cleanly only to terms in the form , 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: becomes , and becomes . 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.
Find the point
Find the tangent to at . The point is .
Find the gradient
, so .
Write the tangent equation
Markers reward the point, the derivative gradient, and the tangent in point-slope form.
- Forgetting to drop the power by one
- , not .
- Misapplying the rule to a constant times
- , not or 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
- . A common slip is using or .
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 marksPaper 2 (complex familiar). For , determine (a) , (b) the gradient at , (c) the equation of the tangent at .Show worked answer →
(a) Power rule term by term: .
(b) .
(c) , so the point is with gradient . Tangent: , so .
Markers reward the term-by-term derivative, the gradient at the point, and the tangent in point-slope form.
Practice4 marksPaper 1 (technique). For (where ), (a) determine by first simplifying, and (b) determine the gradient of the normal at .Show worked answer →
(a) Simplify first: , so .
(b) At the tangent gradient is , so the normal gradient is .
Markers reward simplifying before differentiating, the derivative, and the negative-reciprocal normal gradient.