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# Complex & Imaginary Numbers

## Imaginary numbers

$$i^2 = -1 \quad \therefore i = \sqrt {-1}$$

### Simplifying negative surds

\begin{equation}\begin{split}\sqrt{-2} & = \sqrt{-1 \times 2} \\ & = \sqrt{2}i\end{split}\end{equation}


## Complex numbers

$$\mathbb{C} = \{a+bi : a, b \in \mathbb{R} \}$$

General form: $z=a+bi$  
$\operatorname{Re}(z) = a, \quad \operatorname{Im}(z) = b$

### Addition

If $z_1 = a+bi$ and $z_2=c+di$, then

$$z_1+z_2 = (a+c)+(b+d)i$$

### Subtraction

If $z_1=a+bi$ and $z_2=c+di$, then

$$z_1−z_2=(a−c)+(b−d)i$$

### Multiplication by a real constant

If $z=a+bi$ and $k \in \mathbb{R}$, then

$$kz=ka+kbi$$

### Powers of $i$

- $i^{4n} = 1$
- $i^{4n+1} = i$
- $i^{4n+2} = -1$
- $i^{4n+3} = -i$

For $i^n$, find remainder $r$ when $n \div 4$. Then $i^n = i^r$.

### Multiplying complex expressions

If $z_1 = a+bi$ and $z_2=c+di$, then

$$z_1 \times z_2 = (ac-bd)+(ad+bc)i$$

### Conjugates

$$\overline{z} = a \mp bi$$

##### Properties

- $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$
- $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$
- $\overline{kz} = k \overline{z}, \text{ for } k \in \mathbb{R}$
- $z \overline{z} = = (a+bi)(a-bi) = a^2+b^2 = |z|^2$
- $z + \overline{z} = 2 \operatorname{Re}(z)$

### Modulus

Distance from origin.

$$|{z}|=\sqrt{a^2+b^2} \quad  \therefore z \overline{z} = |z|^2$$

###### Properties

- $|z_1 z_2| = |z_1| |z_2|$
- $|{z_1 \over z_2}| = {|z_1| \over |z_2|}$
- $|z_1 + z_2| \le |z_1 + |z_2|$

### Multiplicative inverse

\begin{equation}\begin{split}z^{-1} & = {1 \over z} \\ & = {{a-bi} \over {a^2+B^2}} \\ & = {\overline{z} \over {|z|^2}}\end{split}\end{equation}

### Dividing complex numbers

$${{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}} \quad \text{(multiplicative inverse)}$$

In practice, rationalise denominator:

$${z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}$$

## Argand planes

- Geometric representation of $\mathbb{C}$
- horizontal $= \operatorname{Re}(z)$; vertical $= \operatorname{Im}(z)$
- Multiplication by $i$ results in an anticlockwise rotation of $\pi \over 2$

\vfil \break

## Complex polynomials

**Include $\pm$ for all solutions, including imaginary**

### Sum of two squares (quadratics)

$$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$$

Complete the square to get to this point.

#### Dividing complex polynomials

$P(z) \div D(z)$ gives quotient $Q(z)$ and remainder $R(z)$:

$$P(z) = D(z)Q(z) + R(z)$$

#### Remainder theorem

Let $\alpha \in \mathbb{C}$. Remainder of $P(z) \div (z - \alpha)$ is $P(\alpha)$

#### Factor theorem

If $a+bi$ is a solution to $P(z)=0$, then:

- $P(a+bi)=0$
- $z-(a+bi)$ is a factor of $P(z)$

#### Sum of two cubes

$$a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$$

## Conjugate root theorem

If $a+bi$ is a solution to $P(z)=0$, then the conjugate $\overline{z}=a-bi$ is also a solution.

## Polar form

\begin{equation}\begin{split}z & =r \operatorname{cis} \theta \\ & = r(\operatorname{cos}\theta+i \operatorname{sin}\theta) \\ & = a + bi \end{split}\end{equation}

- $r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}$
- $\theta=\operatorname{arg}(z)$ (on CAS: `arg(a+bi)`)
- **principal argument** is $\operatorname{Arg}(z) \in (-\pi, \pi]$ (note capital $\operatorname{Arg}$)

Each complex number has multiple polar representations:  
$z=r \operatorname{cis} \theta = r \operatorname{cis} (\theta+2 n\pi$) with $n \in \mathbb{Z}$ revolutions

### Conjugate in polar form

$$(r \operatorname{cis} \theta)^{-1} = r\operatorname{cis} (- \theta)$$

Reflection of $z$ across horizontal axis.

### Multiplication and division in polar form

$$z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)$$

$${z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)$$

## de Moivres' Theorem

$$(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}$$

## Roots of complex numbers

$n$th roots of $z = r \operatorname{cis} \theta$ are

$$z={r^{1 \over n}} \operatorname{cis}({{\theta + 2 k \pi} \over n})$$

Same modulus for all solutions. Arguments are separated by ${2 \pi} \over n$

The solutions of $z^n=a \text{ where } a \in \mathbb{C}$ lie on circle

$$x^2 + y^2 = (|a|^{1 \over n})^2$$

## Sketching complex graphs

### Straight line

- $\operatorname{Re}(z) = c$ or $\operatorname{Im}(z) = c$ (perpendicular bisector)
- $\operatorname{Arg}(z) = \theta$
- $|z+a|=|z+bi|$ where $m={a \over b}$
- $|z+a|=|z+b| \longrightarrow 2(a-b)x=b^2-a^2$

### Circle

$|z-z_1|^2 = c^2 |z_2+2|^2$ or $|z-(a + bi)| = c$

### Locus

$\operatorname{Arg}(z) < \theta$

# Vectors

- **vector:** a directed line segment  
- arrow indicates direction
- length indicates magnitude
- notated as $\vec{a}, \widetilde{A}, \overrightharp{a}$
- column notation: $\begin{bmatrix}
       x \\ y
     \end{bmatrix}$
- vectors with equal magnitude and direction are equivalent


![](graphics/vectors-intro.png){#id .class width=20%} 

## Vector addition

$\boldsymbol{u} + \boldsymbol{v}$ can be represented by drawing each vector head to tail then joining the lines.  
Addition is commutative (parallelogram)

## Scalar multiplication

For $k \in \mathbb{R}^+$, $k\boldsymbol{u}$ has the same direction as $\boldsymbol{u}$ but length is multiplied by a factor of $k$.

When multiplied by $k < 0$, direction is reversed and length is multplied by $k$.

## Vector subtraction

To find $\boldsymbol{u} - \boldsymbol{v}$, add $\boldsymbol{-v}$ to $\boldsymbol{u}$

## Parallel vectors

Same or opposite direction

$$\boldsymbol{u} || \boldsymbol{v} \iff \boldsymbol{u} = k \boldsymbol{v} \text{ where } k \in \mathbb{R} \setminus \{0\}$$ 

## Position vectors

Vectors may describe a position relative to $O$.

For a point $A$, the position vector is $\overrightharp{OA}$

\vfill\eject

## Linear combinations of non-parallel vectors

If two non-zero vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ are not parallel, then:

$$m\boldsymbol{a} + n\boldsymbol{b} = p \boldsymbol{a} + q \boldsymbol{b}\quad \therefore \quad m = p, \> n = q$$

![](graphics/parallelogram-vectors.jpg){#id .class width=20%}
![](graphics/vector-subtraction.jpg){#id .class width=10%}

## Column vector notation

A vector between points $A(x_1,y_1), \> B(x_2,y_2)$ can be represented as $\begin{bmatrix}x_2-x_1\\ y_2-y_1 \end{bmatrix}$

## Component notation

A vector $\boldsymbol{u} = \begin{bmatrix}x\\ y \end{bmatrix}$ can be written as $\boldsymbol{u} = x\boldsymbol{i} + y\boldsymbol{j}$.  
$\boldsymbol{u}$ is the sum of two components $x\boldsymbol{i}$ and $y\boldsymbol{j}$  
Magnitude of vector $\boldsymbol{u} = x\boldsymbol{i} + y\boldsymbol{j}$ is denoted by $|u|=\sqrt{x^2+y^2}$

Basic algebra applies:  
$(x\boldsymbol{i} + y\boldsymbol{j}) + (m\boldsymbol{i} + n\boldsymbol{j}) = (x + m)\boldsymbol{i} + (y+n)\boldsymbol{j}$  
Two vectors equal if and only if their components are equal.

## Unit vector $|\hat{\boldsymbol{a}}|=1$

\begin{equation}\begin{split}\hat{\boldsymbol{a}} & = {1 \over {|\boldsymbol{a}|}}\boldsymbol{a} \\ & = \boldsymbol{a} \cdot {|\boldsymbol{a}|}\end{split}\end{equation}

## Scalar/dot product $\boldsymbol{a} \cdot \boldsymbol{b}$

$$\boldsymbol{a} \cdot \boldsymbol{b} = a_1 b_1 + a_2 b_2$$

**on CAS:** `dotP([a b c], [d e f])`

## Scalar product properties

1. $k(\boldsymbol{a\cdot b})=(k\boldsymbol{a})\cdot \boldsymbol{b}=\boldsymbol{a}\cdot (k{b})$
2. $\boldsymbol{a \cdot 0}=0$
3. $\boldsymbol{a \cdot (b + c)}=\boldsymbol{a \cdot b + a \cdot c}$
4. $\boldsymbol{i \cdot i} = \boldsymbol{j \cdot j} = \boldsymbol{k \cdot k}= 1$
5. If $\boldsymbol{a} \cdot \boldsymbol{b} = 0$, $\boldsymbol{a}$ and $\boldsymbol{b}$ are perpendicular
6. $\boldsymbol{a \cdot a} = |\boldsymbol{a}|^2 = a^2$

For parallel vectors $\boldsymbol{a}$ and $\boldsymbol{b}$:  
$$\boldsymbol{a \cdot b}=\begin{cases}
|\boldsymbol{a}||\boldsymbol{b}| \hspace{2.8em} \text{if same direction}\\
-|\boldsymbol{a}||\boldsymbol{b}| \hspace{2em} \text{if opposite directions}
\end{cases}$$

## Geometric scalar products

$$\boldsymbol{a} \cdot \boldsymbol{b} = |\boldsymbol{a}| |\boldsymbol{b}| \cos \theta$$

where $0 \le \theta \le \pi$

## Perpendicular vectors

If $\boldsymbol{a} \cdot \boldsymbol{b} = 0$, then $\boldsymbol{a} \perp \boldsymbol{b}$ (since $\cos 90 = 0$)

## Finding angle between vectors

**positive direction**

$$\cos \theta = {{\boldsymbol{a} \cdot \boldsymbol{b}} \over {|\boldsymbol{a}| |\boldsymbol{b}|}} = {{a_1 b_1 + a_2 b_2} \over {|\boldsymbol{a}| |\boldsymbol{b}|}}$$

**on CAS:** `angle([a b c], [a b c])` (Action -> Vector -> Angle)

## Angle between vector and axis

Direction of a vector can be given by the angles it makes with $\boldsymbol{i}, \boldsymbol{j}, \boldsymbol{k}$ directions.

For $\boldsymbol{a} = a_1 \boldsymbol{i} + a_2 \boldsymbol{j} + a_3 \boldsymbol{k}$ which makes angles $\alpha, \beta, \gamma$ with positive direction of $x, y, z$ axes:
$$\cos \alpha = {a_1 \over |\boldsymbol{a}|}, \quad \cos \beta = {a_2 \over |\boldsymbol{a}|}, \quad \cos \gamma = {a_3 \over |\boldsymbol{a}|}$$

**on CAS:** `angle([a b c], [1 0 0])` for angle between $a\boldsymbol{i} + b\boldsymbol{j} + c\boldsymbol{k}$ and $x$-axis

## Vector projections

Vector resolute of $\boldsymbol{a}$ in direction of $\boldsymbol{b}$ is magnitude of $\boldsymbol{a}$ in direction of $\boldsymbol{b}$:

$$\boldsymbol{u}={{\boldsymbol{a}\cdot\boldsymbol{b}}\over |\boldsymbol{b}|^2}\boldsymbol{b}=\left({\boldsymbol{a}\cdot{\boldsymbol{b} \over |\boldsymbol{b}|}}\right)\left({\boldsymbol{b} \over |\boldsymbol{b}|}\right)=(\boldsymbol{a} \cdot \hat{\boldsymbol{b}})\hat{\boldsymbol{b}}$$

## Scalar resolute of $\boldsymbol{a}$ on $\boldsymbol{b}$

$$r_s = |\boldsymbol{u}| = \boldsymbol{a} \cdot \hat{\boldsymbol{b}}$$

## Vector resolute of $\boldsymbol{a} \perp \boldsymbol{b}$

$$\boldsymbol{w} = \boldsymbol{a} - \boldsymbol{u} \> \text{ where } \boldsymbol{u} \text{ is projection } \boldsymbol{a} \text{ on } \boldsymbol{b}$$

## Vector proofs

### Concurrent lines

$\ge$ 3 lines intersect at a single point  

### Collinear points

$\ge$ 3 points lie on the same line  
$\implies \vec{OC} = \lambda \vec{OA} + \mu \vec{OB}$ where $\lambda + \mu = 1$. If $C$ is between $\vec{AB}$, then $0 < \mu < 1$  
Points $A, B, C$ are collinear iff $\vec{AC}=m\vec{AB} \text{ where } m \ne 0$

### Useful vector properties

- If $\boldsymbol{a}$ and $\boldsymbol{b}$ are parallel, then $\boldsymbol{b}=k\boldsymbol{a}$ for some $k \in \mathbb{R} \setminus \{0\}$
- If $\boldsymbol{a}$ and $\boldsymbol{b}$ are parallel with at least one point in common, then they lie on the same straight line
- Two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ are perpendicular if $\boldsymbol{a} \cdot \boldsymbol{b}=0$
- $\boldsymbol{a} \cdot \boldsymbol{a} = |\boldsymbol{a}|^2$

## Linear dependence

Vectors $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ are linearly dependent if they are non-parallel and:

$$k\boldsymbol{a}+l\boldsymbol{b}+m\boldsymbol{c} = 0$$
$$\therefore \boldsymbol{c} = m\boldsymbol{a} + n\boldsymbol{b} \quad \text{(simultaneous)}$$

$\boldsymbol{a}, \boldsymbol{b},$ and $\boldsymbol{c}$ are linearly independent if no vector in the set is expressible as a linear combination of other vectors in set, or if they are parallel.

Vector $\boldsymbol{w}$ is a linear combination of vectors $\boldsymbol{v_1}, \boldsymbol{v_2}, \boldsymbol{v_3}$

## Three-dimensional vectors

Right-hand rule for axes: $z$ is up or out of page.

i![](graphics/vectors-3d.png)

## Parametric vectors

Parametric equation of line through point $(x_0, y_0, z_0)$ and parallel to $a\boldsymbol{i} + b\boldsymbol{j} + c\boldsymbol{k}$ is:

\begin{equation}\begin{cases}x = x_o + a \cdot t \\ y = y_0 + b \cdot t \\ z = z_0 + c \cdot t\end{cases}\end{equation}

# Circular functions

Period of $a\sin(bx)$ is ${2\pi} \over b$

Period of $a\tan(nx)$ is $\pi \over n$  
Asymptotes at $x={2k+1)\pi \over 2n} \> \vert \> k \in \mathbb{Z}$

## Reciprocal functions

### Cosecant

![](graphics/csc.png)

$$\operatorname{cosec} \theta = {1 \over \sin \theta} \> \vert \> \sin \theta \ne 0$$

- **Domain** $= \mathbb{R} \setminus {n\pi : n \in \mathbb{Z}}$
- **Range** $= \mathbb{R} \setminus (-1, 1)$
- **Turning points** at $\theta = {{(2n + 1)\pi} \over 2} \> \vert \> n \in \mathbb{Z}$
- **Asymptotes** at $\theta = n\pi \> \vert \> n \in \mathbb{Z}$

### Secant

![](graphics/sec.png)

$$\operatorname{sec} \theta = {1 \over \cos \theta} \> \vert \> \cos \theta \ne 0$$

- **Domain** $= \mathbb{R} \setminus \{{{(2n + 1) \pi} \over 2 } : n \in \mathbb{Z}\}$
- **Range** $= \mathbb{R} \setminus (-1, 1)$
- **Turning points** at $\theta = n\pi \> \vert \> n \in \mathbb{Z}$
- **Asymptotes** at $\theta = {{(2n + 1) \pi} \over 2} \> \vert \> n \in \mathbb{Z}$

### Cotangent

![](graphics/cot.png)

$$\operatorname{cot} \theta = {{\cos \theta} \over {\sin \theta}} \> \vert \> \sin \theta \ne 0$$

- **Domain** $= \mathbb{R} \setminus \{n \pi: n \in \mathbb{Z}\}$
- **Range** $= \mathbb{R}$
- **Asymptotes** at $\theta = n\pi \> \vert \> n \in \mathbb{Z}$

### Symmetry properties

\begin{equation}\begin{split}
  \operatorname{sec} (\pi \pm x) & = -\operatorname{sec} x \\
  \operatorname{sec} (-x) & = \operatorname{sec} x \\
  \operatorname{cosec} (\pi \pm x) & = \mp \operatorname{cosec} x \\
  \operatorname{cosec} (-x) & = - \operatorname{cosec} x \\
  \operatorname{cot} (\pi \pm x) & = \pm \operatorname{cot} x \\
  \operatorname{cot} (-x) & = - \operatorname{cot} x
\end{split}\end{equation}

### Complementary properties

\begin{equation}\begin{split}
  \operatorname{sec} \left({\pi \over 2} - x\right) & = \operatorname{cosec} x \\
  \operatorname{cosec} \left({\pi \over 2} - x\right) & = \operatorname{sec} x \\
  \operatorname{cot} \left({\pi \over 2} - x\right) & = \tan x \\
  \tan \left({\pi \over 2} - x\right) & = \operatorname{cot} x
\end{split}\end{equation}

### Pythagorean identities

\begin{equation}\begin{split}
  1 + \operatorname{cot}^2 x & = \operatorname{cosec}^2 x, \quad \text{where } \sin x \ne 0 \\
  1 + \tan^2 x & = \operatorname{sec}^2 x, \quad \text{where } \cos x \ne 0
\end{split}\end{equation}

## Compound angle formulas

$$\cos(x \pm y) = \cos x + \cos y \mp \sin x \sin y$$
$$\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y$$
$$\tan(x \pm y) = {{\tan x \pm \tan y} \over {1 \mp \tan x \tan y}}$$

## Double angle formulas

\begin{equation}\begin{split}
  \cos 2x &= \cos^2 x - \sin^2 x \\
  & = 1 - 2\sin^2 x \\
  & = 2 \cos^2 x -1
\end{split}\end{equation}

$$\sin 2x = 2 \sin x \cos x$$

$$\tan 2x = {{2 \tan x} \over {1 - \tan^2 x}}$$

## Inverse circular functions

Inverse functions: $f(f^{-1}(x)) = x, \quad f(f^{-1}(x)) = x$  
Must be 1:1 to find inverse (reflection in $y=x$

Domain is restricted to make functions 1:1.

### $\arcsin$

$$\sin^{-1}: [-1, 1] \rightarrow \mathbb{R}, \quad \sin^{-1} x = y, \quad \text{where } \sin y = x \text{ and } y \in [{-\pi \over 2}, {\pi \over 2}]$$

### $\arcos$

$$\cos^{-1} \rightarrow \mathbb{R}, \quad \cos^{-1} x = y, \quad \text{where } \cos y = x \text{ and } y \in [0, \pi]$$

### $\arctan$

$$\tan^{-1}: \mathbb{R} \rightarrow \mathbb{R}, \quad \tan^{-1} x = y, \quad \text{where } \tan y = x \text{ and } y \in \left(-{\pi \over 2}, {\pi \over 2}\right)$$
# Differential calculus

## Limits

$$\lim_{x \rightarrow a}f(x)$$

$L^-$ - limit from below

$L^+$ - limit from above

$\lim_{x \to a} f(x)$ - limit of a point  

- Limit exists if $L^-=L^+$
- If limit exists, point does not.

Limits can be solved using normal techniques (if div 0, factorise)

## Limit theorems

1. For constant function $f(x)=k$, $\lim_{x \rightarrow a} f(x) = k$
2. $\lim_{x \rightarrow a} (f(x) \pm g(x)) = F \pm G$
3. $\lim_{x \rightarrow a} (f(x) \times g(x)) = F \times G$
4. ${\lim_{x \rightarrow a} {f(x) \over g(x)}} = {F \over G}, G \ne 0$

Corollary: $\lim_{x \rightarrow a} c \times f(x)=cF$ where $c=$ constant

## Solving limits for $x\rightarrow\infty$

Factorise so that all values of $x$ are in denominators.

e.g.

$$\lim_{x \rightarrow \infty}{{2x+3} \over {x-2}}={{2+{3 \over x}} \over {1-{2 \over x}}}={2 \over 1} = 2$$


## Continuous functions

A function is continuous if $L^-=L^+=f(x)$ for all values of $x$.

## Gradients of secants and tangents

Secant (chord) - line joining two points on curve

Tangent - line that intersects curve at one point

given $P(x,y) \quad Q(x+\delta x, y + \delta y)$:
gradient of chord joining $P$ and $Q$ is ${m_{PQ}}={\operatorname{rise} \over \operatorname{run}} = {\delta y \over \delta x}$

As $Q \rightarrow P, \delta x \rightarrow 0$. Chord becomes tangent (two infinitesimal points are equal).

Can also be used with functions, where $h=\delta x$.

## First principles derivative

$$f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={dy \over dx}$$

$$m_{\tan}=\lim_{h \rightarrow 0}f^\prime(x)$$



$$m_{\vec{PQ}}=f^\prime(x)$$

first principles derivative:
$${m_{\text{tangent at }P} =\lim_{h \rightarrow 0}}{{f(x+h)-f(x)}\over h}$$

## Gradient at a point

Given point $P(a, b)$ and function $f(x)$, the gradient is $f^\prime(a)$


## Derivatives of $x^n$

$${d(ax^n) \over dx}=anx^{n-1}$$

If $x=$ constant, derivative is $0$

If $y=ax^n$, derivative is $a\times nx^{n-1}$

If $f(x)={1 \over x}=x^{-1}, \quad f^\prime(x)=-1x^{-2}={-1 \over x^2}$

If $f(x)=^5\sqrt{x}=x^{1 \over 5}, \quad f^\prime(x)={1 \over 5}x^{-4/5}={1 \over 5 \times ^5\sqrt{x^4}}$

If $f(x)=(x-b)^2, \quad f^\prime(x)=2(x-b)$

$$f^\prime(x)=\lim_{h \rightarrow 0}{{f(x+h)-f(x)} \over h}$$

## Derivatives of $u \pm v$

$${dy \over dx}={du \over dx} \pm {dv \over dx}$$
where $u$ and $v$ are functions of $x$

## Euler's number as a limit

$$\lim_{h \rightarrow 0} {{e^h-1} \over h}=1$$

## Chain rule for $(f\circ g)$

If $f(x) = h(g(x)) = (h \circ g)(x)$:

$$f^\prime(x) = h^\prime(g(x)) \cdot g^\prime(x)$$

If $y=h(u)$ and $u=g(x)$:

$${dy \over dx} = {dy \over du} \cdot {du \over dx}$$
$${d((ax+b)^n) \over dx} = {d(ax+b) \over dx} \cdot n \cdot (ax+b)^{n-1}$$

Used with only one expression.

e.g. $y=(x^2+5)^7$ - Cannot reasonably expand  
Let $u-x^2+5$ (inner expression)  
${du \over dx} = 2x$  
$y=u^7$  
${dy \over du} = 7u^6$  

## Product rule for $y=uv$

$${dy \over dx} = u{dv \over dx} + v{du \over dx}$$

## Quotient rule for $y={u \over v}$

$${dy \over dx} = {{v{du \over dx} - u{dv \over dx}} \over v^2}$$

$$f^\prime(x)={{v(x)u^\prime(x)-u(x)v^\prime(x)} \over [v(x)]^2}$$

## Logarithms

$$\log_b (x) = n \quad \operatorname{where} \hspace{0.5em} b^n=x$$

Wikipedia:

> the logarithm of a given number $x$ is the exponent to which another fixed number, the base $b$, must be raised, to produce that number $x$

### Logarithmic identities

$\log_b (xy)=\log_b x + \log_b y$  
$\log_b x^n = n \log_b x$  
$\log_b y^{x^n} = x^n \log_b y$

### Index identities

$b^{m+n}=b^m \cdot b^n$  
$(b^m)^n=b^{m \cdot n}$  
$(b \cdot c)^n = b^n \cdot c^n$  
${a^m \div a^n} = {a^{m-n}}$

### $e$ as a logarithm

$$\operatorname{if} y=e^x, \quad \operatorname{then} x=\log_e y$$
$$\ln x = \log_e x$$

### Differentiating logarithms
$${d(\log_e x)\over dx} = x^{-1} = {1 \over x}$$

## Derivative rules

| $f(x)$ | $f^\prime(x)$ |
| ------ | ------------- |
| $\sin x$ | $\cos x$ |
| $\sin ax$ | $a\cos ax$ |
| $\cos x$ | $-\sin x$ |
| $\cos ax$ | $-a \sin ax$ |
| $\tan f(x)$ | $f^2(x) \sec^2f(x)$ |
| $e^x$ | $e^x$ |
| $e^{ax}$ | $ae^{ax}$ |
| $ax^{nx}$ | $an \cdot e^{nx}$ |
| $\log_e x$ | $1 \over x$ |
| $\log_e {ax}$ | $1 \over x$ |
| $\log_e f(x)$ | $f^\prime (x) \over f(x)$ |
| $\sin(f(x))$ | $f^\prime(x) \cdot \cos(f(x))$ |
| $\sin^{-1} x$ | $1 \over {\sqrt{1-x^2}}$ |
| $\cos^{-1} x$ | $-1 \over {sqrt{1-x^2}}$ |
| $\tan^{-1} x$ | $1 \over {1 + x^2}$ |

## Reciprocal derivatives

$${1 \over {dy \over dx}} = {dx \over dy}$$

## Differentiating $x=f(y)$

Find $dx \over dy$. Then ${dx \over dy} = {1 \over {dy \over dx}} \implies {dy \over dx} = {1 \over {dx \over dy}}$.

$${dy \over dx} = {1 \over {dx \over dy}}$$

## Second derivative

$$f(x) \longrightarrow f^\prime (x) \longrightarrow f^{\prime\prime}(x)$$

$$\therefore y \longrightarrow {dy \over dx} \longrightarrow {d({dy \over dx}) \over dx} \longrightarrow {d^2 y \over dx^2}$$

Order of polynomial $n$th derivative decrements each time the derivative is taken

### Points of Inflection

*Stationary point* - point of zero gradient (i.e. $f^\prime(x)=0$)  
*Point of inflection* - point of maximum $|$gradient$|$ (i.e.  $f^{\prime\prime} = 0$)

* if $f^\prime (a) = 0$ and $f^{\prime\prime}(a) > 0$, then point $(a, f(a))$ is a local min (curve is concave up)
* if $f^\prime (a) = 0$ and $f^{\prime\prime} (a) < 0$, then point $(a, f(a))$ is local max (curve is concave down)
* if $f^{\prime\prime}(a) = 0$, then point $(a, f(a))$ is a point of inflection
  + if also $f^\prime(a)=0$, then it is a stationary point of inflection

![](graphics/second-derivatives.png)

## Implicit Differentiation

**On CAS:** Action $\rightarrow$ Calculation $\rightarrow$ `impDiff(y^2+ax=5, x, y)`. Returns $y^\prime= \dots$.

Used for differentiating circles etc.

If $p$ and $q$ are expressions in $x$ and $y$ such that $p=q$, for all $x$ nd $y$, then:

$${dp \over dx} = {dq \over dx} \quad \text{and} \quad {dp \over dy} = {dq \over dy}$$

## Integration

$$\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)$$

$$\int x^n \cdot dx = {x^{n+1} \over n+1} + c$$

- area enclosed by curves
- $+c$ should be shown on each step without $\int$

### Integral laws

$\int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx$  
$\int k f(x) dx = k \int f(x) dx$  

| $f(x)$                          | $\int f(x) \cdot dx$         |
| ------------------------------- | ---------------------------- |
| $k$ (constant) | $kx + c$ |
| $x^n$ | ${x^{n+1} \over {n+1}} + c$ |
| $a x^{-n}$ | $a \cdot \log_e x + c$ |
| ${1 \over {ax+b}}$ | ${1 \over a} \log_e (ax+b) + c$ |
| $(ax+b)^n$ | ${1 \over {a(n+1)}}(ax+b)^{n-1} + c$ |
| $e^{kx}$ | ${1 \over k} e^{kx} + c$ |
| $e^k$ | $e^kx + c$ |
| $\sin kx$ | $-{1 \over k} \cos (kx) + c$ |
| $\cos kx$ | ${1 \over k} \sin (kx) + c$ |
| $\sec^2 kx$ | ${1 \over k} \tan(kx) + c$ |
| $1 \over \sqrt{a^2-x^2}$ | $\sin^{-1} {x \over a} + c \>\vert\> a>0$ |
| $-1 \over \sqrt{a^2-x^2}$ | $\cos^{-1} {x \over a} + c \>\vert\> a>0$ |
| $a \over {a^2-x^2}$ | $\tan^{-1} {x \over a} + c$ |
| ${f^\prime (x)} \over {f(x)}$ | $\log_e f(x) + c$ |
| $g^\prime(x)\cdot f^\prime(g(x)$ | $f(g(x))$ (chain rule)|
| $f(x) \cdot g(x)$ | $\int [f^\prime(x) \cdot g(x)] dx + \int [g^\prime(x) f(x)] dx$ |

Note $\sin^{-1} {x \over a} + \cos^{-1} {x \over a}$ is constant for all $x \in (-a, a)$.

### Definite integrals

$$\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)$$

- Signed area enclosed by: $\> y=f(x), \quad y=0, \quad x=a, \quad x=b$.
- *Integrand* is $f$.
- $F(x)$ may be any integral, i.e. $c$ is inconsequential

#### Properties

$$\int^b_a f(x) \> dx = \int^c_a f(x) \> dx + \int^b_c f(x) \> dx$$

$$\int^a_a f(x) \> dx = 0$$

$$\int^b_a k \cdot f(x) \> dx = k \int^b_a f(x) \> dx$$

$$\int^b_a f(x) \pm g(x) \> dx = \int^b_a f(x) \> dx \pm \int^b_a g(x) \> dx$$

$$\int^b_a f(x) \> dx = - \int^a_b f(x) \> dx$$

### Integration by substitution

$$\int f(u) {du \over dx} \cdot dx = \int f(u) \cdot du$$

Note $f(u)$ must be one-to-one $\implies$ one $x$ value for each $y$ value

e.g. for $y=\int(2x+1)\sqrt{x+4} \cdot dx$:  
let $u=x+4$  
$\implies {du \over dx} = 1$  
$\implies x = u - 4$  
then $y=\int (2(u-4)+1)u^{1 \over 2} \cdot du$  
Solve as a normal integral

#### Definite integrals by substitution

For $\int^b_a f(x) {du \over dx} \cdot dx$, evaluate new $a$ and $b$ for $f(u) \cdot du$.

### Trigonometric integration

$$\sin^m x \cos^n x \cdot dx$$

**$m$ is odd:**  
$m=2k+1$ where $k \in \mathbb{Z}$  
$\implies \sin^{2k+1} x = (\sin^2 z)^k \sin x = (1 - \cos^2 x)^k \sin x$  
Substitute $u=\cos x$

**$n$ is odd:**  
$n=2k+1$ where $k \in \mathbb{Z}$  
$\implies \cos^{2k+1} x = (\cos^2 x)^k \cos x = (1-\sin^2 x)^k \cos x$  
Subbstitute $u=\sin x$

**$m$ and $n$ are even:**  
Use identities:

- $\sin^2x={1 \over 2}(1-\cos 2x)$
- $\cos^2x={1 \over 2}(1+\cos 2x)$
- $\sin 2x = 2 \sin x \cos x$

## Partial fractions

On CAS: Action $\rightarrow$ Transformation $\rightarrow$ `expand/combine`  
or Interactive $\rightarrow$ Transformation $\rightarrow$ `expand` $\rightarrow$ Partial

## Graphing integrals on CAS

In main: Interactive $\rightarrow$ Calculation $\rightarrow$ $\int$ ($\rightarrow$ Definite)  
Restrictions: `Define f(x)=...` $\rightarrow$ `f(x)|x>1` (e.g.)

## Applications of antidifferentiation

- $x$-intercepts of $y=f(x)$ identify $x$-coordinates of stationary points on $y=F(x)$
- nature of stationary points is determined by sign of $y=f(x)$ on either side of its $x$-intercepts
- if $f(x)$ is a polynomial of degree $n$, then $F(x)$ has degree $n+1$

To find stationary points of a function, substitute $x$ value of given point into derivative. Solve for ${dy \over dx}=0$. Integrate to find original function.

## Solids of revolution

Approximate as sum of infinitesimally-thick cylinders

### Rotation about $x$-axis

\begin{align*}
  V &= \int^{x=b}_{x-a} \pi y^2 \> dx \\
    &= \pi \int^b_a (f(x))^2 \> dx
\end{align*}

### Rotation about $y$-axis

\begin{align*}
  V &= \int^{y=b}_{y=a} \pi x^2 \> dy \\
    &= \pi \int^b_a (f(y))^2 \> dy
\end{align*}

### Regions not bound by $y=0$

$$V = \pi \int^b_a f(x)^2 - g(x)^2 \> dx$$  
where $f(x) > g(x)$

## Length of a curve

$$L = \int^b_a \sqrt{1 + ({dy \over dx})^2} \> dx \quad \text{(Cartesian)}$$

$$L = \int^b_a \sqrt{{dx \over dt} + ({dy \over dt})^2} \> dt \quad \text{(parametric)}$$

Evaluate on CAS. Or use Interactive $\rightarrow$ Calculation $\rightarrow$ Line $\rightarrow$ `arcLen`.

## Rates

### Related rates

$${da \over db} \quad \text{(change in } a \text{ with respect to } b)$$

### Gradient at a point on parametric curve

$${dy \over dx} = {{dy \over dt} \div {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$

$${d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \div {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}$$

## Rational functions

$$f(x) = {P(x) \over Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}$$

### Addition of ordinates

- when two graphs have the same ordinate, $y$-coordinate is double the ordinate
- when two graphs have opposite ordinates, $y$-coordinate is 0 i.e. ($x$-intercept)
- when one of the ordinates is 0, the resulting ordinate is equal to the other ordinate

## Fundamental theorem of calculus

If $f$ is continuous on $[a, b]$, then

$$\int^b_a f(x) \> dx = F(b) - F(a)$$

where $F$ is any antiderivative of $f$

## Differential equations

One or more derivatives

**Order** - highest power inside derivative  
**Degree** - highest power of highest derivative  
e.g. ${\left(dy^2 \over d^2 x\right)}^3$: order 2, degree 3

### Verifying solutions

Start with $y=\dots$, and differentiate. Substitute into original equation.

### Function of the dependent variable

If ${dy \over dx}=g(y)$, then ${dx \over dy} = 1 \div {dy \over dx} = {1 \over g(y)}$. Integrate both sides to solve equation. Only add $c$ on one side. Express $e^c$ as $A$.

### Mixing problems

$$\left({dm \over dt}\right)_\Sigma = \left({dm \over dt}\right)_{\text{in}} - \left({dm \over dt}\)_{\text{out}}$$

### Separation of variables

If ${dy \over dx}=f(x)g(y)$, then:

$$\int f(x) \> dx = \int {1 \over g(y)} \> dy$$

### Using definite integrals to solve DEs

Used for situations where solutions to ${dy \over dx} = f(x)$ is not required.

In some cases, it may not be possible to obtain an exact solution.

Approximate solutions can be found by numerically evaluating a definite integral.

### Using Euler's method to solve a differential equation

$${{f(x+h) - f(x)} \over h } \approx f^\prime (x) \quad \text{for small } h$$

$$\implies f(x+h) \approx f(x) + hf^\prime(x)$$