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Introduction to differential calculus: QCE Math Methods Unit 2 Year 11

Syllabus dot point

“Introduction to differential calculus, including the gradient at a point, the derivative as a function, and the power rule for derivatives of polynomial functions”

8 min read

Quick answer

Differential calculus is introduced in Unit 2 through the gradient at a point (limit of average gradients, equal to f′(a)f'(a)), the derivative as a function obtained via the power rule d/dx(xn)=nxn−1d/dx(x^n) = nx^{n-1} extended by linearity to polynomial sums, with applications to tangent lines and stationary points (found by solving f′(x)=0f'(x) = 0 and classified by the sign of f′f').

Jump to a section
  1. What this dot point is asking
  2. The gradient and derivative
  3. Power rule
  4. Increasing and decreasing intervals
  5. Tangents and normals
  6. Stationary points
  7. Exam-style questions

What this dot point is asking

QCAA wants Year 11 students to be introduced to differential calculus through the gradient concept, to define the derivative as a function, to apply the power rule to polynomials, and to classify stationary points using the first derivative. This is the gateway to all the calculus of Units 3 and 4.

The gradient and derivative

The average gradient of ff between two points is the slope of the secant joining them, f(b)−f(a)b−a\dfrac{f(b) - f(a)}{b - a}. The instantaneous gradient at x=ax = a is the limit of average gradients as the second point slides toward the first, lim⁡h→0f(a+h)−f(a)h\displaystyle\lim_{h \to 0}\dfrac{f(a + h) - f(a)}{h}, which equals f′(a)f'(a) and is the slope of the tangent line at that point.

Crucially, the derivative f′(x)f'(x) is itself a function: it takes any input xx and returns the gradient of the tangent there. This shift from "the gradient at one point" to "a gradient function for every point" is the central idea of differential calculus, and it is what makes the power rule so useful, since it gives the whole gradient function in one step.

Power rule

The power rule states that for f(x)=xnf(x) = x^n, the derivative is f′(x)=nxn−1f'(x) = nx^{n-1}: bring the index down as a multiplier and reduce it by one. It holds for any real index, including negative and fractional ones once those are introduced. The rule extends to polynomials through linearity, ddx[af(x)+bg(x)]=af′(x)+bg′(x)\dfrac{d}{dx}[af(x) + bg(x)] = af'(x) + bg'(x), so each term is differentiated separately. A constant has zero gradient, so its derivative is 00, and a constant multiple is preserved, ddx[cf(x)]=cf′(x)\dfrac{d}{dx}[cf(x)] = c f'(x).

Increasing and decreasing intervals

The sign of the derivative describes where a function rises or falls: f′(x)>0f'(x) > 0 on an interval means ff is increasing there, and f′(x)<0f'(x) < 0 means it is decreasing. The boundaries between these intervals are the stationary points, where f′(x)=0f'(x) = 0. Building a sign diagram of f′f' is therefore the systematic way to find and classify turning points and to sketch the overall shape of a polynomial, a method used throughout the calculus content that follows.

Derivative, gradient and stationary points

Find the stationary points of f(x)=x3−3x2+2f(x) = x^3 - 3x^2 + 2 and classify each as a local maximum or minimum.

Step 1: Differentiate using the power rule

Apply the power rule term by term. Bring each index down as a coefficient and reduce the index by one:

f′(x)=3x2−6x.f'(x) = 3x^2 - 6x.

This gives the gradient at any value of xx.

Step 2: Locate the stationary points by solving f′(x)=0f'(x) = 0

Stationary points occur where the tangent is horizontal, i.e. where the gradient is zero. Factor the derivative:

3x2−6x=3x(x−2)=0  ⇒  x=0 or x=2.3x^2 - 6x = 3x(x - 2) = 0 \;\Rightarrow\; x = 0 \text{ or } x = 2.

Find the corresponding yy-values: f(0)=2f(0) = 2 and f(2)=8−12+2=−2f(2) = 8 - 12 + 2 = -2.

Step 3: Classify by the sign of f′f' on each side

Pick a test value on each side of each stationary point and check the sign of f′f'.

For x=0x = 0: f′(−1)=3+6=9>0f'(-1) = 3 + 6 = 9 > 0 (rising before), f′(1)=3−6=−3<0f'(1) = 3 - 6 = -3 < 0 (falling after). The sign changes from ++ to −-, so (0,2)(0, 2) is a local maximum.

For x=2x = 2: f′(1)=−3<0f'(1) = -3 < 0 (falling before), f′(3)=27−18=9>0f'(3) = 27 - 18 = 9 > 0 (rising after). The sign changes from −- to ++, so (2,−2)(2, -2) is a local minimum.

Final answer: Local maximum at (0,2)(0, 2); local minimum at (2,−2)(2, -2).

Tangents and normals

A tangent line touches the curve at one point and shares its gradient there, so the tangent at x=ax = a passes through (a,f(a))(a, f(a)) with gradient f′(a)f'(a), giving y−f(a)=f′(a)(x−a)y - f(a) = f'(a)(x - a). The normal is perpendicular to the tangent at the same point, so its gradient is the negative reciprocal −1f′(a)-\dfrac{1}{f'(a)} (and the normal is vertical, x=ax = a, when the tangent is horizontal). These two lines are a recurring application of the derivative and connect the algebra of gradients to the geometry of the curve.

Stationary points

A stationary point occurs where the tangent is horizontal, that is where f′(x)=0f'(x) = 0. To classify it, examine the sign of f′f' immediately on each side:

  • Positive then negative: the function rises then falls, a local maximum.
  • Negative then positive: the function falls then rises, a local minimum.
  • The same sign on both sides: a stationary point of inflection (the curve flattens momentarily without turning).

This first-derivative test is the standard QCE method, and presenting a clear sign diagram of f′f' is what earns the classification marks. Stationary points are the basis of optimisation and curve sketching met later in the course.

Common errors

Power rule on exe^x, sin⁡x\sin x, ln⁡x\ln x. Power rule applies to xnx^n, not these (introduced in Unit 3).

Forgetting to set f′=0f' = 0 for stationary points. Stationary means zero gradient.

Tangent gradient confused with function value. f(a)f(a) is value; f′(a)f'(a) is gradient.

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.

Practice4 marks
Paper 2 (complex familiar). For f(x)=x3−3x2+2f(x) = x^3 - 3x^2 + 2, determine (a) f′(x)f'(x), (b) the gradient at x=2x = 2, (c) the stationary points and their nature.
Show worked answer →

(a) f′(x)=3x2−6xf'(x) = 3x^2 - 6x.

(b) f′(2)=12−12=0f'(2) = 12 - 12 = 0.

(c) Set f′(x)=0f'(x) = 0: 3x(x−2)=03x(x - 2) = 0, so x=0x = 0 or x=2x = 2, with f(0)=2f(0) = 2 and f(2)=−2f(2) = -2. The sign of f′f' is positive, then negative, then positive, so x=0x = 0 is a local maximum (0,2)(0, 2) and x=2x = 2 is a local minimum (2,−2)(2, -2).

Markers reward the derivative, solving f′(x)=0f'(x) = 0, and classifying by the sign of f′f'.

Practice4 marks
Paper 1 (technique). The curve y=2x2−5x+1y = 2x^2 - 5x + 1 has a tangent at the point where x=2x = 2. (a) Determine the gradient of the tangent. (b) Determine the equation of the tangent.
Show worked answer →

(a) dydx=4x−5\dfrac{dy}{dx} = 4x - 5, so at x=2x = 2 the gradient is 4(2)−5=34(2) - 5 = 3.

(b) The point is (2, 2(4)−10+1)=(2,−1)(2, \ 2(4) - 10 + 1) = (2, -1). Tangent: y−(−1)=3(x−2)y - (-1) = 3(x - 2), so y=3x−7y = 3x - 7.

Markers reward the derivative, the gradient value, the point, and the tangent equation.

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