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Rules of Calculus, Tangents, Rates & Optimization

Differentiation is the cornerstone of SPM Additional Mathematics, appearing heavily across Paper 1 and Paper 2 Section A and B. Master product, quotient, and chain rules, equations of tangents and normals, connected rates of change, small approximations, and second-derivative optimization.

1. Core Differentiation Rules & Formulas

Product & Quotient Rules

$\frac{d}{dx}(uv) = u v' + v u'$
$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}$

In quotient rule, denominator $v$ is squared. Never flip $v u'$ and $u v'$!

Chain Rule & Powers

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
$\frac{d}{dx}[f(x)]^n = n[f(x)]^{n-1} f'(x)$

Remember to multiply by the derivative of the inner bracket $f'(x)$!

Rates & Approximations

$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$
$\delta y \approx \frac{dy}{dx} \cdot \delta x$

$\% \text{ change in } y = \frac{\delta y}{y} \times 100\%$.

Stationary Points & Optimization: At turning points: $\frac{dy}{dx} = 0$. To determine nature: $\frac{d^2y}{dx^2} < 0 \implies \text{Maximum point}$; $\frac{d^2y}{dx^2} > 0 \implies \text{Minimum point}$; $\frac{d^2y}{dx^2} = 0 \implies \text{Test gradient signs on either side (inflection test)}$.

2. SPM Examiner Pitfalls in Differentiation

Trap 1: Forgetting Chain Rule on Brackets:

When differentiating $y = (3x^2 - 5)^4$, writing $4(3x^2 - 5)^3$ is an immediate fail. You MUST multiply by $\frac{d}{dx}(3x^2 - 5) = 6x \implies 24x(3x^2 - 5)^3$.

Trap 2: Differentiating Fraction Terms Individually:

$\frac{d}{dx}\left(\frac{u}{v}\right) \neq \frac{u'}{v'}$. Differentiating top and bottom separately is a fatal blunder. Always apply the formal Quotient Rule $\frac{v u' - u v'}{v^2}$.

Trap 3: Sign of Decreasing Rates of Change:

If a problem states "the radius is decreasing at a rate of $0.2\text{ cm/s}$", you MUST substitute $\frac{dr}{dt} = -0.2\text{ cm/s}$ with a negative sign! Omitting the negative produces wrong signs in the entire connected rate.

3. Progressive Worked Examples with SPM Marking Rubrics

Example 1 • Tangent & Normal [5 Marks]
The curve $y = \frac{4}{2x - 1}$ passes through the point $P(1, 4)$.

(a) Find the equation of the normal to the curve at point $P$. [3 marks]

(b) The normal intersects the $x$-axis at point $Q$. Find the coordinates of $Q$. [2 marks]

$y = 4(2x - 1)^{-1} \implies \frac{dy}{dx} = -4(2x - 1)^{-2}(2) = \frac{-8}{(2x - 1)^2}$. [1m: K1: Differentiate using chain rule]
At $P(1, 4)$: $m_{\text{tangent}} = \frac{-8}{(2(1) - 1)^2} = -8 \implies m_{\text{normal}} = -\frac{1}{-8} = \frac{1}{8}$. [1m: K1: Apply $m_1 m_2 = -1$]
Normal: $y - 4 = \frac{1}{8}(x - 1) \implies 8y - 32 = x - 1 \implies x - 8y + 31 = 0$. [1m: N1: Equation of normal]
(b) At $x$-axis, $y = 0 \implies x - 8(0) + 31 = 0 \implies x = -31$. [1m: K1: Set $y = 0$]
Coordinates of $Q = (-31, 0)$. [1m: N1]
Example 2 • Connected Rates of Change (Inverted Cone) [5 Marks]
Water is poured into an inverted right circular cone of base radius $6\text{ cm}$ and height $12\text{ cm}$ at a constant rate of $18\pi\text{ cm}^3\text{s}^{-1}$.

(a) Show that the volume of water in the cone when depth is $h\text{ cm}$ is $V = \frac{\pi}{12}h^3$. [2 marks]

(b) Calculate the rate of increase of the water level when the depth is $4\text{ cm}$. [3 marks]

(a) By similar triangles: $\frac{r}{h} = \frac{6}{12} = \frac{1}{2} \implies r = \frac{1}{2}h$. [1m: K1: Ratio $r:h$]
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\left(\frac{h}{2}\right)^2 h = \frac{1}{3}\pi\left(\frac{h^2}{4}\right)h = \frac{\pi}{12}h^3$. (Shown) [1m: N1]
(b) $\frac{dV}{dh} = \frac{3\pi}{12}h^2 = \frac{\pi}{4}h^2$. [1m: K1: Differentiate $V(h)$]
Connected rates formula: $\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} \implies 18\pi = \left(\frac{\pi}{4}h^2\right) \cdot \frac{dh}{dt}$. [1m: K1: Substitute $\frac{dV}{dt} = 18\pi$]
At $h = 4$: $18\pi = \frac{\pi}{4}(16) \cdot \frac{dh}{dt} = 4\pi \cdot \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{18\pi}{4\pi} = 4.5\text{ cm s}^{-1}$. [1m: N1: Correct rate with units]
Example 3 • KBAT Maximum Volume Packaging Design [6 Marks]
A rectangular sheet of metal measures $40\text{ cm}$ by $25\text{ cm}$. Equal squares of side length $x\text{ cm}$ are cut out from each of the four corners, and the remaining sides are folded upwards to form an open rectangular storage tray.

(a) Show that the internal capacity volume $V$ of the tray is given by $V = 4x^3 - 130x^2 + 1000x$. [2 marks]

(b) Find the value of $x$ which makes the volume $V$ maximum, and calculate this maximum volume. [4 marks]

(a) Tray dimensions: length $= 40 - 2x$, width $= 25 - 2x$, height $= x$. [1m: K1: Express length and width]
$V = x(40 - 2x)(25 - 2x) = x(1000 - 80x - 50x + 4x^2) = 4x^3 - 130x^2 + 1000x$. (Shown) [1m: N1]
(b) For turning points: $\frac{dV}{dx} = 12x^2 - 260x + 1000 = 0 \implies 4(3x^2 - 65x + 250) = 0$. [1m: K1: Differentiate and set to 0]
$(3x - 50)(x - 5) = 0 \implies x = \frac{50}{3} \approx 16.67$ or $x = 5$. Since width is $25\text{ cm}$, $2x < 25 \implies x < 12.5$. Reject $x = \frac{50}{3}$. [1m: K1: Factorize & reject invalid root]
Check second derivative: $\frac{d^2V}{dx^2} = 24x - 260$. At $x = 5$: $24(5) - 260 = -140 < 0 \implies$ Maximum! [1m: N1: Second derivative test]
$V_{\max} = 5(40 - 10)(25 - 10) = 5(30)(15) = 2,250\text{ cm}^3$. [1m: N1: Correct maximum volume]
Open Form 5 Chapter 2 Worksheet