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Category: O-Level Math

The Cubic Formula

Let $b,c,d$ be constants with $a \neq 0$ .

In this post, we develop a formula to solve the cubic equation

$x^3 + bx^2 + cx + d = 0.$

Problem 1. Using the substitution $x = y + h$ , determine the constants $B,C,D$ in terms of $b,c,d,h$ such that

$x^3 + bx^2 + cx + d = y^3 + By^2 + Cy + D.$

Deduce the value of $h$ such that the right-hand side has no $y^2$ term.

(Click for Solution)

Solution. Making the substitution on the left-hand side,

$\begin{aligned} x^3 + bx^2 + cx + d &= (y+h)^3 + b(y+h)^2 \\ &\phantom{=.} + c(y+h) + d \\ &= (y^3 + 3y^2h + 3yh^2 + h^3) \\ &\phantom{=.} + b(y^2 + 2yh + h^2) \\ &\phantom{=.} + c(y+h) + d \\ &= y^3 + (3h + b)y^2 \\ &\phantom{=.} + (3h^2 + 2bh + c)y \\ &\phantom{=.} + (h^3 + bh^2 + ch + d). \end{aligned}$

Therefore,

$\begin{aligned} B &= 3h + b, \\ C &= 3h^2 + 2bh + c, \\ D &= h^3 + bh^2 + ch + d. \end{aligned}$

Hence, we set $B = 0 \iff h = -b/3$ .

Remark 1. Substituting $x = y - b/3$ ,

$x^3 + bx^2 + cx + d = 0 \quad \iff \quad y^3 + Cy + D = 0.$

Therefore, solving the right-hand side, called the depressed cubic, allows us to solve the left-hand side.

Problem 2 (Viète’s Substitution). Making the substitution

$\displaystyle y = z - \frac C{3z}$

into the right-hand side, obtain an equation that is quadratic in $z^3$ .

(Click for Solution)

Solution. By performing painful algebra,

$\begin{aligned} y^3 + Cy &= y(y^2 + C) \\ &= \left( z - \frac C{3z} \right) \left( \left( z - \frac C{3z} \right)^2 + C \right) \\ &= \left( z - \frac C{3z} \right) \left( z^2 - 2 \cdot z \cdot \frac{C}{3z} + \frac{C^2}{(3z)^2} + C \right) \\ &= \left( z - \frac C{3z} \right) \left( z^2 + \frac{C^2}{9z^2} + \frac C3 \right) \\ &= z^3 -\frac{C^3}{27z^3}. \end{aligned}$

Therefore,

$\begin{aligned} y^3 + Cy + D &= 0\\ z^3 -\frac{C^3}{27z^3} + D &= 0 \\ (z^3)^2 + Dz^3 - \frac{C^3}{27} &= 0. \end{aligned}$

Assume $D^2/4 + C^3/27 \geq 0$ .

Problem 3. Using Problems 1 and 2, solve the equation

$x^3 + bx^2 + cx + d = 0.$

(Click for Solution)

Solution. Using Problem 2, we solve for $z^3$ using the quadratic formula: $m := -D/2$ and $p := -C^2/27$ , so that

$\begin{aligned} m^2 - p &= \left( -\frac D2 \right)^2 - \left( - \frac{C^3}{27} \right) \\ &= \frac{D^2}{4} + \frac{C^2}{27}. \end{aligned}$

yields

$\begin{aligned} z^3 &= m \pm \sqrt{m^2 - p} \\ z^3 &= -\frac D2 \pm \sqrt{\frac{D^2}{4} + \frac{C^3}{27}} \\ z_{\pm} &:= \sqrt[3]{ -\frac D2 \pm \sqrt{\frac{D^2}{4} + \frac{C^3}{27}} }. \end{aligned}$

Substituting, we get two roots $y_{\pm} := z_{\pm}^3 - C^3/(27z_{\pm}^3)$ .

Denote the final root by $y_0$ . By comparing the constant term (i.e. set $y = 0$ ) on both sides of the equation

$\displaystyle (y-y_+)(y-y_-)(y-y_0) = y^3 + Cy + D,$

we get $y_0 = -D/(y_+ y_-)$ . The final solution is then recovered by the substitution in Problem 1:

$\displaystyle x = y_+ - \frac b3,\quad y_- - \frac b3,\quad y_0 - \frac b3.$

—Joel Kindiak, 24 Jan 26, 2056H

April 16, 2026
The Shoelace Formula

Problem 1. Consider the triangle below.

Show that the triangle has area $\frac 12 (x_1 y_2 - x_2 y_1)$ .

(Click for Solution)

Solution. Add dashed lines to draw in “phantom” triangles.

We subtract the area of the three smaller triangles from the larger rectangle:

$\begin{aligned} &\textstyle x_1 y_2 - \frac 12 x_1 y_1 - \frac 12 (x_1 - x_2)(y_2 - y_1) - \frac 12 x_2 y_2 \\ & = \textstyle x_1 y_2 - \frac 12 (x_1 y_1 + (x_1 - x_2)(y_2 - y_1) + x_2 y_2) \\ & = \textstyle x_1 y_2 - \frac 12 (x_1 y_1 + (x_1 y_2 - x_1 y_1 - x_2 y_2 + x_2 y_1) + x_2 y_2) \\ & = \textstyle x_1 y_2 - \frac 12 ( x_1 y_2 + x_2 y_1 ) \\ &= \textstyle \frac 12 (2x_1 y_2 - (x_1 y_2 + x_2 y_1)) \\ &= \textstyle \frac 12 (x_1 y_2 - x_2 y_1). \end{aligned}$

Remark 1. There are many other kinds of triangles, but with enough patience, they can all be shown to yield the same area formula.

Definition 1. Define

$\begin{aligned} \left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| &:= x_1 y_2 - x_2 y_1. \end{aligned}$

In particular, the area of the triangle in Problem 1 can be written as $\displaystyle \frac 12 \left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right|$ .

Problem 2. Show that $\left| \begin{matrix} x_2 & x_1 \\ y_2 & y_1 \end{matrix} \right| = -\left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right|$ . In particular,

$\left| \begin{matrix} x_2 & x_1 \\ y_2 & y_1 \end{matrix} \right| + \left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| = 0.$

(Click for Solution)

Solution. By definition,

$\begin{aligned} \left| \begin{matrix} x_2 & x_1 \\ y_2 & y_1 \end{matrix} \right| &= (x_2y_1 - x_1y_2) \\ &= -(x_1 y_2 - x_1 y_2) \\ &= -\left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right|. \end{aligned}$

Remark 1. In particular, a polygon is said to have positive orientation if its coordinates are traversed in anti-clockwise direction, and have negative orientation if its coordinates are traversed in clock-wise direction.

Problem 3. Consider the triangle below with positive orientation.

Show that the triangle has area

$\displaystyle \frac 12 \left(\left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| + \left| \begin{matrix} x_2 & x_3 \\ y_2 & y_3 \end{matrix} \right| + \left| \begin{matrix} x_3 & x_1 \\ y_3 & y_1 \end{matrix} \right|\right).$

(Click for Solution)

Solution. Draw some “phantom” triangles:

Denote the area of the triangle by $A$ . By Problem 1,

$\begin{aligned}A + \frac 12 \left| \begin{matrix} x_1 & x_3 \\ y_1 & y_3 \end{matrix} \right| &= \frac 12 \left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| + \frac 12 \left| \begin{matrix} x_2 & x_3 \\ y_2 & y_3 \end{matrix} \right|. \end{aligned}$

By Problem 2,

$\begin{aligned}A &= \frac 12 \left(\left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| + \left| \begin{matrix} x_2 & x_3 \\ y_2 & y_3 \end{matrix} \right| - \left| \begin{matrix} x_1 & x_3 \\ y_1 & y_3 \end{matrix} \right| \right) \\ &= \frac 12 \left(\left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| + \left| \begin{matrix} x_2 & x_3 \\ y_2 & y_3 \end{matrix} \right| + \left| \begin{matrix} x_3 & x_1 \\ y_3 & y_1 \end{matrix} \right| \right). \end{aligned}$

Remark 2. In a similar manner, for triangles with other orientations, as long as the vertices are traversed in an anti-clockwise direction, we still recover the same formula as in Problem 3.

Definition 2. Make the notation

$\begin{aligned} \left| \begin{matrix} x_1 & x_2 & \cdots & x_n \\ y_1 & y_2 & \cdots & y_n\end{matrix} \right| &:= \left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| + \left| \begin{matrix} x_2 & x_3 \\ y_2 & y_3 \end{matrix} \right| + \cdots + \left| \begin{matrix} x_{n-1} & x_n \\ y_{n-1} & y_n \end{matrix} \right|. \end{aligned}$

This notation is commonly referred to as the shoelace formula. In particular, the area in Problem 3 can be expressed as

$\displaystyle \frac 12 \left| \begin{matrix} x_1 & x_2 & x_3 & x_1 \\ y_1 & y_2 & y_3 & y_1 \end{matrix} \right|.$

Problem 4. Show that

$\left| \begin{matrix} x_1 & \cdots & x_k \\ y_1 & \cdots & y_k\end{matrix} \right| + \left| \begin{matrix} x_k & \cdots & x_n \\ y_k & \cdots & y_n\end{matrix} \right| = \left| \begin{matrix} x_1 & x_2 & \cdots & x_n \\ y_1 & y_2 & \cdots & y_n\end{matrix} \right|.$

(Click for Solution)

Solution. Patiently expand the right-hand side:

$\begin{aligned} &\left| \begin{matrix} x_1 & \cdots & x_k & \cdots & x_n \\ y_1 & \cdots & y_k & \cdots & y_n\end{matrix} \right| \\ &= \left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| + \cdots + \left| \begin{matrix} x_{k-1} & x_k \\ y_{k-1} & y_k \end{matrix} \right| + \left| \begin{matrix} x_k & x_{k+1} \\ y_k & y_{k+1} \end{matrix} \right| + \cdots + \left| \begin{matrix} x_{n-1} & x_n \\ y_{n-1} & y_n \end{matrix} \right| \\ &= \left| \begin{matrix} x_1 & \cdots & x_k \\ y_1 & \cdots & y_k\end{matrix} \right| + \left| \begin{matrix} x_k & \cdots & x_n \\ y_k & \cdots & y_n\end{matrix} \right| . \end{aligned}$

Problem 5. Consider the quadrilateral below.

Show that the quadrilateral has area

$\displaystyle \frac 12 \left| \begin{matrix} x_1 & x_2 & x_3 & x_4 & x_1 \\ y_1 & y_2 & y_3 & y_4 & y_1 \end{matrix} \right|.$

(Click for Solution)

Solution. Connect $(x_1,y_1)$ to $(x_3,y_3)$ .

The quadrilateral consists of two triangles, and therefore its area is given by the sum of both triangles, whose formulas are given by Problem 4:

$\begin{aligned} & \left| \begin{matrix} x_1 & x_2 & x_3 & x_1 \\ y_1 & y_2 & y_3 & y_1 \end{matrix} \right| + \left| \begin{matrix} x_1 & x_3 & x_4 & x_1 \\ y_1 & y_3 & y_4 & y_1 \end{matrix} \right| \\ &= \left| \begin{matrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{matrix} \right| + \underbrace{ \left| \begin{matrix} x_3 & x_1 \\ y_3 & y_1 \end{matrix} \right| + \left| \begin{matrix} x_1 & x_3 \\ y_1 & y_3 \end{matrix} \right| }_0 + \left| \begin{matrix} x_3 & x_4 & x_1 \\ y_3 & y_4 & y_1 \end{matrix} \right| \\ &= \left| \begin{matrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{matrix} \right| + \left| \begin{matrix} x_3 & x_4 & x_1 \\ y_3 & y_4 & y_1 \end{matrix} \right| \\ &= \left| \begin{matrix} x_1 & x_2 & x_3 & x_4 & x_1 \\ y_1 & y_2 & y_3 & y_4 & y_1 \end{matrix} \right|, \end{aligned}$

where the $0$ comes from Problem 2. Multiplying both sides by $1/2$ yields the formula

$\displaystyle \frac 12 \left| \begin{matrix} x_1 & x_2 & x_3 & x_4 & x_1 \\ y_1 & y_2 & y_3 & y_4 & y_1 \end{matrix} \right|.$

Remark 2. Using mathematical induction and the same strategy in Problem 5, the signed area of an $n$ -gon with vertices

$(x_1, y_1), (x_2,y_2), \dots, (x_n, y_n)$

can be shown to equal

$\displaystyle \frac 12 \left| \begin{matrix} x_1 & x_2 & \cdots & x_n & x_1 \\ y_1 & y_2 & \cdots & y_n & y_1 \end{matrix} \right|.$

Furthermore, this quantity is positive if the vertices are traversed in an anti-clockwise direction and negative if the vertices are traversed in a clockwise direction.

—Joel Kindiak, 24 Jan 26, 1722H

April 15, 2026
Quadri-literacy
Definition 1. A quadrilateral is a four-sided shape.

For example, a rectangle is a quadrilateral whose internal angles are all $90^\circ$ .

Definition 2. A rhombus is a quadrilateral with equal side lengths.

Problem 1. Show that a quadrilateral is a square if and only if it is both a rectangle and a rhombus.
(Click for Solution)

Solution. $(\Leftarrow)$ Suppose a quadrilateral is both a rectangle and a rhombus.
- As a rectangle, all of its angles are $90^\circ$ .
- As a rhombus, all of its side lengths are equal.
Therefore, it must be a square.

$(\Rightarrow)$ Trivial.
Definition 3. A parallelogram is a quadrilateral with two pairs of parallel lines.

Problem 2. Show that the following are equivalent for a given quadrilateral $Q$ :
1. $Q$ is a parallelogram,
2. opposite sides in $Q$ are equal,
3. opposite angles in $Q$ are equal.
(Click for Solution)

Solution. $(1 \Rightarrow 2)$ Consider the parallelogram $Q = ABCD$ .

Draw the diagonal $BD$ . Since $AB \parallel DC$ , by alternate angles,

$\angle ABD = \angle CDB.$

As a common side, $BD = DB$ . Since $AD \parallel BC$ , by alternate angles,

$\angle ADB = \angle CBD.$

By the ASA Criterion, $\Delta ABD \equiv \Delta CDB$ . In particular, $AB = CD$ and $AD = CB$ , so that its opposite sides are equal.

$(2 \Rightarrow 3)$ Draw the diagonal and use the SSS Criterion to conclude that opposite angles equal each other.

$(3 \Rightarrow 1)$ Let $\alpha,\beta,\alpha,\beta$ denote the angles of the quadrilateral.

Since angles in a quadrilateral sum to $360^\circ$ ,

$2(\alpha + \beta) = 360^\circ \quad \Rightarrow \quad \alpha + \beta = 180^\circ.$

Since this equality holds for any pair of interior angles, the parallelogram must have two pairs of parallel sides.

Problem 3. Show that a rectangle is always a parallelogram. Furthermore, show that a parallelogram is a rectangle if and only if it has at least one interior right angle.

(Click for Solution)

Solution. Since all angles in a rectangle is $90^\circ$ , opposite pairs of angles are equal.

By Problem 2, a rectangle is a parallelogram.

$(\Leftarrow)$ Let $90^\circ, \alpha, \beta, \gamma$ denote the interior angles of the parallelogram, labelled anti-clockwise.

By Problem 2, $\alpha = \gamma$ and $\beta = 90^\circ$ . Since angles in a quadrilateral sum to $360^\circ$ ,

$\alpha + 90^\circ + \alpha + 90^\circ = 360^\circ \quad \Rightarrow \quad \alpha = 90^\circ.$

Therefore, all angles equal $90^\circ$ , and the parallelogram is a rectangle.

$(\Rightarrow)$ Trivial.

Problem 4. Show that a rhombus is always a parallelogram. Furthermore, show that a parallelogram is a rhombus if and only if it has at least one pair of equal adjacent sides.

(Click for Solution)

Solution. Since opposite sides in a rhombus are equal, by Problem 2, a rhombus is a parallelogram.

$(\Leftarrow)$ Let $w,x,y,z$ denote the sides of a parallelogram, labelled anti-clockwise.

By Problem 2, $w=y$ and $x=z$ . By hypothesis, suppose $w=x$ without loss of generality. Then $z=x=w=y$ . Therefore, all side lengths are equal, and the parallelogram is a rhombus.

$(\Rightarrow)$ Trivial.

Problem 5. Show that a rectangle is a square if and only if it has at least one pair of equal adjacent sides. Likewise, show that a rhombus is a square if and only if it has at least one interior right angle.
(Click for Solution)

Solution. We first prove the rectangle claim:
- $(\Leftarrow)$ By Problem 3, a rectangle is a parallelogram. By hypothesis and Problem 4, it is a rhombus. By Problem 1, it is a square.
- $(\Rightarrow)$ Trivial.
The rhombus claim follows similarly:
- $(\Leftarrow)$ By Problem 4, a rhombus is a parallelogram. By hypothesis and Problem 2, it is a rectangle. By Problem 1, it is a square.
- $(\Rightarrow)$ Trivial.
Definition 4. A trapezium is a quadrilateral with at least one pair of opposite sides that are parallel.

Problem 6. Show that a parallelogram is always a trapezium. Furthermore, show that a trapezium is a parallelogram if and only if it has at least one pair of equal opposite angles.

(Click for Solution)

Solution. Since a parallelogram has two pairs of equal and parallel sides, it has at least one pair of parallel sides, and is thus a trapezium.

$(\Leftarrow)$ Denote the angles of the trapezium by $\alpha, \beta, \alpha, \theta$ .

Given the pair of parallel sides, interior angles are supplementary, so that

$\alpha +\beta = 180^\circ = \alpha + \theta \quad \Rightarrow \quad \beta = \theta.$

By Problem 2, the trapezium is a parallelogram.

$(\Rightarrow)$ Trivial.

Definition 5. A kite is a quadrilateral with two pairs of adjacent sides that are equal in length.

Problem 7. Show that a kite has at least one pair of equal opposite angles and that its diagonals are perpendicular to each other.

(Click for Solution)

Solution. Consider the kite $ABCD$ below.

As base angles of isosceles triangles,

$\angle CBD = \angle CDB, \quad \angle ABD = \angle ADB.$

Hence,

$\begin{aligned} \angle ABC &= \angle ABD + \angle CBD \\ &= \angle ADB + \angle CDB \\ &= \angle ADC, \end{aligned}$

as required. Using the SAS Criterion, $\Delta ABC \equiv \Delta ADC$ . In particular, $\angle BAR = \angle DAR$ . As a kite, $AB = AD$ . As base angles of an isosceles triangle,

$\angle ABR = \angle ABD = \angle ADB = \angle ADR.$

Using the ASA Criterion, $\Delta ABR \equiv \angle ADR$ , so that $\angle ARB = \angle ARD$ . Since adjacent angles on a straight line are supplementary, $\angle ARB + \angle ARD = 180^\circ$ . Solving, $\angle ARB = 90^\circ$ .

Problem 8. Show that a quadrilateral is a rhombus if and only if it is both a kite and a trapezium.

(Click for Solution)

Solution. $(\Leftarrow)$ Suppose the quadrilateral is both a kite and a trapezium. By Problem 7, it has at least one pair of equal opposite angles. By Problem 6, it is a parallelogram. As a kite, it has at least one pair of equal adjacent sides. By Problem 4, it is a rhombus.

$(\Rightarrow)$ Trivial.

—Joel Kindiak, 24 Jan 26, 1651H
April 15, 2026
Quadratic Identities

Problem 1. Verify the following quadratic identities:

$\begin{aligned} (a+b)^2 &= a^2 + 2ab + b^2, \\ (a-b)^2 &= a^2 - 2ab + b^2, \\ a^2 - b^2 &= (a+b)(a-b). \end{aligned}$

(Click for Solution)

Solution. By the rainbow method,

$\begin{aligned} (a+b)^2 &= (a+b)(a+b) \\ &= a(a+b) + b(a+b) \\ &= a^2 + ab + ba + b^2 \\ &= a^2 + ab + ab + b^2 \\ &= a^2 + 2ab + b^2. \end{aligned}$

Replacing $b$ with $-b$ ,

$\begin{aligned} (a-b)^2 &= (a+(-b))^2 \\ &= a^2 + 2a(-b) + (-b)^2 \\ &= a^2 - 2ab + b^2. \end{aligned}$

Finally,

$\begin{aligned}(a+b)(a-b) &= a(a-b) + b(a-b) \\ &= a^2 - ab + ba - b^2 \\ &= a^2 - ab+ab - b^2 \\ &= a^2 - b^2.\end{aligned}$

Let $b, c$ be integers.

Problem 2. Define $m := -b/2$ and $p:= c$ . Show that the quadratic equation

$x^2 + bx + c = 0$

has solutions given by $x = m \pm \sqrt{m^2 - p}$ .

(Click for Solution)

Solution. Write $b = -2m$ . Using the quadratic formula, the solutions are given by

$\begin{aligned} x &= \frac{-b \pm \sqrt{b^2 - 4 \cdot 1 \cdot c}}{2 \cdot 1} \\ &= \frac{2m \pm \sqrt{(2m)^2 - 4p}}{2} \\ &= \frac{2m \pm \sqrt{4m^2 - 4p}}{2} \\ &= \frac{2m \pm \sqrt 4 \cdot \sqrt{m^2 - p}}{2} \\ &= \frac{2m \pm 2 \sqrt{m^2 - p}}{2} \\ &= m \pm \sqrt{m^2 - p}.\end{aligned}$

Problem 3. Given that the quadratic equation

$x^2 + bx + c = 0.$

has roots $r, s$ , evaluate $b,c$ in terms of $r,s$ . Deduce that

$x^2 + bx + c = (x-r)(x-s),$

and show that $\Delta = 4(s-r)^2$ .

(Click for Solution)

Solution. Using Problem 2, suppose

$r = m - \sqrt{m^2 - p},\quad s = m + \sqrt{m^2 - p}.$

Then $r+s = 2m = -b$ and by Problem 1,

$\begin{aligned} rs &= (m - \sqrt{m^2 - p}) \cdot (m + \sqrt{m^2 - p})\\ &= m^2 - (m^2-p) \\ &= p = c. \end{aligned}$

Therefore, $b = -(r+s)$ and $c = rs$ . Hence,

$\begin{aligned} x^2 + bx + c &= x^2 -(r+s)x + rs \\ &= x^2 -rx-xs + rs \\ &= (x-r)x-(x-r)s \\ &= (x-r)(x-s). \end{aligned}$

It follows that $\Delta = 4 (m^2 - p) = 4(s-r)^2$ .

Now suppose $c$ is a positive integer.

Problem 4. Suppose a quadratic equation of the form

$x^2 - x - c = 0$

has integer solutions with larger solution $r$ . Evaluate $c$ in terms of $r$ .

(Click for Solution)

Solution. By Problem 3,

$-(r+s) = -1\quad \Rightarrow \quad s = 1-r = -(r-1)$

and $c = -rs = -r(1-r) = r(r-1)$ .

Problem 5. Show that a quadratic equation of the form

$x^2 - x + c = 0$

has no real roots.

(Click for Solution)

Solution. Since $c \geq 1$ , by calculating the discriminant of the quadratic equation,

$\begin{aligned} \Delta &= (-1)^2 - 4 \cdot 1 \cdot c \\ &= 1 - 4c \leq 1 - 4 \cdot 1 \\ &= -3 < 0. \end{aligned}$

Therefore, the quadratic has no real roots.

Problem 6. Determine the values of $r > 0$ such that

$x^2 - x + r = 0$

has real roots.

(Click for Solution)

Solution. We require $\Delta \geq 0$ :

$(-1)^2 - 4 \cdot 1 \cdot r \geq 0 \quad \iff \quad r \leq 1/4.$

Therefore, we require $0 < r \leq 1/4$ .

Problem 7. Given non-negative numbers $a, b$ , show that

$\displaystyle \sqrt{ab} \leq \frac{a+b}{2}.$

This inequality is a special case of the arithmetic mean-geometric mean (AM-GM) inequality.

(Click for Solution)

Solution. Using Problem 1,

$\begin{aligned} 0 &\leq (\sqrt{a}-\sqrt{b})^2 \\ &= (\sqrt{a})^2 - 2\sqrt{a}\sqrt{b} + (\sqrt{b})^2 \\ &= a - 2\sqrt{ab} + b. \end{aligned}$

By algebruh, $2\sqrt{ab}\leq a+b$ . Dividing by $2$ yields the desired result.

—Joel Kindiak, 24 Jan 26, 0213H

April 14, 2026
Further Calculus

If you won’t further your studies into a STEM discipline, then the tools and techniques in this post would not be terribly relevant for you. But for the aspiring STEM student, we are now going to do calculus with trigonometric functions and exponential functions.

The derivative of $y = \sin(x)$ is an exceedingly challenging idea to compute without more technical tool of limits outside the high school syllabus. To conceive of a meaningful derivation sacrifices either rigour or intuitiveness. Nevertheless, for this post, I will present the intuitive idea that appeared to me as I lay in my bed, fast asleep.

Denote $f(x) = \sin(x)$ . Derivatives intuitively measures the gradient of the tangent to the curve. Therefore, in the diagram below, I have computed the gradient of the tangent to each point on the curve $y = f(x)$ . I colour-coded each point $(t, f(t))$ using the gradient $f'(t)$ at that point on a scale from very-red (i.e. gradient of $-1$ ) to very-green (i.e. gradient of $+1$ ). Hence, the graph goes from green to red, then back to green at the very end.

If now we plot the points of the new graph $y = f'(x)$ according to the colour-coding, the first quarter of the new graph lies in the top-half green section. The second and third quarters of the new graph lie in the bottom-half red section. And finally, the fourth quarter of the new graph lie in the top half-green section once again.

Recalling our discussions on trigonometric graphs, $y=f'(x)$ looks very much like the graph of $y = \cos(x)$ . This is, in fact, mathematically true.

Theorem 1. $\displaystyle \frac{\mathrm d}{\mathrm dx}(\sin(x)) = \cos(x)$ .

Proof. Omitted. Or relegated to a study in more advanced calculus. In the language of limits, we need to use the limit definition of the derivative to show that

$\displaystyle \lim_{h \to 0} \frac{\sin(x + h) - \sin(x)}{ h } = \cos(x).$

Then the core result is proven using the limit

$\displaystyle \lim_{h \to 0} \frac{\sin(h)}{h} = 1.$

Remark 1. Strictly speaking, we have put the cart before the horse—I couldn’t colour the sine graph unless I already knew that the derivative of $\sin(x)$ is $\cos(x)$ . Nevertheless, the goal of this post is to motivate the result, then relegating a formal proof elsewhere in the blog.

You might wonder—in that case, why not obtain all other derivatives in this manner? Other than the fact that I need to slave-drive ChatGPT harder than I already do, it turns out that the existing differentiation technology, if we accept them to remain true for non-polynomials, empowers us to obtain (almost) all of these other derivatives.

For instance, the chain rule helps us compute the derivative of $\cos(x)$ , since the complementary angle identities tell us that

$\begin{aligned} \cos(x) &= \sin(\pi/2 - x), \\ \sin(x) &= \cos(\pi/2 - x). \end{aligned}$

Example 1. Show that $\displaystyle \frac{\mathrm d}{\mathrm dx}(\cos(x)) = -{\sin(x)}$ .

Solution. Using the complementary angle identities, the chain rule, and Theorem 1,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(\cos(x)) &= \frac{\mathrm d}{\mathrm dx}(\sin(\pi/2 - x)) \\ &= \cos(\pi/2 - x) \cdot \frac{\mathrm d}{\mathrm dx} (\pi/2 - x) \\ &= \sin( x) \cdot \left( \frac{\mathrm d}{\mathrm dx} (\pi/2) - \frac{\mathrm d}{\mathrm dx} (x) \right) \\ &= \sin( x) \cdot ( 0 - 1 ) \\ &= -{\sin(x)}. \end{aligned}$

These two results, coupled with the quotient rule, help us compute the derivative of $\tan(x)$ , since $\tan(x) = \sin(x)/{\cos(x)}$ . These reasons and more motivated our study on trigonometric identities.

Example 2. Show that $\displaystyle \frac{\mathrm d}{\mathrm dx}(\tan(x)) = \sec^2(x)$ .

Solution. Using the quotient rule and previous theorems,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(\tan(x)) &= \frac{\mathrm d}{\mathrm dx} \left( \frac{\sin(x)}{\cos(x)} \right) \\ &= \frac{\frac{\mathrm d}{\mathrm dx}(\sin(x)) \cdot \cos(x) - \sin(x) \cdot \frac{\mathrm d}{\mathrm dx}(\cos(x)) }{\cos^2(x)} \\ &= \frac{ \cos(x) \cdot \cos(x) - \sin(x) \cdot (-{\sin(x)}) }{\cos^2(x)} \\ &= \frac{ \cos^2(x) + \sin^2(x) }{\cos^2(x)} \\ &= \frac 1{\cos^2(x)} \\ &= \sec^2(x). \end{aligned}$

There are many more commonly-used derivatives in this exercise here. We now state the integral versions of these three results.

Theorem 4. The following integrals hold:

$\begin{aligned} \int \sin(x)\, \mathrm dx &= -{\cos(x)} + C, \\ \int \cos(x)\, \mathrm dx &= \sin(x) + C, \\ \int \sec^2(x)\, \mathrm dx &= \tan(x) + C. \end{aligned}$

Proof. We prove the first result to illustrate the power of thinking of integration as reverse differentiation. Using Theorem 2 and linearity,

$\displaystyle -\int \sin(x)\, \mathrm dx= \int -{\sin(x)}\, \mathrm dx = \cos(x) + C.$

Therefore dividing by $-1$ on both sides,

$\displaystyle \int \sin(x)\, \mathrm dx = -{\cos(x)} + C',$

where $C' := -C$ .

We have to discuss the exponential family too. Once again, just like with the derivative of $\sin(x)$ , the derivative of $e^x$ requires more technical real-analytic tools to properly establish. Nevertheless, here is an experimentally-inspired attempt at motivating the result.

Let’s first ask ourselves: how can we experimentally compute the gradient $f'(t)$ of the tangent $L_t$ at a point $P(t, f(t))$ on the graph of $y = f(x)$ ? By definition, this tangent must pass through the point $Q(t+1, f(t) + f'(t))$ .

Draw a line $M_t$ parallel to the tangent passing through $R(t+1, f(t))$ . We leave it as an exercise to check that its $x$ -intercept $x_0$ is given by the expression

$\displaystyle x_0 = (t+1)-\frac{ f(t) }{ f'(t) }.$

In particular, if we know that $M_t$ has an $x$ -intercept of $x_0$ , then we can compute $f'(t)$ via

$\displaystyle f'(t) = \frac{ f(t) }{(t+1)-x_0}.$

Now particularise to $f(x) = e^x$ , which we can draw by sampling input values:

Notice that the $x$ -intercept of $M_t$ is exactly $x_0 = t$ . This result is not a bug—it’s a feature of the graph of $y = e^x$ , and it works for any $t$ . In particular, using our “experimental” calculation of $f'(t)$ , we get

$\displaystyle f'(t) = \frac{f(t)}{(t+1) - t} = f(t) = e^t.$

This result is a consequence of the famous differentiation result for exponentials.

Theorem 2. $\displaystyle \frac{\mathrm d}{\mathrm dx}(e^x) = e^x$ .

Proof. Omitted.

Remark 2. Once again, we have assumed Theorem 2 in our illustration. A complete proof requires a rigorous definition of the real exponential, then deducing its various properties, and concluding by using the limit definition of the derivative. In the language of limits, the heart of the proof is the limit result

$\displaystyle \lim_{h \to 0} \frac{e^h - 1}{h} = 1.$

Recall that $e^x$ has a foil: the logarithm. More precisely, whenever well-defined,

$e^{\ln(x)} = x,\quad x > 0.$

Example 3. Show that $\displaystyle \frac{\mathrm d}{\mathrm dx}(\ln(x)) = 1/x$ for $x > 0$ . The domain restriction $x > 0$ allows the expression $\ln(x)$ to make sense.

Solution. Using the chain rule and Theorem 2,

$\begin{aligned} e^{\ln(x)} &= x \\ \frac{\mathrm d}{\mathrm dx} (e^{\ln(x)}) &= \frac{\mathrm d}{\mathrm dx}(x) \\ e^{\ln(x)} \cdot \frac{\mathrm d}{\mathrm dx}(\ln(x)) &= 1 \\ \frac{\mathrm d}{\mathrm dx}(\ln(x)) &= \frac 1{ e^{\ln(x)} } = \frac 1x. \end{aligned}$

Remark 1. For $x < 0$ , we have $-x > 0$ , so the chain rule yields

$\displaystyle \frac{\mathrm d}{\mathrm dx} (\ln(-x)) = \frac 1{-x} \cdot (-1) = \frac 1x.$

Therefore, the integral version of these results are given by

$\displaystyle \int \frac 1x\, \mathrm dx = \begin{cases} \ln(x) + C_1, & x > 0, \\ \ln(-x) + C_2, & x < 0.\end{cases}$

Since the absolute value is defined by

$|x| = \begin{cases} x, & x \geq 0, \\ -x, & x < 0, \end{cases}$

we can abbreviate the integral result as follows:

$\displaystyle \int \frac 1x\, \mathrm dx = \ln |x| + C,\quad x \neq 0.$

In particular, the absolute value symbol $| \cdot |$ is essential in our final answer.

Definition 1. Define $-\pi/2 \leq \sin^{-1}(x) \leq \pi/2$ for $-1 \leq x \leq 1$ by

$\sin^{-1}(x) = y \quad \iff \quad \sin(y) = x.$

Similarly, define $0 \leq \cos^{-1}(x) \leq \pi$ for $-1 \leq x \leq 1$ by

$\cos^{-1}(x) = y \quad \iff \quad \cos(y) = x.$

Likewise, define $-\pi/2 < \tan^{-1}(x) < \pi/2$ for real $x$ by

$\tan^{-1}(x) = y \quad \iff \quad \tan(y) = x.$

In particular,

$\begin{aligned}\sin(\sin^{-1}(x)) &= x, \\ \cos(\cos^{-1}(x)) &= x, \\ \tan(\tan^{-1}(x)) &= x. \end{aligned}$

The domain restrictions allow the expressions $\sin^{-1}(x)$ , $\cos^{-1}(x)$ , $\tan^{-1}(x)$ to make sense, similar to how the domain restriction $x > 0$ allows the expression $\ln(x)$ to make sense.

Example 4. Show that $\displaystyle \int \frac 1{\sqrt{1-x^2}}\, \mathrm dx = \sin^{-1}(x) + C$ .

Solution. For the first result, use the chain rule and Theorem 1:

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(\sin(\sin^{-1}(x))) &= \frac{\mathrm d}{\mathrm dx}(x) \\ \cos(\sin^{-1}(x)) \cdot \frac{\mathrm d}{\mathrm dx}(\sin^{-1}(x)) &= 1 . \end{aligned}$

Since $-1 < x < 1$ , $-\pi/2 < \sin^{-1}(x) < \pi/2$ , so that $0 < \cos(\sin^{-1}(x)) \leq 1$ .

By the Pythagorean identity,

$\begin{aligned} \cos(\sin^{-1}(x)) &= \sqrt{ 1 - \sin^2(\sin^{-1}(x)) } \\ &={ \sqrt{ 1 - ( \sin(\sin^{-1}(x)) )^2 } } \\ &= { \sqrt{ 1 - x^2 } }. \end{aligned}$

Therefore,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(\sin^{-1}(x)) &= \frac 1{\cos(\sin^{-1}(x))} = \frac 1{\sqrt{1-x^2}}.\end{aligned}$

By definition,

$\displaystyle \int \frac 1{\sqrt{1-x^2}}\, \mathrm dx = \sin^{-1}(x) + C.$

Example 5. Show that $\displaystyle \int \frac 1{1+x^2}\, \mathrm dx = \tan^{-1}(x) + C$ .

Solution. Follow the strategy in Example 4 by using the chain rule and Example 2:

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(\tan(\tan^{-1}(x))) &= \frac{\mathrm d}{\mathrm dx}(x) \\ \sec^2(\tan^{-1}(x)) \cdot \frac{\mathrm d}{\mathrm dx}(\tan^{-1}(x)) &= 1. \end{aligned}$

Using the Pythagorean identity,

$\begin{aligned} \sec^2(\tan^{-1}(x)) &= 1 + \tan^2(\tan^{-1}(x)) \\ &= 1 +( \tan(\tan^{-1}(x)) )^2 \\ &= 1 + x^2. \end{aligned}$

Therefore,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(\tan^{-1}(x)) = \frac 1{\sec^2(\tan^{-1}(x))} = \frac 1{1 + x^2}.$

By definition,

$\displaystyle \int \frac 1{ 1+x^2 }\, \mathrm dx = \tan^{-1}(x) + C.$

We conclude with one rather peculiar result regarding $\cos^{-1}(x)$ .

Example 6. Given that $-1 \leq x \leq 1$ , show that

$\displaystyle \frac{\mathrm d}{\mathrm dx}(\cos^{-1}(x)) = - \frac 1{\sqrt{1-x^2}}.$

Solution. By the complementary angle identities,

$\begin{aligned}\cos^{-1}(x) &= \cos^{-1}(\sin(\sin^{-1}(x))) \\ &= \cos^{-1}(\cos(\pi/2 - \sin^{-1}(x))) \\ &= \pi/2 - \sin^{-1}(x).\end{aligned}$

Differentiating on both sides,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(\cos^{-1}(x)) &= \frac{\mathrm d}{\mathrm dx}(\pi/2) - \frac{\mathrm d}{\mathrm dx}(\sin^{-1}(x)) \\ &= 0 -\frac 1{\sqrt{ 1 - x^2 }} \\ &= -\frac 1{\sqrt{ 1 - x^2 }}. \end{aligned}$

We have barely scratched the surface of calculus. There is much more that can be explored in many depths. At a high level, the area interpretation of integration allows mathematicians and statisticians to model probabilities—our conceived notions of randomness. We explore a special sub-branch of this topic next time.

—Joel Kindiak, 23 Jan 26, 2045H

April 13, 2026
Applied Integration

What was the point of integration? To compute areas. Which sounds strange. Don’t we already have meaningful formulas for common areas?

Example 1. Consider the graph $y = x$ below, with $0 < a < b$ .

Define $F(x) = \displaystyle \int x\, \mathrm dx$ . Show that the shaded region has area

$F(b) - F(a).$

Solution. Using integration,

$F(x) = \displaystyle \int x\, \mathrm dx = \textstyle \frac{1}{2}x^2 + C.$

In particular,

$\begin{aligned} F(b) - F(a) &= \textstyle \left( \frac 12 b^2 + C \right) - \left( \frac 12 a^2 + C \right) \\ &= \textstyle \frac 12 b^2 - \frac 12 a^2. \end{aligned}$

Since the shaded region is a trapezium, it has an area given by

$\begin{aligned} (\text{area}) &= \textstyle \frac 12 \cdot (b+a) \cdot (b-a) \\ &= \textstyle \frac 12 \cdot (b^2 - a^2) \\ &= \textstyle \frac 12 b^2 - \frac 12 a^2 \\ &= F(b) - F(a). \end{aligned}$

Example 2. Consider the graph $y = x^2$ below, with $b > 0$ .

Define $\displaystyle F(x) = \int x^2\, \mathrm dx$ . The Greek mathematician Archimedes calculated the blue region to have an area of $\frac 16 a^3$ . Show that the area of $R_{0,a}$ is given by

$F(a) - F(0).$

Deduce that the area of $R_{a,b}$ is given by $F(b) - F(a)$ .

Solution. Using integration,

$F(x) = \displaystyle \int x^2\, \mathrm dx = \textstyle \frac{1}{3}x^3 + C.$

In particular, $F(a) - F(0) = \frac 13 a^3$ .

On the other hand, $R_{0,a}$ has area

$\begin{aligned} (\text{area of } R_{0,a}) &= \textstyle \frac 12 \cdot a \cdot a^2 - \frac 16 a^3 \\ &= \textstyle \frac 12 a^3 - \frac 16 a^3 \\ &= \textstyle \frac 13 a^3 = F(a) - F(0). \end{aligned}$

The region $R_{a,b}$ can be thought of as the “leftover” region of $R_{0,b}$ after removing $R_{0,a}$ , and hence has area

$(F(b) - F(0)) - (F(a) - F(0)) = F(b) - F(a).$

This pattern turns out to be true in more general settings. We will simply state this general pattern, and omit its proof (referencing it for more a more advanced study of calculus).

Theorem 1. Consider the graph $y = f(x)$ below, and suppose it lies above the $x$ -axis.

Define $F(x) = \displaystyle \int f(x)\, \mathrm dx$ . Then the area of the shaded region is given by

$\displaystyle F(b) - F(a).$

Proof. This result is known as the famous fundamental theorem of calculus, which is rigorously proven elsewhere in the blog.

Definition 1. Given any function $G(x)$ , make the notation

$[G(x)]_a^b := G(b) - G(a).$

Suppose we know the functions $f(x), F(x)$ such that

$\displaystyle F'(x) = f(x) \quad \iff \quad \int f(x)\, \mathrm dx = F(x) + C.$

Then we define the definite integral of $f(x)$ from $a$ to $b$ by

$\displaystyle \int_a^b f(x)\, \mathrm dx := [F(x)]_a^b \equiv F(b) - F(a).$

In particular, if $y = f(x)$ lies above the $x$ -axis, then

$\displaystyle \int_a^b f(x)\, \mathrm dx$

denotes the area of the region in Theorem 1.

Example 3. For $n \neq -1$ , evaluate $\displaystyle \int_0^1 x^n\, \mathrm dx$ .

Solution. Using integration,

$\displaystyle \int x^n\, \mathrm dx = \frac{x^{n+1}}{n+1} +C.$

Hence,

$\displaystyle \int_0^1 x^n\, \mathrm dx = \left[ \frac{x^{n+1}}{n+1} \right]_0^1 = \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1} = \frac 1{n+1}.$

More generally, for any real $t$ ,

$\displaystyle \int_0^t x^n\, \mathrm dx = \frac {t^{n+1}}{n+1}.$

Using the definition of the definite integral, we can recover several important integration properties.

Theorem 2. Let $a,b,c,\alpha$ be real constants with $a \leq b \leq c$ . Whenever well-defined, the following definite integral properties hold:

$\begin{aligned} \int_a^b(f(x)+g(x))\, \mathrm dx &= \int_a^b f(x)\, \mathrm dx + \int_a^b g(x)\, \mathrm dx, \\ \int_a^b(\alpha \cdot f(x))\, \mathrm dx &= \alpha \cdot \int_a^b f(x)\, \mathrm dx, \\ \int_a^c f(x)\, \mathrm dx &= \int_a^b f(x)\, \mathrm dx + \int_b^c f(x)\, \mathrm dx, \\ \int_a^a f(x)\, \mathrm dx &= 0, \\ \int_b^a f(x)\, \mathrm dx &= -\int_a^b f(x)\, \mathrm dx. \end{aligned}$

Proof. We will prove just the third result and relegate the rest of the results as exercises. Suppose $F(x)$ satisfies

$\displaystyle F(x) = \int f(x)\, \mathrm dx.$

Then

$\begin{aligned} \int_a^b f(x) \, \mathrm dx + \int_b^c f(x) \, \mathrm dx &= [F(x)]_a^b + [F(x)]_b^c \\ &= (F(b) - F(a)) + (F(c) - F(b)) \\ &= F(c) - F(a) \\ &= \int_a^c f(x) \, \mathrm dx. \end{aligned}$

Example 4. Evaluate the exact area of the following region.

Solution. Since the area under the graph is calculated using an integral,

$\begin{aligned}(\text{area}) &= \int_0^1(x^2 + 1)\, \mathrm dx \\ &= \int_0^1 x^2\, \mathrm dx + \int_0^1 x^0\, \mathrm dx \\ &= \frac 1{2+1} + \frac 1{0+1} \\ &= \frac 43\, \text{units}^2.\end{aligned}$

Remark 1. Paradoxically, a rigorous treatment of calculus first proves Theorem 2 using a more fundamental definition of integration, then uses Theorem 2 and other real-analytic tools to prove Theorem 1.

Remark 2. The theory of integration is an incredibly deep rabbit hole, arguably deeper than that of differentiation. Its uses are deep and far-reaching due to its connections with probability (which influences basically almost every area of life). However, to keep things simple, we will restrict our attention to simple computations of integrals.

The rest of this post could evolve into a mere hodge-podge of integration drills, but perhaps we can think about Theorem 1 a little more closely.

Example 5. Consider the graph of $y = 1-x^2$ below.

Evaluate $\displaystyle \int_0^1 (1 - x^2)\, \mathrm dx$ and $\displaystyle \int_0^1 (x^2 - 1)\, \mathrm dx$ .

Solution. Using the linearity of integration

$\begin{aligned} \int_0^1(1-x^2 )\, \mathrm dx &= \int_0^1 x^0\, \mathrm dx - \int_0^1 x^2\, \mathrm dx \\ &= \frac 1{0+1} - \frac 1{2+1} \\ &= \frac 23.\end{aligned}$

By Theorem 2,

$\begin{aligned} \int_0^1(x^2 - 1)\, \mathrm dx &= -\int_0^1(1-x^2 )\, \mathrm dx = -\frac 23.\end{aligned}$

If, instead, we graphed $y = x^2 - 1$ , the corresponding shaded region would lie below the $x$ -axis.

Furthermore, it would have an area of $2/3$ units².

Hence, strictly speaking, the integral only accounts for the signed area, which is positive if $y = f(x)$ lies above the $x$ -axis, and negative if $y = f(x)$ lies below the $x$ -axis.

In many ways, integration was formulated to answer problems in physics.

Definition 2. Sir Isaac Newton used calculus to formulate the connections between displacement $s(T)$ , velocity $v(T)$ , and acceleration $a(T)$ at time $T$ as follows:

$\begin{aligned} s(T) &:= s(0) + \int_0^T v(t)\, \mathrm dt, \\ v(T) &:= v(0) + \int_0^T a(t)\, \mathrm dt. \end{aligned}$

We remark that these definitions agree with the usual velocity-time graph (for displacement) and the acceleration-time graph (for velocity), where we calculate the desired quantities by evaluating the areas under the graphs (hence, corresponding with the integral formulation)

He was interested in the special case when $a(t)$ is a constant number, namely $a(t) \approx -9.81 =: g$ , known as the gravitational acceleration near the surface of Earth.

Theorem 3. Given constants $s_0, v_0, a_0$ , suppose

$s(0) = s_0, \quad v(0) = v_0, \quad a(t) = a_0$ .

Then

$v(T) = v_0 + a_0 T, \quad s(T) = s_0 + v_0 T + \frac 12 a_0 T^2.$

Furthermore, $v(T)^2 = v_0^2 + 2a_0 \cdot (s(T) - s_0)$ .

Proof. Integrating $a(t)$ ,

$\displaystyle \int a(t)\, \mathrm dt = \int a_0\, \mathrm dt = a_0t + C.$

Therefore,

$\begin{aligned} v(T) &= v(0) + \int_0^T a(t)\, \mathrm dt \\ &= v_0 + [a_0x]_0^T \\ &= v_0 + (a_0 T - a_0 \cdot 0) \\ &= v_0 + a_0T. \end{aligned}$

Similarly,

$\begin{aligned} \int v(t)\, \mathrm dt &= \int (v_0+a_0t)\, \mathrm dt \\ &= \int v_0\, \mathrm dt + a_0 \int t\, \mathrm dt \\ &= (v_0t + C_1) + \left(a_0 \cdot \frac{t^2}{2} + C_2 \right) \\ &= \textstyle v_0t + \frac 12 a_0t^2 + C, \end{aligned}$

where $C := C_1 + C_2$ . Therefore,

$\begin{aligned} s(T) &= s(0) + \int_0^T v(t)\, \mathrm dt \\ &= \textstyle s_0 + \left[ v_0t + \frac 12 a_0t^2 \right]_0^T \\ &= \textstyle s_0 + \left( v_0T + \frac 12 a_0T^2 \right) - \left( v_0\cdot 0 + \frac 12 a_0 \cdot 0^2 \right) \\ &= \textstyle s_0 + v_0T + \frac 12 a_0T^2 . \end{aligned}$

In particular,

$\begin{aligned} v(T)^2 &= ( v_0 + a_0T )^2 \\ &= \textstyle v_0^2 + 2v_0a_0T + a_0^2 T^2 \\ &= \textstyle v_0^2 + 2a_0 {\left( v_0T + \frac 12 a_0 T^2 \right)} \\ &= v_0^2 + 2a_0 \cdot (s(T) - s_0). \end{aligned}$

Theorem 3 lists out several common laws of kinematics, in particular, for an object moving in a straight line at constant acceleration. If instead we started by knowing the displacement $s(t)$ , we can recover the velocity and acceleration of the particle using differentiation.

Theorem 4. The displacement $s(t)$ , velocity $v(t)$ , and acceleration $a(t)$ of a particle moving in a straight line are related by the equations:

$\displaystyle v(t) = \frac{\mathrm ds}{\mathrm dt},\quad a(t) = \frac{\mathrm dv}{\mathrm dt}.$

In particular, $\displaystyle a(t) = \frac{\mathrm d^2 s}{\mathrm dt^2}$ . This connection arises ubiquitously in physics due to Newton’s second law.

Proof. Write $\displaystyle F(x) = \int v(x)\, \mathrm dx$ so that $F'(x) = v(x)$ and

$s(t) = s(0) + F(t) - F(0).$

Differentiating on both sides,

$\begin{aligned} \frac{\mathrm ds}{\mathrm dt} = \frac{\mathrm d}{\mathrm dt}(s(t)) &= \frac{\mathrm d}{\mathrm dt}(s(0) + F(t) - F(0)) \\ &= \frac{\mathrm d}{\mathrm dt}(s(0)) + \frac{\mathrm d}{\mathrm dt}(F(t)) - \frac{\mathrm d}{\mathrm dt}(F(0)) \\ &= 0 + F'(t) - 0 \\ &= F'(t) \\ &= v(t). \end{aligned}$

The argument holds similarly for $a(t)$ .

Similarly, these connections agree with the usual displacement-time graph (for velocity) and the velocity-time graph (for acceleration), where we calculate the desired quantities at a specific time by evaluating the gradient of the tangent at that point.

We have only scratched the surface regarding applied calculus, and you can explore even more in adjacent STEM fields like physics and economics.

For now, we need to answer a crucial question. So far, we have only discussed calculus regarding the simple-enough polynomials. But can we discuss calculus on the trigonometric functions like $\sin(x)$ and $\cos(x)$ , and the exponential family $e^x$ and $\ln(x)$ of functions?

The answer is yes, sort of. A rigorous treatment takes a lot more effort into the nuts and bolts of calculus. Nevertheless, we can still appreciate these formulas from a visually intuitive perspective, and I think there is still a lot to enjoy from viewing it that way.

—Joel Kindiak, 23 Jan 26, 1937H

April 10, 2026
An Absurd Puzzle

Problem 1. Evaluate

$\displaystyle S := \frac 1{\sqrt 1 + \sqrt 2} + \frac 1{\sqrt 2 + \sqrt 3} + \cdots + \frac 1{\sqrt{2024} + \sqrt{2025}}.$

(Click for Solution)

Solution. By rationalising the denominator,

$\begin{aligned} \frac 1{\sqrt n + \sqrt{n+1}} &= \frac 1{\sqrt n + \sqrt{n+1}} \cdot \frac{ \sqrt n - \sqrt{n+1} }{ \sqrt n - \sqrt{n+1} } \\ &= \frac{ \sqrt n - \sqrt{n+1} }{ (\sqrt n)^2 - (\sqrt{n+1})^2 } \\ &= \frac{ \sqrt n - \sqrt{n+1} }{ n - (n+1) } \\ &= \frac{ \sqrt n - \sqrt{n+1} }{ -1 } \\ &= \sqrt{n+1} - \sqrt{n}. \end{aligned}$

Hence, summing the terms,

$\begin{aligned} S &=\frac 1{\sqrt 1 + \sqrt 2} + \frac 1{\sqrt 2 + \sqrt 3} + \cdots + \frac 1{\sqrt{2023} + \sqrt{2024}} + \frac 1{\sqrt{2024} + \sqrt{2025}} \\ &= \sqrt 2 + \sqrt 3 + \cdots + \sqrt{2024} + \sqrt{2025} \\ &\phantom{=} - \sqrt{1} - \sqrt 2 - \cdots - \sqrt{2023} - \sqrt{2024} \\ &= \sqrt{2025} - 1 \\ &= 45 - 1 = 44. \end{aligned}$

Problem 2. Evaluate

$\displaystyle T := \frac 1{1 \times 2} + \frac 1{2 \times 3} + \cdots + \frac 1{n(n+1)}$

in terms of $n$ .

(Click for Solution)

Solution. For any $k$ ,

$\begin{aligned} \frac 1{k(k+1)} &= \frac {(k+1)-k}{k(k+1)} \\ &= \frac{k+1}{k(k+1)} - \frac k{k(k+1)} \\ &= \frac 1k - \frac 1{k+1}. \end{aligned}$

Hence,

$\begin{aligned} T &= \frac 1{1 \times 2} + \frac 1{2 \times 3} + \cdots + \frac 1{(n-1)n} + \frac 1{n(n+1)} \\ &= \frac 11 + \frac 12 + \cdots + \frac 1{n-1} + \frac 1n \\ &\phantom{=} - \frac 12 - \frac 13 - \cdots - \frac 1n - \frac 1{n+1} \\ &= 1 - \frac 1{n+1} = \frac{n}{n+1}. \end{aligned}$

Remark 1. In both problems, we used the method of differences or a telescoping sum (or more fancifully, the fundamental theorem of discrete calculus), which states that if $u_n = v_{n+1} - v_n$ , then

$u_1 + u_2 + \cdots + u_n = v_{n+1} - v_1.$

—Joel Kindiak, 19 Jan 26, 2204H

April 9, 2026
The Construction Toolkit

Definition 1. A rhombus is a shape made up of four straight line segments whose side lengths are all equal.

In the rhombus ABCD above, we call AC and BD the diagonals of the rhombus that intersect at R.

Problem 1. Show that the diagonals of a rhombus are perpendicular bisectors of each other.

(Click for Solution)

Solution. Since angles in a triangle are supplementary,

$\begin{aligned} \angle ABD + \angle ADB + \angle BAD &= 180^\circ, \\ \angle CBD + \angle CDB + \angle BCD &= 180^\circ. \end{aligned}$

Since

$\begin{aligned} \angle ABC &= \angle ABD + \angle CBD,\\ \angle ADC &= \angle ADB + \angle CDB, \end{aligned}$

summing the equations yields

$\begin{aligned} \angle ABC + \angle ADC + \angle BAD + \angle BCD &= 360^\circ. \end{aligned}$

Since $BA = BC$ , $\Delta BAC$ is isosceles, so that $\angle BAC = \angle BCA$ . Similarly,

$\angle ABD = \angle ADB,\quad \angle CBD = \angle CDB, \quad \angle DAC = \angle DCA.$

Therefore,

$\begin{aligned} \angle ABC &= \angle ABD + \angle CBD \\ &= \angle ADB + \angle CDB \\ &= \angle ADC. \end{aligned}$

Similarly, $\angle BAD = \angle BCD$ . Hence,

$\begin{aligned} \angle ABC + \angle ADC + \angle BAD + \angle BCD &= 360^\circ \\ \angle ABC + \angle ABC + \angle BAD + \angle BAD &= 360^\circ \\ 2 \times (\angle ABC + \angle BAD) &= 360^\circ \\ \angle ABC + \angle BAD &= 180^\circ. \end{aligned}$

Since the interior angles sum to $180^\circ$ , $AD \parallel BC$ . By alternate angles,

$\angle ADB = \angle CBD = \angle CDB.$

Using previous data, we have $AD = CD$ ,

$\begin{aligned} \angle ADR &= \angle ADB = \angle CDB = \angle CDR, \\ \angle RAD &= \angle CAD = \angle ACD = \angle RCD. \end{aligned}$

By the ASA Criterion, $\Delta ADR \equiv \Delta CDR$ . Similarly,

$\Delta ADR \equiv \Delta CDR \equiv \Delta CBR \equiv \Delta ABR.$

Therefore, $AR = CR$ and $BR = DR$ . Furthermore, since adjacent angles on a straight line are supplementary,

$\begin{aligned} 2 \angle ARB &= \angle ARB + \angle ARB \\ &= \angle ARB + \angle CRB \\ &= 180^\circ, \end{aligned}$

so that $\angle ARB = 90^\circ$ , and similarly,

$\angle ARB = \angle CRB = \angle CRD = \angle ARD = 90^\circ.$

Therefore, $AC$ and $BD$ are perpendicular bisectors of each other.

Problem 2. Let AB be the line segment below, and let r > 0 be a length such that

AB/2 < r ≤ AB.

Here, C_A is a circle with centre A and radius r. Define C_B likewise. Let P, Q denote the intersections between C_A, C_B.

Show that the line passing through P, Q is the perpendicular bisector of AB.

(Click for Solution)

Solution. As radii of the same circle,

$AP = AQ = r = BP = BQ.$

Therefore, $APBQ$ forms a rhombus. By Problem 1, $PQ$ lies on the perpendicular bisector of $AB$ .

Problem 3. Let OA, OB be the line segments below, and let r > 0 be any length.

Here, C_O denotes the circle with centre O and radius r. Suppose C_O intersects OA at P and OB at Q. Define C_A, C_B similarly, and denote their intersection by R.

Show that ∠AOR = ∠BOR. In this case, we call the line passing through O, R the angle bisector of ∠AOB.

(Click for Solution)

Solution. As radii of the same circle,

$PO = QO = r = PR = QR.$

Therefore, $OPRQ$ forms a rhombus. By Problem 1, $PQ$ and $OR$ are perpendicular bisectors of each other.

Denote their intersection by $S$ . Then $PS = QS$ , $OP = OQ$ , and $OS = OS$ . By the SSS Criterion, $\Delta POS = \Delta QOS$ . In particular,

$\angle AOR = \angle POS = \angle QOS = \angle BOR,$

as required.

—Joel Kindiak, 19 Jan 26, 1400H

April 8, 2026
Baby Integration

Let’s introduce baby integration as reverse-differentiation, which, comically, I have criticised elsewhere on my blog.

Given functions $f(x)$ and $F(x)$ , write

$\displaystyle \int f(x)\, \mathrm dx = F(x) \quad \iff \quad \frac{\mathrm d}{\mathrm dx}(F(x)) = f(x).$

We call the left-hand equation the indefinite integral of $f(x)$ with respect to $x$ .

(Wrong) Example 1. Recall that $\displaystyle \frac{\mathrm d}{\mathrm dx}(x^n) = n x^{n-1}$ . Therefore,

$\displaystyle \int n x^{n-1}\, \mathrm dx = x^n.$

Now strictly speaking, our integrals are wrong.

Lemma 1. Suppose $f(x)$ and $F(x)$ are functions such that

$\displaystyle \frac{\mathrm d}{ \mathrm dx}(F(x)) = f(x).$

Then, for any real number $C$ ,

$\displaystyle \frac{\mathrm d}{ \mathrm dx}(F(x) + C) = f(x).$

Proof. Using the linearity of differentiation,

$\begin{aligned} \frac{\mathrm d}{ \mathrm dx}(F(x) + C) &= \frac{\mathrm d}{ \mathrm dx}(F(x)) + \frac{\mathrm d}{ \mathrm dx}(C) \\ &= f(x) + 0 \\ &= f(x). \end{aligned}$

Therefore, we use the following definition for indefinite integrals.

Definition 1. Given functions $f(x)$ and $F(x)$ , write

$\displaystyle \int f(x)\, \mathrm dx = F(x) +C \quad \iff \quad \frac{\mathrm d}{\mathrm dx}(F(x)) = f(x),$

where $C$ is an arbitrary constant of integration.

Example 1. Recall that $\displaystyle \frac{\mathrm d}{\mathrm dx}(x^n) = n x^{n-1}$ . Therefore, for $n \neq 0$ ,

$\displaystyle \int n x^{n-1}\, \mathrm dx = x^n + C.$

Now, Definition 1 is far from complete, but works well enough for most secondary-school assessments.

Theorem 1. Indefinite integration is linear: given functions $f(x), g(x)$ and any real constant $c$ ,

$\begin{aligned} \int(f(x) + g(x))\, \mathrm dx &= \int f(x)\, \mathrm dx + \int g(x)\, \mathrm dx, \\ \int cf(x)\, \mathrm dx &= c \int f(x)\, \mathrm dx. \end{aligned}$

Proof. Write

$\displaystyle \int f(x)\, \mathrm dx = F(x) + C_1,\quad \int g(x)\, \mathrm dx = G(x) + C_2$

so that

$\displaystyle \frac{\mathrm d}{\mathrm dx}(F(x) + C_1) = f(x),\quad \frac{\mathrm d}{\mathrm dx}(G(x) + C_1) = g(x).$

By the linearity of differentiation,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(F(x) + C_1 + G(x) + C_2) &= \frac{\mathrm d}{\mathrm dx}(F(x) + C_1) + \frac{\mathrm d}{\mathrm dx}(G(x) + C_2) \\ &= f(x) + g(x). \end{aligned}$

For the second result, the linearity of differentiation yields

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(c\cdot (F(x) + C_1)) &= c \cdot \frac{\mathrm d}{\mathrm dx}(F(x) + C_1) = c f(x). \end{aligned}$

By Definition 1,

$\begin{aligned} \int( c f(x) )\, \mathrm dx &= c \cdot ( F(x) + C_1 ) = c \int f(x)\, \mathrm dx . \end{aligned}$

Example 2. For any $n \neq -1$ , evaluate $\displaystyle \int x^n\, \mathrm dx$ . In particular, for any constant $a$ , evaluate

$\displaystyle \int a\, \mathrm dx,\quad \int \sqrt x\, \mathrm dx,\quad \int \frac 1{x^2}\, \mathrm dx.$

Solution. From Example 1, if $m \neq 0$ , then

$\displaystyle \int mx^{m-1}\, \mathrm dx = x^m + C.$

Hence, writing $m:=n+1 \iff n := m-1$ , for $n \neq -1$ ,

$\displaystyle \int (n+1)x^n\, \mathrm dx = x^{n+1} + C.$

By the linearity of integration,

$\displaystyle (n+1)\int x^n\, \mathrm dx = x^{n+1} + C.$

Therefore,

$\displaystyle \int x^n\, \mathrm dx = \frac{x^{n+1}}{n+1} + \frac{C}{n+1}.$

Since $C' := C/(n+1)$ is still an arbitrary real constant,

$\displaystyle \int x^n\, \mathrm dx = \frac{x^{n+1}}{n+1} + C'.$

The other results follow from linearity and the law of exponents:

$\begin{aligned} \int a\, \mathrm dx &= \int a x^0\, \mathrm dx = a \int x^0 \, \mathrm dx = a \cdot \frac x1 + C = ax + C, \\ \int \sqrt x\, \mathrm dx &= \int x^{1/2}\, \mathrm dx = \frac{x^{3/2}}{3/2} + C = \frac 23 \cdot x\sqrt x + C, \\ \int \frac 1{x^2}\, \mathrm dx &= \int x^{-2}\, \mathrm dx = \frac{x^{-1}}{-1} + C = -\frac 1x + C. \end{aligned}$

Example 3. Evaluate $\displaystyle \int (x^2 + 1)^2\, \mathrm dx$ .

Solution. Expanding $(x^2 + 1)^2$ yields

$(x^2 + 1)^2 = (x^2)^2 + 2x^2 + 1^2 = x^4 + 2x^2 + 1.$

Since integration is linear,

$\begin{aligned} \int (x^2 + 1)^2 \, \mathrm dx &= \int (x^4 + 2x^2 + 1)\, \mathrm dx \\ &= \int x^4\, \mathrm dx + 2 \int x^2\, \mathrm dx + \int 1\, \mathrm dx \\ &= \frac{ x^5 }{ 5 } + 2 \cdot \frac{ x^3 }{ 3 } + x + C \\ &= \textstyle \frac 15 x^5 + \frac 23 x^3 + x + C. \end{aligned}$

Remark 1. Where possible, it is good practice to present your final answer as a sum of terms of the form $Ax^nf(x)$ , where $A,n$ are constants with $n$ being a positive integer and $f(x)$ is not a polynomial.

Example 3 should suggest that computing integrals exactly would require a lot more effort than computing derivatives exactly. That is correct. For instance, we don’t actually yet know how to calculate

$\displaystyle \int \frac 1x\, \mathrm dx = \int x^{-1}\, \mathrm dx,$

since Example 2 only works for $n \neq -1$ .

Eventually, we will see that

$\displaystyle \int \frac 1x\, \mathrm dx = \ln|x| + C,$

which should surprise us—how did the logarithm magically appear there? For another fun result,

$\displaystyle \int \frac 1{ 1 + x^2 }\, \mathrm dx = \tan^{-1}(x) + C.$

What else could we do with integration?

Example 4. Let $a, b$ be real constants. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx}((ax+b)^n)$ . Hence, evaluate

$\displaystyle \int (ax+b)^n\, \mathrm dx.$

Solution. Using the chain rule,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}((ax+b)^n) &= n(ax + b)^{n-1} \cdot \frac{\mathrm d}{\mathrm dx}(ax+b) \\ &= n(ax+b)^{n-1} \cdot \left(a \cdot \frac{\mathrm d}{\mathrm dx}(x) + \frac{\mathrm d}{\mathrm dx}(b) \right) \\ &= n(ax+b)^{n-1} \cdot (a\cdot 1 + 0) \\ &= an(ax+b)^{n-1}.\end{aligned}$

By Definition 1 and the linearity of integration,

$\begin{aligned}\int an(ax+b)^{n-1}\, \mathrm dx &= (ax+b)^n + C \\ an\int (ax+b)^{n-1}\, \mathrm dx &= (ax+b)^n + C \\\int (ax+b)^{n-1}\, \mathrm dx &= \frac {(ax+b)^n}{an} + \frac{C}{an} \\ &= \frac {(ax+b)^n}{an} + C', \end{aligned}$

where $C' := C/(an)$ is an arbitrary real constant of integration.

Using the same idea in Example 4, we have the following special form of integration.

Theorem 2. If $\displaystyle \frac{\mathrm d}{\mathrm dx}(F(x)) = f(x)$ , then for real constants $a, b$ ,

$\displaystyle \int f(ax+b)\, \mathrm dx = \frac 1a \cdot F(ax+b) + C.$

Proof. Using the chain rule,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(F(ax+b)) &= f(ax + b) \cdot \frac{\mathrm d}{\mathrm dx}(ax+b) \\ &= a \cdot f(ax+b).\end{aligned}$

By Definition 1 and the linearity of integration,

$\begin{aligned}\int a f(ax+b)\, \mathrm dx &= F(ax+b) + C \\ a \int f(ax+b)\, \mathrm dx &= (ax+b)^n + C \\ \int f(ax+b)\, \mathrm dx &= \frac {1}{a} \cdot F(ax+b) + \frac{C}{a} \\ &= \frac { 1 }{ a } \cdot F(ax+b) + C', \end{aligned}$

where $C' := C/a$ is an arbitrary real constant of integration.

Remark 1. Theorem 2 is a very special case of integrating by substitution, which is, in spirit, simply the reverse of the usual chain rule in differentiation.

Example 5. Differentiate $x(x^2+1)^3$ . Hence, using Example 3, evaluate

$\displaystyle \int (x^2+1)^3\, \mathrm dx.$

Solution. Using the product rule and the chain rule,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(x(x^2+1)^3) &= \frac{\mathrm d}{\mathrm dx}(x) \cdot (x^2 + 1)^3 + \frac{\mathrm d}{\mathrm dx}((x^2+1)^3) \cdot x \\ &= 1 \cdot (x^2 + 1)^3 + 3(x^2+1)^2 \cdot 2x \cdot x \\ &= (x^2 + 1)^3 + 6x^2(x^2+1)^2 \\ &= (x^2 + 1)^3 + 6((x^2 + 1) - 1)(x^2+1)^2 \\ &= (x^2 + 1)^3 + 6((x^2+1)^3-(x^2+1)^2) \\ &= 7(x^2 + 1)^3 - 6(x^2+1)^2. \end{aligned}$

Integrating on both sides and using linearity,

$\begin{aligned} x(x^2 + 1)^3 + C_1 &= 7 \int(x^2+1)^3\, \mathrm dx - 6 \int (x^2 + 1)^2\, \mathrm dx.\end{aligned}$

Doing algebra and using Example 3,

$\begin{aligned} \int(x^2+1)^3\, \mathrm dx &= \frac 17 \cdot (x(x^2 + 1)^3 + C_1) + \frac 67 \int (x^2 + 1)^2\, \mathrm dx \\ &= \frac 17 \cdot (x(x^2 + 1)^3 + C_1) + \frac 67 \left( \frac 15 x^5 + \frac 23 x^3 + x + C_2 \right) \\ &= \textstyle \frac 17 \cdot x(x^2 + 1)^3 + \frac 6{35} x^5 + \frac 47 x^3 + \frac 67 x + C', \end{aligned}$

where $C' := \frac 17 C_1 + \frac 67 C_2$ is an arbitrary real constant of integration.

Remark 2. Example 5 is a guided example of integrating by parts, which is, in spirit, simply the reverse of the usual product rule in differentiation.

There is a lot more to say about integration, especially since we have not even discussed calculus involving the non-algebraic functions like the trigonometric functions $\sin(x)$ , $\cos(x)$ , $\tan(x)$ , as well as the exponential function $e^x$ and its inverse the natural logarithm $\ln(x)$ .

Nevertheless, we will push these discussions to the final post on integration, and pursue the essentials of the topic first. These functions have massive uses in the STEM fields, but for now we want to cover the big-idea bases of calculus that do not necessarily require them.

As such, we will apply integration to computing areas, as well as discuss some basic physics.

—Joel Kindiak, 18 Jan 25, 1200H

April 7, 2026
Differentiating Products

Previously, we explored the mighty chain rule:

$\displaystyle \frac{\mathrm d}{\mathrm dx}( f( g(x) ) ) = f'( g(x) ) \cdot g'(x).$

By considering $f(x) = x^2$ and $g(x) = h(x)$ , we have

$\displaystyle \frac{\mathrm d}{\mathrm dx}( h(x)^2 ) = 2 \cdot h(x) \cdot h'(x).$

Using this result, we showed that

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}( ( f(x) + g(x) )^2 ) &= \frac{\mathrm d}{\mathrm dx}(f(x)^2) + \frac{\mathrm d}{\mathrm dx}(g(x)^2) \\ &\phantom{==} + 2 \cdot (f'(x) \cdot g(x) + g'(x) \cdot f(x) ).\end{aligned}.$

This result is responsible for the product rule.

Theorem 1 (Product Rule). For functions $f(x), g(x)$ with derivatives $f'(x), g'(x)$ ,

$\displaystyle \frac{\mathrm d}{\mathrm dx} ( f(x) \cdot g(x) ) = f'(x) \cdot g(x) + g'(x) \cdot f(x).$

Proof. We have the result

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}\left((f(x) + g(x))^2 \right) &= \frac{\mathrm d}{\mathrm dx}(f(x)^2 ) + \frac{\mathrm d}{\mathrm dx}(g(x)^2 ) \\ &\phantom{==} + 2 \cdot (f'(x) \cdot g(x) + g'(x) \cdot f(x) ).\end{aligned}$

On the other hand, using vanilla algebra,

$(f(x) + g(x))^2 = f(x)^2 + 2 \cdot f(x) \cdot g(x) + g(x)^2.$

Hence,

$2 \cdot f(x) \cdot g(x) = (f(x) + g(x))^2 - f(x)^2 - g(x)^2.$

Using linearity,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx} ( 2 \cdot f(x) \cdot g(x) ) &= \frac{\mathrm d}{\mathrm dx}\left( (f(x) + g(x))^2 - (f(x))^2 - (g(x))^2 \right) \\ 2 \cdot \frac{\mathrm d}{\mathrm dx} ( f(x) \cdot g(x) ) &= \frac{\mathrm d}{\mathrm dx}\left( (f(x) + g(x))^2 \right) - \frac{\mathrm d}{\mathrm dx}(f(x)^2 ) - \frac{\mathrm d}{\mathrm dx}(g(x)^2 ) \\ 2 \cdot \frac{\mathrm d}{\mathrm dx} ( f(x) \cdot g(x) ) &= 2 \cdot (f'(x) \cdot g(x) + g'(x) \cdot f(x) ) \\ \frac{\mathrm d}{\mathrm dx} ( f(x) \cdot g(x) ) &= f'(x) \cdot g(x) + g'(x) \cdot f(x). \end{aligned}$

Previously, we have also proven that

$\displaystyle \frac{\mathrm d}{\mathrm dx} \left( \frac 1{h(x)} \right) = -\frac{h'(x)}{h(x)^2}.$

Together with the product rule, we can prove the quotient rule.

Theorem 2 (Quotient Rule). For $g(x) \neq 0$ ,

$\displaystyle \frac{ \mathrm d }{ \mathrm dx } \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}.$

Proof. Using the product rule,

$\begin{aligned} \frac{ \mathrm d }{ \mathrm dx } \left( \frac{f(x)}{g(x)} \right) &= \frac{ \mathrm d }{ \mathrm dx } \left( f(x) \cdot \frac 1{g(x)} \right) \\ &= f'(x) \cdot \frac 1{g(x)} + \frac{\mathrm d}{\mathrm dx} \left( \frac 1{g(x)} \right) \cdot f(x) \\ &= f'(x) \cdot \frac 1{g(x)} + \left( -\frac{g'(x)}{g(x)^2} \right) \cdot f(x) \\ &= \frac {f'(x) \cdot g(x)}{g(x)^2} - \frac{g'(x) \cdot f(x)}{g(x)^2} \\ &= \frac {f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2} .\end{aligned}$

Most differentiation problems becomes simply applying these results one after another, ensuring very careful algebraic calculations.

Example 1. Given that $f(x) \neq 0$ , $g(x) \neq 0$ , and $\displaystyle y = \frac{f(x)}{g(x)}$ , show that

$\displaystyle \frac{ \mathrm dy }{ \mathrm dx } = 0 \quad \iff \quad \frac{f'(x)}{ f(x) } = \frac{g'(x)}{g(x)}.$

Solution. Using the quotient rule,

$\displaystyle \begin{aligned} \frac{ \mathrm dy }{ \mathrm dx } &= \frac {f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}. \end{aligned}$

Since $g(x) \neq 0$ ,

$\displaystyle \frac{ \mathrm dy }{ \mathrm dx } = 0 \quad \iff \quad f'(x) \cdot g(x) - g'(x) \cdot f(x) = 0.$

Since $f(x) \neq 0$ ,

$\displaystyle f'(x) \cdot g(x) - g'(x) \cdot f(x) = 0 \quad \iff \quad \frac{f'(x)}{ f(x) } = \frac{g'(x)}{g(x)}.$

The equation $\displaystyle \frac{ \mathrm dy }{ \mathrm dx } = 0$ is called the zero derivative condition, which plays a vital role in optimisation applications.

Oddly enough, when compared with the chain rule, there is really not much else to be discussed about these two rules until we start differentiating other kinds of functions, like the trigonometric functions, exponential functions, and logarithm functions.

So for now, we switch gears and discuss turning points, applying it in the context of optimisation.

—Joel Kindiak, 8 Jan 26, 2036H

April 6, 2026

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