SPM Additional Mathematics: Complete Mathematical Proofs and Formula Derivations Guide

Overview and Compilation Details

• Purpose and Scope:

  • This study guide provides definitive mathematical proofs and systematic derivations of core formulas and theorems within the SPM Additional Mathematics syllabus (covering both Form 4 and Form 5 curriculums).
  • It serves as a comprehensive reference for educators and students to understand the conceptual origins and mechanical progressions behind standard mathematical formulas rather than relying on rote memorization.

• Author and Editorial Acknowledgements:

  • Author: Cikgu Muhammad Fazdhly bin Abdul Muttalib, Guru Cemerlang Matematik Tambahan, based in Ipoh, Perak.
  • Reviewers and Collaborators:
  • En Ku Haslizam bin Ku Azmi (Guru Cemerlang Matematik)
  • En Haris Fadzli bin Awang (AMSG Administrator)
  • En Aizuddin bin Yusoff (AMSG Administrator)
  • En Noor Ishak bin Mohd Salleh (AMSG Administrator)
  • En Clemente Chock (AMSG Administrator)
  • Release Information: First released on 28 June 2021 as a continuous charitable contribution (amal jariah) to aid candidates sitting for the SPM examinations.

• Underlying References:

  • The derivations synthesize personal pedagogical expertise, selected international mathematical references, and local Malaysian educational texts spanning IPTA publications, STPM syllabi, KBSM textbooks, and KSSM secondary school textbooks.

Functions: Inverse Function Verification and Composite Functions

• Theoretical Invertibility Condition:

  • Two functions $f(x)$ and $g(x)$ are strictly inverse to one another if and only if their compositions in both directions yield the identity function:     $fg(x) = f[g(x)] = x$$gf(x) = g[f(x)] = x$
  • If either composite condition evaluates to anything other than $x$, the function $g(x)$ cannot be considered the inverse function of $f(x)$.

• Counterexample Problem:

  • Given $f(x) = x^2$ for $x \ge 0$ and $g(x) = -\sqrt{x}$ for $x \ge 0$.
  • Determine $fg(x)$ and $gf(x)$, and conclude whether $g(x)$ is the inverse function of $f(x)$.

• Step-by-Step Composition Derivation:

  • Evaluating $fg(x)$:     $fg(x) = f[g(x)]$$fg(x) = (-\sqrt{x})^2$$fg(x) = x$
  • Evaluating $gf(x)$:     $gf(x) = g[f(x)]$$gf(x) = -\sqrt{x^2}$$gf(x) = -x$

• Conclusion:

  • Because $gf(x) = -x \ne x$, the function $g(x)$ is not the inverse function of $f(x)$.

Quadratic Functions and Equations

• Derivation of the Quadratic Formula:

  • Given the general quadratic equation:     $ax^2 + bx + c = 0$     where $a$, $b$, and $c$ are constants, with $a \ne 0$.
  • Factor out the coefficient $a$ or divide through by $a$:     $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$
  • Apply the method of completing the square by adding and subtracting $\,\left(\frac{b}{2a}\right)^2$:     $\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} = 0$
  • Expand and isolate the squared binomial:     $\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} = 0$$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a}$
  • Establish a common denominator of $4a^2$ on the right-hand side:     $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$
  • Take the square root of both sides:     $x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$$x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}$
  • Solve explicitly for $x$ to yield the quadratic formula:     $x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

• Derivation of the Sum of Roots (SOR) and Product of Roots (POR):

  • Given that the roots of $ax^2 + bx + c = 0$ are $\alpha$ and $\beta$.
  • Divide the quadratic equation by $a$ to obtain the monic standard form:     $x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \quad \text{--- (1)}$
  • Express the quadratic equation in terms of its linear root factors:     $(x - \alpha)(x - \beta) = 0$
  • Expand the linear factors:     $x^2 - \alpha x - \beta x + \alpha\beta = 0$$x^2 - (\alpha + \beta)x + \alpha\beta = 0 \quad \text{--- (2)}$
  • Equate corresponding coefficients between equation (1) and equation (2):
  • Equating the coefficient of $x$:       $-(\alpha + \beta) = \frac{b}{a} \implies \alpha + \beta = -\frac{b}{a}$
  • Equating the constant term:       $\alpha\beta = \frac{c}{a}$

Laws of Logarithms

• The Product Law of Logarithms:

  • Objective: Prove that $\log_a(mn) = \log_a(m) + \log_a(n)$.
  • Let:     $m = a^x \implies \log_a(m) = x$$n = a^y \implies \log_a(n) = y$
  • Multiply the two terms using the index law for multiplication:     $mn = a^x \times a^y$$mn = a^{x+y}$
  • Apply the fundamental definition of a logarithm with base $a$:     $\log_a(mn) = x + y$
  • Substitute back the expressions for $x$ and $y$:     $\log_a(mn) = \log_a(m) + \log_a(n)$

• The Quotient Law of Logarithms:

  • Objective: Prove that $\log_a\left(\frac{m}{n}\right) = \log_a(m) - \log_a(n)$.
  • Let:     $m = a^x \implies \log_a(m) = x$$n = a^y \implies \log_a(n) = y$
  • Divide the two terms using the index law for division:     $\frac{m}{n} = \frac{a^x}{a^y}$$\frac{m}{n} = a^{x-y}$
  • Apply the fundamental definition of a logarithm with base $a$:     $\log_a\left(\frac{m}{n}\right) = x - y$
  • Substitute back $x$ and $y$:     $\log_a\left(\frac{m}{n}\right) = \log_a(m) - \log_a(n)$

• The Change of Base Formula:

  • Objective: For positive real numbers $a$, $b$, and $c$, where $a \ne 1$ and $c \ne 1$, prove:     $\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$
  • Let:     $\log_a(b) = x$
  • Convert into equivalent index/exponential form:     $a^x = b$
  • Take the logarithm of base $c$ on both sides:     $\log_c(a^x) = \log_c(b)$
  • Apply the power rule of logarithms on the left-hand side:     $x\log_c(a) = \log_c(b)$
  • Isolate $x$ by dividing by $\log_c(a)$:     $x = \frac{\log_c(b)}{\log_c(a)}$
  • Since $x = \log_a(b)$, substitution gives:     $\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$

Arithmetic and Geometric Progressions

• Sum of the First $n$ Terms of an Arithmetic Progression ($S_n$):

  • Let an arithmetic progression have first term $a$ and common difference $d$.
  • The sum of the first $n$ terms is written in ascending order of terms:     $S_n = T_1 + T_2 + T_3 + \dots + T_n$$S_n = a + [a + d] + [a + 2d] + \dots + [a + (n-1)d] \quad \text{--- (1)}$
  • Rewrite the exact sum in descending order of terms:     $S_n = [a + (n-1)d] + [a + (n-2)d] + [a + (n-3)d] + \dots + a \quad \text{--- (2)}$
  • Add equations (1) and (2) term by term:     $2S_n = [2a + (n-1)d] + [2a + (n-1)d] + [2a + (n-1)d] + \dots + [2a + (n-1)d]$
  • Because there are exactly $n$ identical terms in the summation:     $2S_n = n[2a + (n-1)d]$
  • Divide by $2$:     $S_n = \frac{n}{2}[2a + (n-1)d]$

• Alternative Form of $S_n$ Using the Last Term ($l$):

  • Beginning from the derived formula:     $S_n = \frac{n}{2}[2a + (n-1)d]$
  • Decompose the term $2a$ into $a + a$:     $S_n = \frac{n}{2}[a + a + (n-1)d]$
  • Recognize that $T_n = a + (n-1)d = l$, where $l$ is the last term of the sequence:     $S_n = \frac{n}{2}[a + T_n]$$S_n = \frac{n}{2}[a + l]$

• Sum of the First $n$ Terms of a Geometric Progression ($S_n$ for $r > 1$):

  • Let a geometric progression have first term $a$ and common ratio $r$.
  • Express the sum of $n$ terms:     $S_n = a + ar + ar^2 + \dots + ar^{n-1} \quad \text{--- (1)}$
  • Multiply equation (1) across by the common ratio $r$:     $rS_n = ar + ar^2 + ar^3 + \dots + ar^n \quad \text{--- (2)}$
  • Subtract equation (1) from equation (2):     $rS_n - S_n = (ar + ar^2 + ar^3 + \dots + ar^n) - (a + ar + ar^2 + \dots + ar^{n-1})$
  • All intermediate terms cancel out:     $S_n(r - 1) = ar^n - a$$S_n(r - 1) = a(r^n - 1)$
  • Divide through by $(r - 1)$ for $r > 1$:     $S_n = \frac{a(r^n - 1)}{r - 1}$

• Sum of the First $n$ Terms of a Geometric Progression ($S_n$ for $r < 1$):

  • Subtract equation (2) from equation (1):     $S_n - rS_n = (a + ar + ar^2 + \dots + ar^{n-1}) - (ar + ar^2 + ar^3 + \dots + ar^n)$
  • Factor out $S_n$ on the left and $a$ on the right:     $S_n(1 - r) = a - ar^n$$S_n(1 - r) = a(1 - r^n)$
  • Divide through by $(1 - r)$ for $r < 1$:     $S_n = \frac{a(1 - r^n)}{1 - r}$

• Sum to Infinity ($S_\infty$) of a Geometric Progression:

  • Consider the sum formula for $|r| < 1$:     $S_n = \frac{a(1 - r^n)}{1 - r}$
  • As the number of terms approaches infinity ($n \to \infty$), the term $r^n$ decays towards zero:     $\lim_{n \to \infty} r^n = 0$(For example, taking a fraction such as $0.399999 \approx 0$ when raised to an extremely high power)
  • Therefore, $(1 - r^n) \to (1 - 0) = 1$.
  • Substituting this limit into the sum yields:     $S_\infty = \frac{a(1)}{1 - r}$$S_\infty = \frac{a}{1 - r}, \quad |r| < 1$

Coordinate Geometry

• Division of a Line Segment (Section Formula):

• Problem Definition: Point $P(x, y)$ divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m : n$.

• Derivation via Ratio of Displacements (Method 1):

  • Comparing horizontal displacements:       $\frac{x - x_1}{x_2 - x} = \frac{m}{n}$$n(x - x_1) = m(x_2 - x)$$nx - nx_1 = mx_2 - mx$$nx + mx = nx_1 + mx_2$$x(m + n) = nx_1 + mx_2$$x = \frac{nx_1 + mx_2}{m + n}$
  • Comparing vertical displacements:       $\frac{y - y_1}{y_2 - y} = \frac{m}{n}$$n(y - y_1) = m(y_2 - y)$$ny - ny_1 = my_2 - my$$ny + my = ny_1 + my_2$$y(m + n) = ny_1 + my_2$$y = \frac{ny_1 + my_2}{m + n}$
  • Consequently, the coordinates of $P$ are:       $(x, y) = \left(\frac{nx_1 + mx_2}{m + n}, \frac{ny_1 + my_2}{m + n}\right)$

• Alternative Derivation via Fractional Vector Displacement (Method 2):

  • Position $x$ calculated from $x_1$ added to the fraction $\frac{m}{m+n}$ of the total run $(x_2 - x_1)$:       $x = x_1 + \frac{m}{m + n}(x_2 - x_1)$$x = \frac{x_1(m + n) + m(x_2 - x_1)}{m + n}$$x = \frac{mx_1 + nx_1 + mx_2 - mx_1}{m + n}$$x = \frac{nx_1 + mx_2}{m + n}$

  • Similarly, for the vertical coordinate $y$:       $y = y_1 + \frac{m}{m + n}(y_2 - y_1)$$y = \frac{y_1(m + n) + m(y_2 - y_1)}{m + n}$$y = \frac{my_1 + ny_1 + my_2 - my_1}{m + n}$$y = \frac{ny_1 + my_2}{m + n}$

  • Yielding the identical coordinate formula:       $(x, y) = \left(\frac{nx_1 + mx_2}{m + n}, \frac{ny_1 + my_2}{m + n}\right)$

  • Gradients of Parallel Lines:

• Let straight lines $L_1$ and $L_2$ be parallel.

• Let $\theta_1$ and $\theta_2$ denote the inclination angles made by $L_1$ and $L_2$ with the positive direction of the $x$-axis, respectively.

• By the geometric property of corresponding angles between parallel transversals:     $\theta_1 = \theta_2$

• Applying the tangent definition of gradient ($m = \tan(\theta)$):     $\tan(\theta_1) = \tan(\theta_2)$$m_1 = m_2$

  • Gradients of Perpendicular Lines:

• Let two straight lines $L_1$ and $L_2$ intersect perpendicularly at point $P$, with inclination angles $\theta_1$ and $\theta_2$ relative to the positive $x$-axis.

• The geometric relation between the angles is:     $\theta_2 = \theta_1 + 90^\circ$

• Taking the tangent of both sides:     $\tan(\theta_2) = \tan(\theta_1 + 90^\circ)$

• Using the trigonometric identity for an angle shifted into the second quadrant (where $\tan(\theta + 90^\circ) = -\frac{1}{\tan(\theta)}$):     $\tan(\theta_2) = -\frac{1}{\tan(\theta_1)}$

• Substituting gradients $m_1 = \tan(\theta_1)$ and $m_2 = \tan(\theta_2)$:     $m_2 = -\frac{1}{m_1}$$m_1 m_2 = -1$

  • Area of a Triangle on a Cartesian Plane (Shoelace Formula):

• Let vertices of triangle $ABC$ be defined as $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.

• Construct boundary perpendiculars onto the axes to divide the encompassing region into geometric components: triangle $ABE$, trapezium $ACDE$, and triangle $BCD$.

• Calculate component areas:

  • Area of triangle $ABE$ ($A_1$):       $A_1 = \frac{1}{2}(x_1 - x_2)(y_1 - y_2) = \frac{1}{2}(x_1 y_1 - x_1 y_2 - x_2 y_1 + x_2 y_2)$
  • Area of trapezium $ACDE$ ($A_2$):       $A_2 = \frac{1}{2}[(y_1 - y_2) + (y_3 - y_2)](x_3 - x_1)$$A_2 = \frac{1}{2}(x_3 y_1 - x_1 y_1 + x_3 y_3 - x_1 y_3 - 2x_3 y_2 + 2x_1 y_2)$
  • Area of triangle $BCD$ ($A_3$):       $A_3 = \frac{1}{2}(x_3 - x_2)(y_3 - y_2) = \frac{1}{2}(x_3 y_3 - x_3 y_2 - x_2 y_3 + x_2 y_2)$

• The total area of $\triangle ABC$ is given by combining the composite regions ($A_1 + A_2 - A_3$):     $\text{Area} = A_1 + A_2 - A_3$$\text{Area} = \frac{1}{2} |x_1 y_2 + x_2 y_3 + x_3 y_1 - x_2 y_1 - x_3 y_2 - x_1 y_3|$

Solution of Triangles

• The Sine Rule:

  • Let triangle $ABC$ have side lengths $AB = c$, $AC = b$, and altitude $AD = t$ perpendicular to $BC$.
  • In right-angled triangle $ABD$:     $\sin(B) = \frac{t}{c} \implies t = c\sin(B) \quad \text{--- (1)}$
  • In right-angled triangle $ACD$:     $\sin(C) = \frac{t}{b} \implies t = b\sin(C) \quad \text{--- (2)}$
  • Equate equations (1) and (2):     $b\sin(C) = c\sin(B)$
  • Divide through by $\sin(B)\sin(C)$:     $\frac{b}{\sin(B)} = \frac{c}{\sin(C)}$

• The Cosine Rule (Acute Angle Case):

  • In triangle $PQR$, let side lengths be $PQ = r$, $PR = q$, and $QR = p$.
  • Drop a perpendicular altitude $PS = t$ onto base $QR$. Divide $QR$ into segments $QS = x$ and $SR = p - x$.
  • In right-angled triangle $PQS$:     $t^2 = r^2 - x^2 \quad \text{--- (1)}$$\cos(Q) = \frac{x}{r} \implies x = r\cos(Q) \quad \text{--- (2)}$
  • In right-angled triangle $PSR$, apply Pythagoras' theorem:     $q^2 = t^2 + (p - x)^2$$q^2 = t^2 + p^2 - 2px + x^2 \quad \text{--- (3)}$
  • Substitute equations (1) and (2) into (3):     $q^2 = (r^2 - x^2) + p^2 - 2p(r\cos(Q)) + x^2$$q^2 = p^2 + r^2 - 2pr\cos(Q)$

• The Cosine Rule (Obtuse Angle Case):

  • Let angle $Q$ be obtuse. Extend base $RQ$ beyond $Q$ to meet the perpendicular altitude $PS = t$ at point $S$.
  • Let $SQ = x$; the total base of the large right-angled triangle $PSR$ is $SR = p + x$.
  • In right-angled triangle $PQS$:     $t^2 = r^2 - x^2 \quad \text{--- (1)}$$\cos(180^\circ - Q) = \frac{x}{r} \implies -\cos(Q) = \frac{x}{r} \implies x = -r\cos(Q) \quad \text{--- (2)}$
  • In right-angled triangle $PSR$:     $q^2 = t^2 + (p + x)^2$$q^2 = t^2 + p^2 + 2px + x^2 \quad \text{--- (3)}$
  • Substitute equations (1) and (2) into (3):     $q^2 = (r^2 - x^2) + p^2 + 2p(-r\cos(Q)) + x^2$$q^2 = p^2 + r^2 - 2pr\cos(Q)$

• Area of a Triangle Using Sine:

  • Let triangle $PQR$ have side lengths $PQ = r$, $PR = q$, with base $PR$.
  • Construct perpendicular altitude $QS = t$ to the line containing $PR$.
  • The standard geometric area formula is:     $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}qt \quad \text{--- (1)}$
  • From right-angled triangle $PQS$:     $\sin(P) = \frac{t}{r} \implies t = r\sin(P) \quad \text{--- (2)}$
  • Substitute equation (2) into equation (1):     $\text{Area} = \frac{1}{2}qr\sin(P)$

Circular Measure

• Arc Length Formula ($s = j\theta$):

  • Consider sector $OAB$ of a circle with center $O$, radius $j$, and central angle $\theta$ measured in radians.
  • Form a ratio of the arc length to total circumference relative to the central angle compared to a full revolution ($2\pi\text{ rad}$):     $\frac{\text{Arc length}}{\text{Circumference}} = \frac{\text{Subtended central angle}}{2\pi}$$\frac{s}{2\pi j} = \frac{\theta}{2\pi}$
  • Cancel $2\pi$ from both denominators:     $s = j\theta$

• Area of a Sector Formula ($L = \frac{1}{2}j^2\theta$):

  • Form a ratio of the sector area to total circular area relative to the central angle compared to a full revolution:     $\frac{\text{Area of sector}}{\text{Area of circle}} = \frac{\text{Subtended central angle}}{2\pi}$$\frac{L}{\pi j^2} = \frac{\theta}{2\pi}$
  • Cancel $\pi$ from both sides and cross-multiply $j^2$:     $L = \frac{1}{2}j^2\theta$

Differentiation: Foundational Rules from First Principles

• Inductive Derivation of the Power Rule:

  • Observing progressive polynomials and their first derivatives:
  • For $y = 6x$, $\frac{dy}{dx} = 6 = 6(1)x^{1-1}$
  • For $y = 6x^2$, $\frac{dy}{dx} = 12x = 6(2)x^{2-1}$
  • For $y = 6x^3$, $\frac{dy}{dx} = 18x^2 = 6(3)x^{3-1}$
  • By mathematical induction across positive integers $n$:     $\text{If } y = 6x^n, \quad \frac{dy}{dx} = 6nx^{n-1}$

• Sum Rule from First Principles:

  • Objective: Prove that $\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$.
  • Define $y = f(x) + g(x)$.
  • Introduce small increments $\delta x$ and $\delta y$:     $y + \delta y = f(x + \delta x) + g(x + \delta x)$$\delta y = [f(x + \delta x) + g(x + \delta x)] - [f(x) + g(x)]$$\delta y = [f(x + \delta x) - f(x)] + [g(x + \delta x) - g(x)]$
  • Divide across by $\delta x$:     $\frac{\delta y}{\delta x} = \frac{f(x + \delta x) - f(x)}{\delta x} + \frac{g(x + \delta x) - g(x)}{\delta x}$
  • Take the limit as $\delta x \to 0$:     $\frac{dy}{dx} = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} + \lim_{\delta x \to 0} \frac{g(x + \delta x) - g(x)}{\delta x}$$\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$

• Product Rule from First Principles (Functional Notation):

  • Objective: Prove that $\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$.
  • Let $y = u(x)v(x)$.
  • Express the incremented function:     $y + \delta y = u(x + \delta x)v(x + \delta x)$$\delta y = u(x + \delta x)v(x + \delta x) - u(x)v(x)$
  • Insert null terms by adding and subtracting $u(x + \delta x)v(x)$:     $\delta y = u(x + \delta x)v(x + \delta x) - u(x + \delta x)v(x) + u(x + \delta x)v(x) - u(x)v(x)$$\delta y = u(x + \delta x)[v(x + \delta x) - v(x)] + v(x)[u(x + \delta x) - u(x)]$
  • Divide by $\delta x$ and apply limits as $\delta x \to 0$:     $\frac{\delta y}{\delta x} = u(x + \delta x)\left[\frac{v(x + \delta x) - v(x)}{\delta x}\right] + v(x)\left[\frac{u(x + \delta x) - u(x)}{\delta x}\right]$$\frac{dy}{dx} = \left(\lim_{\delta x \to 0} u(x + \delta x)\right) v'(x) + v(x) u'(x)$
  • Because $u(x)$ is continuous, $\lim_{\delta x \to 0} u(x + \delta x) = u(x)$, giving:     $\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$

• Alternative Product Rule Derivation (Increment Notation):

  • Let $y = uv$.
  • Incrementing the variables:     $y + \delta y = (u + \delta u)(v + \delta v)$$y + \delta y = uv + u\,\delta v + v\,\delta u + \delta u\,\delta v$$\delta y = u\,\delta v + v\,\delta u + \delta u\,\delta v$
  • Divide through by $\delta x$:     $\frac{\delta y}{\delta x} = u\frac{\delta v}{\delta x} + v\frac{\delta u}{\delta x} + \delta u\frac{\delta v}{\delta x}$
  • Evaluate the limit as $\delta x \to 0$ (noting that $\delta u \to 0$ as $\delta x \to 0$):     $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} + (0)\frac{dv}{dx}$$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$

• Quotient Rule from First Principles (Functional Notation):

  • Objective: Prove that $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}$, given $v(x) \ne 0$.
  • Let $y = \frac{u(x)}{v(x)}$.
  • Increment the equation:     $y + \delta y = \frac{u(x + \delta x)}{v(x + \delta x)}$$\delta y = \frac{u(x + \delta x)}{v(x + \delta x)} - \frac{u(x)}{v(x)}$$\delta y = \frac{v(x)u(x + \delta x) - u(x)v(x + \delta x)}{v(x)v(x + \delta x)}$
  • Add and subtract $u(x)v(x)$ within the numerator:     $\delta y = \frac{v(x)[u(x + \delta x) - u(x)] - u(x)[v(x + \delta x) - v(x)]}{v(x)v(x + \delta x)}$
  • Divide by $\delta x$:     $\frac{\delta y}{\delta x} = \frac{1}{v(x)v(x + \delta x)} \left[ v(x)\left(\frac{u(x + \delta x) - u(x)}{\delta x}\right) - u(x)\left(\frac{v(x + \delta x) - v(x)}{\delta x}\right) \right]$
  • Take limits as $\delta x \to 0$:     $\frac{dy}{dx} = \frac{1}{v(x)v(x)} [v(x)u'(x) - u(x)v'(x)]$$\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}$

• Alternative Quotient Rule Derivation (Increment Notation):

  • Let $y = \frac{u}{v}$.
  • Set up incremented expressions:     $y + \delta y = \frac{u + \delta u}{v + \delta v}$$\delta y = \frac{u + \delta u}{v + \delta v} - \frac{u}{v}$$\delta y = \frac{v(u + \delta u) - u(v + \delta v)}{v(v + \delta v)}$$\delta y = \frac{v\,\delta u - u\,\delta v}{v(v + \delta v)}$
  • Divide across by $\delta x$:     $\frac{\delta y}{\delta x} = \frac{1}{v(v + \delta v)} \left[v\frac{\delta u}{\delta x} - u\frac{\delta v}{\delta x}\right]$
  • As $\delta x \to 0$, $\delta v \to 0$:     $\frac{dy}{dx} = \frac{1}{v(v + 0)} \left[v\frac{du}{dx} - u\frac{dv}{dx}\right]$$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

• Chain Rule Derivation Using Limits:

  • Given $y = f(u)$ and $u = g(x)$.
  • Express the ratio of finite increments:     $\frac{\delta y}{\delta x} = \frac{\delta y}{\delta u} \times \frac{\delta u}{\delta x}$
  • Take limits on both sides as $\delta x \to 0$ (noting that $\delta u \to 0$ as $\delta x \to 0$):     $\lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \left(\lim_{\delta u \to 0} \frac{\delta y}{\delta u}\right) \times \left(\lim_{\delta x \to 0} \frac{\delta u}{\delta x}\right)$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

Integration: Rules, Areas, and Volumes of Revolution

• Inductive Derivation of the Power Rule for Indefinite Integrals:

  • Evaluating integral sequences systematically:
  • $\int 6x\,dx = 3x^2 + c = \frac{6}{1 + 1}x^{1+1} + c$
  • $\int 6x^2\,dx = 2x^3 + c = \frac{6}{2 + 1}x^{2+1} + c$
  • $\int 6x^3\,dx = \frac{3}{2}x^4 + c = \frac{6}{3 + 1}x^{3+1} + c$
  • Inductive generalization for any real number $n \ne -1$:     $\text{If } y = 6x^n, \quad \int 6x^n\,dx = \frac{6}{n+1}x^{n+1} + c$
  • Mathematical Constraints:
  • $n \ne -1$ (since division by zero occurs if $n = -1$).
  • $c$ represents an arbitrary constant (pemalar sembarangan).

• Integration of Linear Composite Functions $(ax + b)^n$:

  • Objective: Prove that $\int (ax + b)^n\,dx = \frac{(ax + b)^{n+1}}{a(n + 1)} + c$ for $n \ne -1$.
  • Use algebraic substitution:     $\text{Let } u = ax + b$
  • Differentiate $u$ with respect to $x$:     $\frac{du}{dx} = a \implies dx = \frac{du}{a}$
  • Substitute into the integral:     $\int (ax + b)^n\,dx = \int u^n \left(\frac{du}{a}\right)$$= \frac{1}{a} \int u^n\,du$
  • Apply the power rule of integration:     $= \frac{1}{a} \left[\frac{u^{n+1}}{n + 1}\right] + c$
  • Substitute back $u = ax + b$:     $\int (ax + b)^n\,dx = \frac{(ax + b)^{n+1}}{a(n + 1)} + c$

• Area Under a Curve along the $x$-Axis as a Riemann Sum:

  • Consider the continuous curve $y = f(x)$ bounded from $x = a$ to $x = b$.
  • Subdivide the interval into $n$ rectangular strips of uniform width:     $\delta x = \frac{b - a}{n}$
  • Each strip has height $y_i$ and individual area $\delta A_i = y_i \delta x$.
  • Total approximate area of $n$ strips:     $L_n = \sum_{i=1}^n y_i \delta x$
  • Taking the limit as $n \to \infty$ (which forces $\delta x \to 0$):     $L = \lim_{\delta x \to 0} \sum_{i=1}^n y_i \delta x = \int_a^b y\,dx$

• Area Under a Curve along the $y$-Axis as a Riemann Sum:

  • Consider the continuous curve $x = f(y)$ bounded from $y = a$ to $y = b$.
  • Subdivide the vertical interval into $n$ horizontal strips of uniform width:     $\delta y = \frac{b - a}{n}$
  • Each strip has length $x_i$ and individual area $\delta A_i = x_i \delta y$.
  • Total approximate area:     $L_n = \sum_{i=1}^n x_i \delta y$
  • Taking the limit as $n \to \infty$ ($\delta y \to 0$):     $L = \lim_{\delta y \to 0} \sum_{i=1}^n x_i \delta y = \int_a^b x\,dy$

• Volume of Revolution Rotated $360^\circ$ About the $x$-Axis:

  • When the planar region bounded by $y = f(x)$, $x = a$, and $x = b$ rotates through $360^\circ$ around the $x$-axis, the volume is approximated by summing $n$ discrete cylindrical discs.
  • Each disc has thickness $\delta x = \frac{b - a}{n}$ and circular radius $y_i$.
  • The volume of a single discrete cylindrical disc is:     $\delta I_i = \pi y_i^2 \delta x$
  • Sum of all $n$ cylindrical discs:     $I_n = \sum_{i=1}^n \pi y_i^2 \delta x = \pi \sum_{i=1}^n y_i^2 \delta x$
  • As $n \to \infty$ ($\delta x \to 0$):     $I = \lim_{\delta x \to 0} \sum_{i=1}^n \pi y_i^2 \delta x = \pi \int_a^b y^2\,dx$

• Volume of Revolution Rotated $360^\circ$ About the $y$-Axis:

  • When the planar region bounded by $x = f(y)$, $y = a$, and $y = b$ rotates through $360^\circ$ around the $y$-axis, the solid is partitioned into $n$ horizontal cylindrical discs.
  • Each disc has thickness $\delta y = \frac{b - a}{n}$ and radius $x_i$.
  • Volume of a single disc:     $\delta I_i = \pi x_i^2 \delta y$
  • Summing the volume of $n$ discs:     $I_n = \pi \sum_{i=1}^n x_i^2 \delta y$
  • In the continuous limit where $n \to \infty$ ($\delta y \to 0$):     $I = \lim_{\delta y \to 0} \sum_{i=1}^n \pi x_i^2 \delta y = \pi \int_a^b x^2\,dy$