← Unit 2: Calculus and further functions

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How is differential calculus introduced in QCE Math Methods Unit 2?

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

A focused answer to the QCE Math Methods Unit 2 subject-matter point on differential calculus. The gradient as the slope of a tangent, the derivative as a function, the power rule $d/dx(x^n) = nx^{n-1}$, and applications to tangent lines and stationary points; foundation for Unit 3 calculus.

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What this dot point is asking

QCAA wants Year 11 students to introduce differential calculus via the gradient concept, define the derivative, apply the power rule to polynomials, and classify stationary points using the first derivative.

The gradient and derivative

Average gradient between two points: (f(b)βˆ’f(a))/(bβˆ’a)(f(b) - f(a)) / (b - a).

Instantaneous gradient at x=ax = a: limit as hβ†’0h \to 0 of (f(a+h)βˆ’f(a))/h(f(a + h) - f(a))/h, equals fβ€²(a)f'(a).

The derivative fβ€²(x)f'(x) is a function: at each xx, it gives the gradient of the tangent.

Power rule

For f(x)=xnf(x) = x^n: fβ€²(x)=nxnβˆ’1f'(x) = n x^{n-1}.

Linearity: d/dx[af(x)+bg(x)]=afβ€²(x)+bgβ€²(x)d/dx[a f(x) + b g(x)] = a f'(x) + b g'(x).

Constants: derivative of a constant is 0.

Tangents and normals

Tangent. Line at x=ax = a with gradient fβ€²(a)f'(a) passing through (a,f(a))(a, f(a)): yβˆ’f(a)=fβ€²(a)(xβˆ’a)y - f(a) = f'(a)(x - a).

Normal. Perpendicular: gradient βˆ’1/fβ€²(a)-1/f'(a).

Stationary points

Where fβ€²(x)=0f'(x) = 0. Classify by sign of fβ€²f' around the point:

  • Positive to negative: local maximum.
  • Negative to positive: local minimum.
  • Same sign on both sides: stationary inflection.

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 fβ€²=0f' = 0 for stationary points. Stationary means zero gradient.

Tangent gradient confused with function value. f(a)f(a) is value; fβ€²(a)f'(a) is gradient.

In one sentence

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

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

Year 11 SAC4 marksFor $f(x) = x^3 - 3x^2 + 2$, find (a) $f'(x)$, (b) the gradient at $x = 2$, (c) the stationary points.
Show worked answer β†’

(a) fβ€²(x)=3x2βˆ’6xf'(x) = 3x^2 - 6x.

(b) fβ€²(2)=12βˆ’12=0f'(2) = 12 - 12 = 0.

(c) fβ€²(x)=0f'(x) = 0: 3x(xβˆ’2)=03x(x - 2) = 0, x=0x = 0 or x=2x = 2.

At x=0x = 0: f(0)=2f(0) = 2. At x=2x = 2: f(2)=8βˆ’12+2=βˆ’2f(2) = 8 - 12 + 2 = -2.

Sign of fβ€²f': positive for x<0x < 0, negative for 0<x<20 < x < 2, positive for x>2x > 2. So x=0x = 0 is a local max (f(0)=2f(0) = 2); x=2x = 2 is a local min (f(2)=βˆ’2f(2) = -2).

Markers reward the derivative, the stationary points by solving fβ€²(x)=0f'(x) = 0, and classification using the sign of fβ€²f'.

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