# Mathematics: Introduction to Calculus

$\newcommand{\ddx}[1]{\frac{d #1}{dx}}$

• Calculus is the study of continous change
• Calculus is used to find the gradient of a graph, the maximum and minimum values, the area under or over a graph, inflection points, etc.

## Derivation

• A derivation is a function which generalises some property of another function
• the derivative of $$f(x)$$ (say “f of x”) is $$f\prime (x)$$ (say “f prime of x”)

### Rules (Copy into notes)

• $$f\prime (x)=\ddx{y}x=y\prime$$
• $$f\prime(a)=0$$ if $$a$$ is a number/constant
• $$f\prime(ax)=a$$ if $$a$$ is a number/constant
• $$f\prime(ax^n )=anx^{n-1}$$ if $$a$$ and $$n$$ are a numbers/constants
• $$f\prime(e^x )=e^x$$
• $$f\prime(a^x )=\ln(a)a^x$$
• $$ln(x)=\frac{1}{\ln(x)}\cdot \frac{1}{x}$$

## First Principle

• The first principle is used to calculate the derivative of a graph

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

#### Example of First Principle

###### Find the first derivative of $$f(x)=x^4+5x+3$$
1. Define f(x)

$$f(x)=x^4+5x+3$$

1. define f(x+h)

$$f(x+h)=(x+h)^4+5(x+h)+3$$

1. State the First Principle

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

1. Substitute f(x) and f(x+h) into First Principle

$$=\displaystyle {\lim_{h\to 0}\frac{(x+h)^4+5(x+h)+3-(x^4+5x+3)}{h}}$$

1. Expand f(x) and f(x+h) where possible

$$=\frac{x^4+4x^3h+6x^2h^2+4xh^3+h^4+5x+5h+3-x^4-5x-3}{h}$$

1. Eliminate as many variables as possible from the numerator

$$\require{cancel}=\frac{\cancel{x^4}+4x^3h+6x^2h^2+4xh^3+h^4+\cancel{5x}+5h\cancel{+3}\cancel{-x^4}\cancel{-5x}\cancel{-3}}{h}$$ $$=\frac{4x^{3}h+6x^2 h^2 +4xh^3 +h^4 +5h}{h}$$

1. factorise numerator to eliminate h

$$=\frac{\cancel{h}(4x^3 +6x^2 h +4xh^2 +h^3 +5)}{\cancel{h}}$$ $$=4x^3 +6x^2 h +4xh^2 +h^3 +5) 1. Substitute \(h=0$$

$$\require{enclose}(4x^3) \cancel{+(6x^2\cdot 0)} +\cancel{+(4x\cdot (0)^2)} \cancel{+(0)^3} +5$$ $$\color{teal}{\enclose{roundedbox}{=4x^3 +5}}$$

$$\color{green}{\enclose{roundedbox}{\therefore \ddx{}(x^4+5x+3)=4x^3 +5}}$$

## Shortcuts for the First Princple

• Here’s some convenient shortcuts for people who don’t want to waste time doing all the steps of the First Principle
• Use these unless a question SPECIFICALLY asks for First Principle
• Note that anything except $x$, $y$ and $t$ represent NUMBERS
• Remember, differentiation is just finding the gradient of a curve!!!
1. $\ddx{}(a) = 0$
2. $\ddx{}ax=a$
3. $\ddx{}ax^n =(a\cdot n)x^{n-1}$
4. $\ddx{}(A+B)=\ddx{}(A)+\ddx{}(B)$
5. $\ddx{}(A-B)=\ddx{}(A)-\ddx{}(B)$
6. $(AB)\prime = A\prime B+B\prime A$
7. $\ddx{}\frac{A}{B}=\frac{A\prime B - B\prime A}{B^2}$
8. $\ddx{}((f(x))^n )= n(f\prime (x))^{n-1}$
9. $\ddx{y}=\frac{dy}{du}\cdot\ddx{u}$
10. $\ddx{}af(x)=a(f\prime (x))$
$$\ddx{} \Leftarrow\Rightarrow$$ differentiate with respect to $x$ $$\ddx{y} \Leftarrow\Rightarrow\ddx{}(y)\Leftarrow\Rightarrow y\prime\Leftarrow\Rightarrow$$ differentiate $y$ with respect to $x$

## Graphing Derivatives

$$f(x)$$ $$\ddx{}(f(x))$$
Turning Points/Standing Points $$x$$-Intercepts
Points of Inflection Turning Points
Horizontal Points of Inflection Turning points ON THE X-AXIS

## Chain Rule

• The chain rule is used when one function is acting on another (e.g. $$f(g(x))$$), and you need to find the derivative of the whole thing

$$\color{lightblue}{f(g(x))\prime=f\prime(g(x))g\cdot\prime(x)}$$

### Steps

$$\color{lightgreen}{\text{Example question: find }f\prime(x)\text{ where }f(x)=(3x^2 +5)^{10}}$$

1. Bring the exponent to the front (Parentheses stay the same)

$$10(3x^2 +5)^{10}$$

1. Subtract 1 from exponent

$$10(3x^2 +5)^{9}$$

1. Multiply by derivative of inside

$$10(3x^2 +5)^{9}\cdot (6x)$$

1. Solve and answer the question

$$\therefore f\prime(x)=60x\left(3x^2+5\right)^9$$

## Product Rule (UWU Rule)

• The product rule is used to find the product of two differentials when given the original multipliers, for example:

$$\color{lightgreen}{\text{Find }f\prime(x)\text{ where }f(x)=(2x+5)(4x^2 +5)}$$

• The product rule is as follows:

$$\color{red}{\text{If }f(x)=u(x)\cdot v(x)\text{, then }f\prime(x)=u(x)\prime v(x)+v(x)\prime u(x)}$$

### Steps

1. Find $$u\prime(x)$$ and $$v\prime (x)$$

$$u\prime (x)=2$$ $$v\prime (x)=8x$$

1. Find $$u\prime v$$ and $$v\prime u$$

$$u\prime v = 2\cdot (4x^2 +5) =8x^2 +10$$

$$v\prime u = 8x\cdot (2x+5) = 16x^2 + 40x$$

1. Solve and answer the question

$$f\prime(x)=8x^2 +10+16x^2 + 40x$$

$$\therefore f\prime(x)=24x^2 +40x+10$$

## Quotient Rule

• The Quotient rule is used to find the quotient of two differentials when given the divisor and dividend, for example:

$$\color{lightgreen}{\text{Find }f\prime(x)\text{ where }f(x)=\frac{2x+5}{4x^2 +5}}$$

Yes I used the same functions. Deal with it 😎

- Pranav

### Example Question

$$\color{lightgreen}{\text{Find }f\prime(x)\text{ where }f(x)=\frac{2x+5}{4x^2 +5}}$$

1. Find $u(x), u\prime (x), v(x), and v\prime (x)$

$u(x)=2x+5$

$u\prime (x)=2$

$v(x)=4x^2 +5$

$v\prime (x)=8x$

2. Substitute values into the Quotient formula

$= \frac{u\prime v - v\prime u}{v^2}$

$f\prime (x)=\frac{2(4x^2+5) - 8(2x+5)}{(4x^2 +5)^2}$

3. Solve/Simplify

$2(4x^2 +5)-8(2x+5)=8x^2 -16x-30$

$\therefore f\prime (x)=\frac{8x^2 -16x-30}{(4x^2 +5)^2}$

## Differentiating Sine, Cosine, and Tangent equations

### TL;DR

• $\ddx{sin(x)}=cos(x)$
• $\ddx{cos(x)}=-sin(x)$
• $\ddx{tan(x)}=sec ^2 (x)$

(WHAT A PLOT TWIST) - Pranav Sharma

### The Long Version

• When differentiating Sin, Cos and Tan, it’s important to remember that $\ddx{y}=\frac{dy}{du}\cdot \frac{du}{dx}$
• The First Principles (I hate it too, but it’s occasionally useful) can be used to find the derivatives of Sin, Cos, and Tan.

#### First derivative of Sine ($f(x)=sin(x)$)

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

• So in the case of $f(x)=\sin(x)$, we have:

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

• From this, we can use the Sum-Difference identity $\sin(A+B)=\sin A\cos B+\sin B\cos A$ to get

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

$=\displaystyle {\lim_{h\to 0}} \frac{\sin x(\cos h-1)+\sin h\cos x}{h}$ $=\displaystyle {\lim_{h\to 0}} (\frac{\sin x(\cos h-1)}{h}+\frac{\sin h\cos x}{h})$

$=(\sin x)\displaystyle {\lim_{h\to 0}} \frac{\cos h-1}{h}+\cos(x)\displaystyle {\lim_{h\to 0}} \frac{\sin h}{h}$

• Based on our limits, and some lad named L’Hopital, we know that $\displaystyle {\lim_{h\to 0}} \frac{\sin h}{h}=1$, and $\displaystyle {\lim_{h\to 0}} \frac{\cos h-1}{h}=0$
• Therefore:

$f\prime (x)=(\sin x)\displaystyle {\lim_{h\to 0}} \frac{\cos h-1}{h}+\cos(x)\displaystyle {\lim_{h\to 0}} \frac{\sin h}{h}=1(\cos x)+0$ $\therefore f\prime (x)=\cos(x)$

##### Pranav Sharma
###### Site Owner

Year 12 Student, site owner and developer.

##### Jackson Taylor
###### Post Writer

I’m a Year 12 student at Sydney Tech.

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