5.1

Limits & Rates of Change

  • The Limit Concept: The expression $\lim_{x \to a} f(x) = L$ means that as $x$ gets arbitrarily close to $a$ (from both left and right, with $x \ne a$), the function values $f(x)$ get arbitrarily close to $L$.
  • Indeterminate Form $\frac{0}{0}$ (Algebraic Resolution): When direct substitution yields $\frac{0}{0}$, factorise the numerator and denominator, cancel common non-zero factors, and evaluate the remaining limit: $$\lim_{x \to a} \frac{P(x)}{Q(x)} = \lim_{x \to a} \frac{(x - a) P_1(x)}{(x - a) Q_1(x)} = \frac{P_1(a)}{Q_1(a)}$$
  • Average vs. Instantaneous Rate: $$\text{Average Rate on } [a, b] = \frac{f(b) - f(a)}{b - a}$$ $$\text{Instantaneous Rate at } x = a = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$

Example 1

Evaluating Limits and Instantaneous Rates

(a) Evaluate $\lim_{x \to 3} \frac{x^2 - 2x - 3}{x - 3}$.
(b) A particle moves along a line with displacement $s(t) = 4t^2 - t$ (metres) after $t$ seconds. Find its average velocity between $t = 1$ and $t = 3$, and its instantaneous velocity at $t = 1$ using limits.

  1. (a) Direct substitution yields $\frac{3^2 - 2(3) - 3}{3 - 3} = \frac{0}{0}$. Factorise the numerator: $$\lim_{x \to 3} \frac{(x - 3)(x + 1)}{x - 3} = \lim_{x \to 3}(x + 1) = 3 + 1 = 4$$
  2. (b) Average velocity: $$s(1) = 4(1)^2 - 1 = 3\text{ m}, \quad s(3) = 4(3)^2 - 3 = 33\text{ m}$$ $$\text{Average velocity} = \frac{s(3) - s(1)}{3 - 1} = \frac{33 - 3}{2} = 15\text{ m s}^{-1}$$ Instantaneous velocity at $t = 1$: $$s(1+h) = 4(1+h)^2 - (1+h) = 4(1 + 2h + h^2) - 1 - h = 3 + 7h + 4h^2$$ $$\lim_{h \to 0} \frac{s(1+h) - s(1)}{h} = \lim_{h \to 0} \frac{3 + 7h + 4h^2 - 3}{h} = \lim_{h \to 0}(7 + 4h) = 7\text{ m s}^{-1}$$

Answers: (a) $4$; (b) Average $= 15\text{ m s}^{-1}$, Instantaneous $= 7\text{ m s}^{-1}$.

5.1

The Derivative from First Principles

  • Definition of the Derivative Function: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
  • Geometrical Interpretation: $\frac{f(x+h) - f(x)}{h}$ is the gradient of the secant through $(x, f(x))$ and $(x+h, f(x+h))$. The limit as $h \to 0$ gives the exact gradient of the tangent at $x$.
  • Step-by-Step Algebraic Procedure:
    1. Substitute $(x+h)$ into the function and expand brackets fully.
    2. Subtract the original expression $f(x)$. The constant terms without $h$ must cancel out.
    3. Factor out $h$ from the numerator and cancel with the denominator $h$.
    4. Evaluate the limit by setting $h = 0$.
Derivative as Limit of Secants (First Principles) Shows curve y = f(x) with secant line between P(x, f(x)) and Q(x+h, f(x+h)). As h approaches 0, Q slides toward P and the secant rotates to become the tangent line with slope f'(x). x y y = f(x) h = Δx Δy = f(x+h) − f(x) Secant slope = Δy / h Tangent at P: slope = f'(x) as h → 0 P(x, f(x)) Q(x+h, f(x+h)) x x + h
Figure 5.1: As $h \to 0$, the secant line through $P$ and $Q$ rotates to become the tangent line at $P$, defining the derivative $f'(x)$.

Example 2

Differentiating from First Principles

Find the derivative of $f(x) = 2x^2 - 5x + 3$ from first principles.

  1. Write $f(x+h)$: $$f(x+h) = 2(x+h)^2 - 5(x+h) + 3 = 2(x^2 + 2xh + h^2) - 5x - 5h + 3 = 2x^2 + 4xh + 2h^2 - 5x - 5h + 3$$
  2. Subtract $f(x)$: $$f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 - 5x - 5h + 3) - (2x^2 - 5x + 3) = 4xh + 2h^2 - 5h$$
  3. Divide by $h$: $$\frac{f(x+h) - f(x)}{h} = \frac{h(4x + 2h - 5)}{h} = 4x + 2h - 5 \quad (h \ne 0)$$
  4. Take the limit as $h \to 0$: $$f'(x) = \lim_{h \to 0}(4x + 2h - 5) = 4x - 5$$

Answer: $f'(x) = 4x - 5$.

5.3

The Power Rule & Polynomial Differentiation

  • Power Rule: For any real exponent $n \in \mathbb{R}$: $$\frac{d}{dx}(x^n) = n x^{n-1}, \quad \frac{d}{dx}(a x^n) = a n x^{n-1}$$
  • Constants: The derivative of any constant $c$ is zero: $\frac{d}{dx}(c) = 0$.
  • Linearity (Term-by-Term): $$\frac{d}{dx}[u(x) \pm v(x)] = u'(x) \pm v'(x)$$
  • Converting Indices Before Differentiating: $$\frac{1}{x^k} = x^{-k}, \quad \sqrt[k]{x} = x^{1/k}, \quad \frac{1}{\sqrt[k]{x^m}} = x^{-m/k}, \quad x\sqrt{x} = x^{3/2}$$

Example 3

Differentiating with Fractional and Negative Powers

Differentiate $f(x) = 5x^4 - \frac{4}{x^3} + 6\sqrt{x} - \frac{2}{\sqrt[3]{x}} + 7$ with respect to $x$.

  1. Rewrite in power form: $$f(x) = 5x^4 - 4x^{-3} + 6x^{1/2} - 2x^{-1/3} + 7$$
  2. Differentiate each term: $$\frac{d}{dx}(5x^4) = 5(4x^3) = 20x^3$$ $$\frac{d}{dx}(-4x^{-3}) = -4(-3x^{-4}) = +12x^{-4}$$ $$\frac{d}{dx}(6x^{1/2}) = 6\left(\frac{1}{2}x^{-1/2}\right) = 3x^{-1/2}$$ $$\frac{d}{dx}(-2x^{-1/3}) = -2\left(-\frac{1}{3}x^{-4/3}\right) = +\frac{2}{3}x^{-4/3}$$ $$\frac{d}{dx}(7) = 0$$
  3. Combine terms: $$f'(x) = 20x^3 + \frac{12}{x^4} + \frac{3}{\sqrt{x}} + \frac{2}{3\sqrt[3]{x^4}}$$

Answer: $f'(x) = 20x^3 + 12x^{-4} + 3x^{-1/2} + \frac{2}{3}x^{-4/3}$.

5.3

Pre-Differentiation Algebra (Expanding & Splitting Fractions)

  • Why Pre-Differentiation Algebra is Essential: In AA SL, before learning the product and quotient rules (Topic 5.4), expressions must be simplified algebraically into sums of separate powers $ax^n$.
  • Method 1 (Expanding Products): Expand brackets completely using the distributive law: $$(2x - 3)(x^2 + 4) = 2x^3 - 3x^2 + 8x - 12$$
  • Method 2 (Splitting Monomial Denominators): Divide each numerator term individually by the denominator term: $$\frac{A + B - C}{D} = \frac{A}{D} + \frac{B}{D} - \frac{C}{D}$$ $$\frac{6x^4 - 4x^2 + 5}{2x^2} = \frac{6x^4}{2x^2} - \frac{4x^2}{2x^2} + \frac{5}{2x^2} = 3x^2 - 2 + \frac{5}{2}x^{-2}$$
  • Fatal Error: NEVER differentiate the numerator and denominator separately! $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) \ne \frac{f'(x)}{g'(x)}$.

Example 4

Expanding Products and Splitting Denominators

Differentiate with respect to $x$:
(a) $f(x) = (3x - 2)(2x^2 + 5)$
(b) $g(x) = \frac{8x^5 - 6x^3 + 4x - 1}{2x^2}$

  1. (a) Expand the brackets first: $$f(x) = 6x^3 + 15x - 4x^2 - 10 = 6x^3 - 4x^2 + 15x - 10$$ Now differentiate term-by-term: $$f'(x) = 6(3x^2) - 4(2x) + 15(1) - 0 = 18x^2 - 8x + 15$$
  2. (b) Divide each term by $2x^2$: $$g(x) = \frac{8x^5}{2x^2} - \frac{6x^3}{2x^2} + \frac{4x}{2x^2} - \frac{1}{2x^2} = 4x^3 - 3x + 2x^{-1} - \frac{1}{2}x^{-2}$$ Now differentiate: $$g'(x) = 4(3x^2) - 3(1) + 2(-1x^{-2}) - \frac{1}{2}(-2x^{-3}) = 12x^2 - 3 - 2x^{-2} + x^{-3}$$ $$g'(x) = 12x^2 - 3 - \frac{2}{x^2} + \frac{1}{x^3}$$

Answers: (a) $f'(x) = 18x^2 - 8x + 15$; (b) $g'(x) = 12x^2 - 3 - \frac{2}{x^2} + \frac{1}{x^3}$.

5.1, 5.3

Tangent Lines & Horizontal Tangents

  • Tangent Gradient: The gradient of the tangent to the curve $y = f(x)$ at $x = x_1$ is: $$m = f'(x_1)$$
  • Complete Tangent Line Equation: Use point-slope form with point $(x_1, y_1)$ where $y_1 = f(x_1)$: $$y - y_1 = m(x - x_1)$$
  • Horizontal Tangents (Stationary Points): A horizontal line has gradient $m = 0$. Solve: $$f'(x) = 0$$
  • Curve Coordinates vs. Derivative: To find the coordinates of a point on the curve, substitute the solved $x$-value back into $y = f(x)$, not $f'(x)$.
Tangent Slopes and Stationary Points Shows a cubic curve with horizontal tangents f'(x)=0 at turning points, positive slope f'(x)>0 where the curve rises, and negative slope f'(x)<0 where the curve falls. x y y = f(x) f'(x) > 0 (Rising) Horizontal Tangent: f'(x) = 0 (Local Max) f'(x) < 0 (Falling) Horizontal Tangent: f'(x) = 0 (Local Min) f'(x) > 0
Figure 5.2: Geometric relationship between the derivative $f'(x)$ and curve behavior: rising ($f' > 0$), turning point stationary points ($f' = 0$), and falling ($f' < 0$).

Example 5

Finding the Tangent Equation and Stationary Coordinates

Consider the curve $y = x^3 - 3x^2 - 9x + 5$.
(a) Find the equation of the tangent line to the curve at $x = 2$. Give your answer in the form $y = mx + c$.
(b) Find the coordinates of the points on the curve where the tangent is horizontal.

  1. (a) Find the point $(x_1, y_1)$ on the curve: $$y_1 = f(2) = (2)^3 - 3(2)^2 - 9(2) + 5 = 8 - 12 - 18 + 5 = -17 \implies (2, -17)$$ Differentiate to find the gradient function: $$\frac{dy}{dx} = 3x^2 - 6x - 9$$ Evaluate at $x = 2$: $$m = 3(2)^2 - 6(2) - 9 = 12 - 12 - 9 = -9$$ Apply point-slope formula: $$y - (-17) = -9(x - 2) \implies y + 17 = -9x + 18 \implies y = -9x + 1$$
  2. (b) Set $\frac{dy}{dx} = 0$: $$3x^2 - 6x - 9 = 0 \implies 3(x^2 - 2x - 3) = 0 \implies (x - 3)(x + 1) = 0$$ $$x = 3 \quad \text{or} \quad x = -1$$ Find corresponding $y$-coordinates: $$\text{At } x = 3: \quad y = (3)^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22 \implies (3, -22)$$ $$\text{At } x = -1: \quad y = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10 \implies (-1, 10)$$

Answers: (a) $y = -9x + 1$; (b) Points: $(3, -22)$ and $(-1, 10)$.

Tier 1: Core Fluency & Differentiation Techniques (Questions 1–6)

  1. 1. Indeterminate Limits by Factoring: Evaluate each limit algebraically:
    (a) $\lim_{x \to 5} \frac{x^2 - 25}{x - 5}$
    (b) $\lim_{x \to -3} \frac{x^2 + 7x + 12}{x + 3}$

    Show solution

    (a) Direct substitution gives $\frac{0}{0}$. Factorise the difference of squares: $$\lim_{x \to 5} \frac{(x - 5)(x + 5)}{x - 5} = \lim_{x \to 5}(x + 5) = 5 + 5 = \mathbf{10}$$ (b) Factorise the numerator quadratic: $$\lim_{x \to -3} \frac{(x + 3)(x + 4)}{x + 3} = \lim_{x \to -3}(x + 4) = -3 + 4 = \mathbf{1}$$

  2. 2. Limits with Radicals (Conjugate Multiplication): Evaluate the limit:
    $$\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}$$

    Show solution

    Direct substitution yields $\frac{0}{0}$. Multiply numerator and denominator by the conjugate $(\sqrt{x + 9} + 3)$: $$\lim_{x \to 0} \frac{(\sqrt{x + 9} - 3)(\sqrt{x + 9} + 3)}{x(\sqrt{x + 9} + 3)} = \lim_{x \to 0} \frac{(x + 9) - 9}{x(\sqrt{x + 9} + 3)}$$ $$= \lim_{x \to 0} \frac{x}{x(\sqrt{x + 9} + 3)} = \lim_{x \to 0} \frac{1}{\sqrt{x + 9} + 3} = \frac{1}{\sqrt{9} + 3} = \mathbf{\frac{1}{6}}$$

  3. 3. First Principles Quadratic Derivation: Use $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ to prove that the derivative of $f(x) = 3x^2 + 2x$ is $f'(x) = 6x + 2$.

    Show solution

    $$f(x+h) = 3(x+h)^2 + 2(x+h) = 3(x^2 + 2xh + h^2) + 2x + 2h = 3x^2 + 6xh + 3h^2 + 2x + 2h$$ $$f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 + 2x + 2h) - (3x^2 + 2x) = 6xh + 3h^2 + 2h$$ $$\frac{f(x+h) - f(x)}{h} = \frac{h(6x + 3h + 2)}{h} = 6x + 3h + 2 \quad (h \ne 0)$$ $$f'(x) = \lim_{h \to 0}(6x + 3h + 2) = \mathbf{6x + 2}$$

  4. 4. Standard Polynomial Power Rule: Differentiate with respect to $x$:
    (a) $f(x) = 7x^5 - 4x^3 + 9x - 12$
    (b) $g(x) = \frac{1}{4}x^8 - \frac{2}{3}x^6 + 5x^2 - x$

    Show solution

    (a) Term-by-term power rule: $$f'(x) = 7(5x^4) - 4(3x^2) + 9(1) - 0 = \mathbf{35x^4 - 12x^2 + 9}$$ (b) Differentiate: $$g'(x) = \frac{1}{4}(8x^7) - \frac{2}{3}(6x^5) + 5(2x) - 1 = \mathbf{2x^7 - 4x^5 + 10x - 1}$$

  5. 5. Negative & Fractional Indices: Convert to index form and differentiate:
    (a) $y = \frac{5}{x^4} - \frac{2}{\sqrt{x}} + 3x^{2/3}$
    (b) $y = \frac{4}{\sqrt[3]{x^2}} - \frac{1}{2x^5}$

    Show solution

    (a) Rewrite: $y = 5x^{-4} - 2x^{-1/2} + 3x^{2/3}$. $$\frac{dy}{dx} = 5(-4x^{-5}) - 2\left(-\frac{1}{2}x^{-3/2}\right) + 3\left(\frac{2}{3}x^{-1/3}\right) = \mathbf{-20x^{-5} + x^{-3/2} + 2x^{-1/3}}$$ $$= -\frac{20}{x^5} + \frac{1}{\sqrt{x^3}} + \frac{2}{\sqrt[3]{x}}$$ (b) Rewrite: $y = 4x^{-2/3} - \frac{1}{2}x^{-5}$. $$\frac{dy}{dx} = 4\left(-\frac{2}{3}x^{-5/3}\right) - \frac{1}{2}(-5x^{-6}) = \mathbf{-\frac{8}{3}x^{-5/3} + \frac{5}{2}x^{-6}} = -\frac{8}{3\sqrt[3]{x^5}} + \frac{5}{2x^6}$$

  6. 6. Pre-Differentiation Algebra (Expanding & Splitting): Differentiate with respect to $x$:
    (a) $f(x) = (2x - 5)(3x^2 + 4)$ by expanding first.
    (b) $g(x) = \frac{6x^4 - 8x^2 + 3}{x^2}$ by dividing each term by the denominator.

    Show solution

    (a) Expand brackets: $$f(x) = 6x^3 + 8x - 15x^2 - 20 = 6x^3 - 15x^2 + 8x - 20$$ $$f'(x) = 6(3x^2) - 15(2x) + 8(1) - 0 = \mathbf{18x^2 - 30x + 8}$$ (b) Split denominator: $$g(x) = \frac{6x^4}{x^2} - \frac{8x^2}{x^2} + \frac{3}{x^2} = 6x^2 - 8 + 3x^{-2}$$ $$g'(x) = 12x - 0 + 3(-2x^{-3}) = 12x - 6x^{-3} = \mathbf{12x - \frac{6}{x^3}}$$

Tier 2: Tangents, Normals & Stationary Points (Questions 7–11)

  1. 7. Complete Equation of Tangent Line: Find the equation of the tangent line to the curve $y = 2x^2 - 5x + 3$ at $x = 2$. Give your answer in the form $y = mx + c$.

    Show solution

    Step 1: Find the $y$-coordinate on the curve: $$y = 2(2)^2 - 5(2) + 3 = 8 - 10 + 3 = 1 \implies \text{Point } (2, 1)$$ Step 2: Differentiate: $$\frac{dy}{dx} = 4x - 5 \implies \text{At } x = 2: \quad m = 4(2) - 5 = 3$$ Step 3: Line equation: $$y - 1 = 3(x - 2) \implies y - 1 = 3x - 6 \implies \mathbf{y = 3x - 5}$$

  2. 8. Equation of Normal Line: Find the equation of the normal to the curve $y = x^3 - 4x + 1$ at $x = 1$.

    Show solution

    Step 1: Point on the curve: $$y = (1)^3 - 4(1) + 1 = 1 - 4 + 1 = -2 \implies (1, -2)$$ Step 2: Tangent gradient: $$\frac{dy}{dx} = 3x^2 - 4 \implies m_{\text{tangent}} = 3(1)^2 - 4 = -1$$ Step 3: Normal gradient (perpendicular, $m_1 m_2 = -1$): $$m_{\text{normal}} = -\frac{1}{-1} = 1$$ Step 4: Equation of normal: $$y - (-2) = 1(x - 1) \implies y + 2 = x - 1 \implies \mathbf{y = x - 3}$$

  3. 9. Stationary Points (Horizontal Tangents): Find the coordinates of the two points on the curve $y = 2x^3 - 3x^2 - 12x + 7$ where the tangent is horizontal.

    Show solution

    Set $\frac{dy}{dx} = 0$: $$\frac{dy}{dx} = 6x^2 - 6x - 12 = 0 \implies 6(x^2 - x - 2) = 0 \implies (x - 2)(x + 1) = 0$$ $$x = 2 \quad \text{or} \quad x = -1$$ Substitute into the original curve equation: $$\text{At } x = 2: \quad y = 2(8) - 3(4) - 12(2) + 7 = 16 - 12 - 24 + 7 = -13 \implies \mathbf{(2, -13)}$$ $$\text{At } x = -1: \quad y = 2(-1) - 3(1) - 12(-1) + 7 = -2 - 3 + 12 + 7 = 14 \implies \mathbf{(-1, 14)}$$

  4. 10. Tangents Parallel to a Given Line: Find the coordinates of the points on the curve $y = x^3 - 3x^2 - 6x + 4$ where the tangent is parallel to the line $y = 3x - 5$.

    Show solution

    Parallel tangents have equal gradient, so $\frac{dy}{dx} = 3$: $$3x^2 - 6x - 6 = 3 \implies 3x^2 - 6x - 9 = 0 \implies 3(x^2 - 2x - 3) = 0$$ $$(x - 3)(x + 1) = 0 \implies x = 3 \quad \text{or} \quad x = -1$$ Find corresponding $y$-coordinates: $$\text{At } x = 3: \quad y = (3)^3 - 3(3)^2 - 6(3) + 4 = 27 - 27 - 18 + 4 = -14 \implies \mathbf{(3, -14)}$$ $$\text{At } x = -1: \quad y = (-1)^3 - 3(-1)^2 - 6(-1) + 4 = -1 - 3 + 6 + 4 = 6 \implies \mathbf{(-1, 6)}$$

  5. 11. Tangent Intersections & Triangle Area: The tangent to the curve $y = 4 - x^2$ at the point $P(1, 3)$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. Find the coordinates of $A$ and $B$, and calculate the area of triangle $OAB$ (where $O$ is the origin).

    Show solution
    Tangent Line and Triangle OAB Area Shows curve y = 4 - x^2 with tangent line y = -2x + 5 touching at P(1, 3) and intersecting axes at A(2.5, 0) and B(0, 5), enclosing shaded triangle OAB. x y O Area = 6.25 y = 4 − x² Tangent: y = −2x + 5 B(0, 5) A(2.5, 0) P(1, 3) base = 2.5 ht = 5
    Figure 11.1: Geometric configuration of tangent at $P(1, 3)$ intersecting the axes at $A(2.5, 0)$ and $B(0, 5)$ to form triangle $OAB$.

    Step 1: Find the tangent equation at $P(1, 3)$: $$\frac{dy}{dx} = -2x \implies m = -2(1) = -2$$ $$y - 3 = -2(x - 1) \implies y - 3 = -2x + 2 \implies y = -2x + 5$$ Step 2: Find intercepts: $$x\text{-intercept } A \ (y = 0): \quad 0 = -2x + 5 \implies x = 2.5 \implies A(2.5, 0)$$ $$y\text{-intercept } B \ (x = 0): \quad y = 5 \implies B(0, 5)$$ Step 3: Triangle area: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.5 \times 5 = \mathbf{6.25\text{ units}^2}$$

Tier 3: Examination Modeling & Advanced Applications (Questions 12–15)

  1. 12. Determining Unknown Parameters (Simultaneous Equations): A curve has equation $y = ax^2 + bx$, where $a$ and $b$ are constants. The curve passes through the point $(2, 14)$, and the gradient of the tangent at $x = 3$ is $15$. Find the values of $a$ and $b$.

    Show solution

    Condition 1 (Curve passes through $(2, 14)$): $$14 = a(2)^2 + b(2) \implies 4a + 2b = 14 \implies 2a + b = 7 \quad \text{--- (1)}$$ Condition 2 (Gradient at $x = 3$ is $15$): $$\frac{dy}{dx} = 2ax + b$$ $$\text{At } x = 3: \quad 2a(3) + b = 15 \implies 6a + b = 15 \quad \text{--- (2)}$$ Subtract (1) from (2): $$(6a + b) - (2a + b) = 15 - 7 \implies 4a = 8 \implies \mathbf{a = 2}$$ Substitute $a = 2$ into (1): $$2(2) + b = 7 \implies 4 + b = 7 \implies \mathbf{b = 3}$$ The equation is $y = 2x^2 + 3x$.

  2. 13. Second Derivatives & Rate of Change of Gradient: For the function $f(x) = 2x^4 - 6x^3 + 5x - 8$:
    (a) Find $f'(x)$ and $f''(x)$.
    (b) Find the values of $x$ for which the rate of change of the gradient is zero (i.e. $f''(x) = 0$).

    Show solution

    (a) Differentiate once: $$f'(x) = 8x^3 - 18x^2 + 5$$ Differentiate again: $$f''(x) = \frac{d}{dx}(8x^3 - 18x^2 + 5) = \mathbf{24x^2 - 36x}$$ (b) Set $f''(x) = 0$: $$24x^2 - 36x = 0 \implies 12x(2x - 3) = 0$$ $$\mathbf{x = 0 \quad \text{or} \quad x = 1.5}$$

  3. 14. Kinematics Application (Displacement, Velocity, Acceleration): The displacement of a particle moving in a straight line after $t$ seconds ($t \ge 0$) is given by $s(t) = t^3 - 6t^2 + 9t + 4$ metres.
    (a) Find expressions for velocity $v(t)$ and acceleration $a(t)$.
    (b) Find the initial velocity ($t = 0$).
    (c) Find the times when the particle is momentarily at rest.
    (d) Calculate the particle's acceleration when it is first at rest.

    Show solution

    (a) Velocity: $v(t) = s'(t) = \mathbf{3t^2 - 12t + 9}$.
    Acceleration: $a(t) = v'(t) = s''(t) = \mathbf{6t - 12}$.

    (b) At $t = 0$: $v(0) = 3(0)^2 - 12(0) + 9 = \mathbf{9\text{ m s}^{-1}}$.

    (c) Momentarily at rest when $v(t) = 0$: $$3t^2 - 12t + 9 = 0 \implies 3(t^2 - 4t + 3) = 0 \implies (t - 1)(t - 3) = 0$$ $$\mathbf{t = 1\text{ s} \quad \text{and} \quad t = 3\text{ s}}$$ (d) The particle is first at rest at $t = 1\text{ s}$: $$a(1) = 6(1) - 12 = \mathbf{-6\text{ m s}^{-2}}$$

  4. 15. Contextual Rate of Change (Reservoir Drainage): The volume $V$ of water (in cubic metres) remaining in a reservoir $t$ hours after a drain valve is opened is given by $V(t) = 50(10 - t)^2$ for $0 \le t \le 10$.
    (a) Expand the expression for $V(t)$ and find $\frac{dV}{dt}$.
    (b) Calculate the rate of flow of water out of the reservoir at $t = 2$ and at $t = 8$ hours.
    (c) State the physical meaning of the negative sign in $\frac{dV}{dt}$.

    Show solution

    (a) Expand $V(t)$: $$V(t) = 50(100 - 20t + t^2) = 5000 - 1000t + 50t^2$$ Differentiate: $$\frac{dV}{dt} = \mathbf{-1000 + 100t\text{ m}^3\text{ h}^{-1}}$$ (b) $$\text{At } t = 2: \quad \frac{dV}{dt} = -1000 + 100(2) = \mathbf{-800\text{ m}^3\text{ h}^{-1}}$$ $$\text{At } t = 8: \quad \frac{dV}{dt} = -1000 + 100(8) = \mathbf{-200\text{ m}^3\text{ h}^{-1}}$$ (c) The negative sign indicates that the volume of water in the reservoir is decreasing over time (water is draining out).

First Principles Definition

Memorize
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Formal definition of derivative function. Tested as algebraic derivation in Paper 1.

Power Rule

Formula Booklet
$$f(x) = x^n \implies f'(x) = n x^{n-1}$$

Valid for all real exponents $n \in \mathbb{Q}$. In formula booklet Topic 5.

Constant & Linearity

Memorize
$$\frac{d}{dx}(c) = 0, \quad \frac{d}{dx}[af(x)] = a f'(x)$$

Constants differentiate to zero. Multiplicative constants factor through.

Complete Tangent Equation

Memorize
$$y - y_1 = m(x - x_1), \quad m = f'(x_1)$$

Point-slope equation using curve coordinate $(x_1, y_1)$ and derivative slope $m$.

Horizontal Tangent Condition

Memorize
$$f'(x) = 0$$

Tangents are horizontal when gradient is zero. Stationary points satisfy this.

Kinematics Relations

Memorize
$$v(t) = s'(t) = \frac{ds}{dt}, \quad a(t) = v'(t) = s''(t)$$

Velocity is the derivative of displacement; acceleration is derivative of velocity.

Exam Pitfalls to Avoid

  • Evaluating before differentiating: Substituting a value $x = a$ into $f(x)$ yields a number, whose derivative is zero. Differentiate $f(x) \to f'(x)$ first, then evaluate $f'(a)$.
  • Negative exponent subtraction: $\frac{d}{dx}(x^{-3}) = -3x^{-4}$, NOT $-3x^{-2}$. Subtracting 1 from $-3$ gives $-4$.
  • Differentiating quotients before splitting: $\frac{d}{dx}\left(\frac{x^3 + 1}{x}\right) \ne \frac{3x^2}{1}$. Always split into $x^2 + x^{-1}$ first to get $2x - x^{-2}$.
  • Stationary point $y$-coordinates: After solving $f'(x) = 0$ for $x$, substitute $x$ into the original function $f(x)$ to get the $y$-coordinate. Substituting into $f'(x)$ always gives 0.

Section A: Diagnostic Questions

Check fundamental differentiation rules.

Score: 0 / 4

1. If $f(x) = 4x^3 - 6x^2 + 7x - 5$, then $f'(x)$ is:

Solution: Differentiating term-by-term: $4(3x^2) - 6(2x) + 7(1) - 0 = 12x^2 - 12x + 7$.

2. The derivative of $g(x) = \frac{6}{x^2}$ is:

Solution: $g(x) = 6x^{-2} \implies g'(x) = 6(-2x^{-3}) = -12x^{-3} = -\frac{12}{x^3}$.

3. The gradient of $y = x^3 - 5x + 2$ at $x = 2$ is:

Solution: $\frac{dy}{dx} = 3x^2 - 5$. At $x = 2$: $3(2)^2 - 5 = 12 - 5 = 7$.

4. At what value of $x$ does $f(x) = 2x^2 - 8x + 3$ have a horizontal tangent?

Solution: $f'(x) = 4x - 8 = 0 \implies 4x = 8 \implies x = 2$.

Section B: Examination Practice

Multi-step questions with markscheme breakdowns.

Paper 1 (Non-Calculator) [6 marks]

Let $f(x) = \frac{1}{3}x^3 - x^2 - 8x + 5$.

(a) Find $f'(x)$. [2 marks]
(b) Find the gradient of the tangent to the curve at $x = 1$. [2 marks]
(c) Determine the values of $x$ for which the tangent to the curve is parallel to the line $y = 7x - 4$. [2 marks]
View Markscheme
(a)
M1 for power rule on at least two terms.
$$f'(x) = x^2 - 2x - 8 \quad \text{[A1]}$$ (b)
M1 for substituting $x = 1$ into $f'(x)$:
$$f'(1) = (1)^2 - 2(1) - 8 = -9 \quad \text{[A1]}$$ (c)
Parallel to $y = 7x - 4 \implies m = 7$.
M1 for setting $f'(x) = 7$:
$$x^2 - 2x - 8 = 7 \implies x^2 - 2x - 15 = 0 \implies (x - 5)(x + 3) = 0$$ $$x = 5 \text{ or } x = -3 \quad \text{[A1]}$$
Paper 2 (Calculator Allowed) [6 marks]

A curve has equation $y = k x^2 + \frac{16}{x}$, where $x \ne 0$ and $k$ is a constant. The gradient of the tangent at $x = 2$ is $10$.

(a) Show that $k = 3.5$. [3 marks]
(b) Find the $x$-coordinate of the point on the curve where the tangent is horizontal. Give your answer to 3 significant figures. [3 marks]
View Markscheme
(a)
$y = k x^2 + 16 x^{-1}$
M1 for differentiating: $\frac{dy}{dx} = 2kx - \frac{16}{x^2}$
M1 for substituting $x = 2$ and setting equal to $10$:
$$2k(2) - \frac{16}{(2)^2} = 10 \implies 4k - 4 = 10 \implies 4k = 14$$ $$k = 3.5 \quad \text{[A1]}$$ (b)
Horizontal tangent $\implies \frac{dy}{dx} = 0$:
M1 for $7x - \frac{16}{x^2} = 0 \implies 7x^3 = 16$
M1 for solving $x = \sqrt[3]{\frac{16}{7}}$:
$$x \approx 1.32 \quad \text{[A1]}$$

Calculus Unit

Next Lessons in Calculus

5.1, 5.3 Limits and First Derivatives Rates of change, first principles limits, and power rule for polynomials. 5.2, 5.7, 5.8 Increasing, Decreasing & Stationary Points Monotonicity tests, second derivatives, local extrema, and inflection points. 5.4, 5.6 Tangents, Normals & Rules Product, quotient, and chain rules with tangent and normal line equations. 5.9 Kinematics with Derivatives Motion in a line: displacement, velocity, acceleration, and total distance. 5.5, 5.10, 5.11 Integration & Area Anti-differentiation, definite integrals, and area enclosed between curves.