RadMaths home page $\text{Note created in February,2021}$ This note will be updated once in a while :)
Practice questions are being uploaded :) You will find them in nested buttons :) Chapters with trailing ‘+test‘s have questions :)
ALL QUOTES ARE COPIED FROM BRILLIANT WIKI WITHOUT CHANGE.

Calculus with only one variable—$y=f(x)$

Chapters:
1.Limits
2.Differentiation and derivatives
3.Integrals
4.Infinite sums

1. Limits

off we go!

First we must know what a limit is:

Brilliant wiki:
The limit of a function at a point $a$ in its domain (if it exists) is the value that the function approaches as its argument approaches $a$.

Formally, we have the ‘epsilon-delta’ definition:

Brilliant wiki:
The limit of $f(x)$ as $x$ approaches $x_0$ is $L$, i.e.
$\lim _{x\to x_0} f(x)=L$ if, for every $\epsilon \gt 0$, there exists $\delta \gt 0$ such that, for all $x$, $0\lt |x-x_0|\lt \delta ~~ \Longrightarrow ~~|f(x)-L| \lt \epsilon.$

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Limits have notation $\lim$. The limit of function $f(x)$ at $x=a$ is $\displaystyle \lim _{x\to a} f(x)$.
And here are the properties of limits:
$\text{Let} \lim _{x\to a} f(x)=L_f, ~\lim _{x\to a} g(x)=L_g,~\text{and let}~ f(x)=F,~g(x)=G.$ Then $\color{#D61F06} \begin{aligned} &(1)~~~\lim _{x\to a} F\pm G =L_f \pm L_g. \\ &(2)~~~\lim _{x\to a} FG =L_f \cdot L_g. \\ &(3)~~~~\text{If} ~L_g\neq 0,~\text{then} \lim _{x \to a} \frac{F}{G} =\frac{L_f}{L_g}. ~~~\text{This is to say,} ~G ~\text{can equal zero.}\\ &(4)~~~\lim _{x\to a} F^k=L_f ^k,~ \text{where}~k~\text{is a constant.} \end{aligned}$

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But what if I want to find the limit of $f(x)=\lfloor x \rfloor$ at $x=1$?
Time to introduce the one-sided limit!

Brilliant wiki:
A one-sided limit only considers values of a function that approaches a value from either above or below.

That is to say we look at the limit from two directions:
the left side and the right side :) The limit from the right side is $\displaystyle \lim _{x\to 1^+} \lfloor x \rfloor$, while the one from the left side is $\displaystyle \lim _{x\to 1^-} \lfloor x \rfloor.$
See the $^+$? This means we are approaching the limit from the positive side, i.e. the right side.
Obviously, $\displaystyle \lim _{x\to 1^+} \lfloor x \rfloor =1$ and $\displaystyle \lim _{x\to 1^-} \lfloor x \rfloor =0$.
If a two-sided limit doesn’t have its two sides equal, we say the limit doesn’t exist.

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Final question: what if the limit approaches infinity?
Answer: if the limit approaches $\pm\infty$, then we say the limit DOES NOT EXIST. If the limit approaches an indeterminate value as below: $\frac00,~~\frac{\infty}{\infty},~~\infty -\infty ,~~\infty \cdot 0,~~0^0,~~\infty ^0,~~1^\infty$ Then we compute the limit by other means. Multiples of these indeterminate forms are obviously indeterminate as well so I won’t waste time writing more :)
ONE COMMON MISCONCEPTION:

Brilliant wiki:
The fractions $\dfrac10$ and $\dfrac{\infty}{0}$ are not indeterminate. If the denominator is positive, the limit is $\infty$. If the denominator is negative, the limit is $-\infty$.

Read the next chapter to find out more.

There are three good ways to compute limits. Let’s start with the easiest one first:

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$\huge \text{1. Limits by substitution}$ Since the limit $\displaystyle \lim_{x\to a} f(x)$ computes the value of $f(x)$ when $x$ is infinitely close to $a$, under a common situation, we can just substitute $x$ with $a$.
$\color{#D61F06}\text{Rule I:} \lim _{x\to a}f(x)=f(a) \text{ when event A is real.}$ We shall explore event $A$ later.
Have a try first: compute $\lim _{x\to 1}x^2$.
Solution: plug $x=1$ into $x^2$ to get $\lim_{x\to 1} x^2=1^2=1$.

Then let’s explore event $A$ by computing the below limits:
$\color{#3D99F6}\lim _{x\to 2} \frac{x-1}{x-1} ~~~ \text{and} ~~~\lim_{x\to 1} \frac{x-1}{x-1}.$ We can plug $x=2$ in $\frac{1}{x-1}$ to get $\lim =1$, but when we do it with $x=1$, we get $\frac{0}{0}$.
Boom! Calculator explodes! Here we see how the limit cannot be computed by substitution. This is because it doesn’t fulfill event $A$:
$\color{#D61F06}\text{For }\lim_{x\to a}f(x),\text{ The value } f(a) \text{ does exist, meaning the function has to be well defined, or continuous, at }x=a.$ For limits that cannot be found by substitution, we have other means of computing them: $\huge \text{2. Limits by factoring}$ See Brilliant wiki :) $\huge \text{3. Limits by rationalization}$ See Brilliant wiki :)

See Brilliant wiki :)

2. Differentiation and derivatives

off again!

First, we must understand when a mathematical expression is able to be differentiated:
The mathematical definition of which is: if $\displaystyle \lim _{\Delta x\rightarrow 0} \frac{f(a+\Delta x)-f(a)}{\Delta x}$ exists and converges, then the function $f(x)$ is able to be differentiated at $x=a$. This limit has notation $f’(a)$ and is the derivative of $f(x)$ at $x=a$.

Brilliant wiki:
A function $f$ is differentiable at a point $x_0$ if
$$
1)$f$ is continuous at $x_0$ and
2)the slope of tangent at $x_0$ is well defined.

In a more simple way: if $f(x)$ is smooth at $x=a$, it is able to be differentiated at $x=a$. a ‘smooth’ function at x=a P.S. interesting fact - when we refer to a variable of change, we usually notate it as $\Delta -$ or $d-$.

In the previous chapter we have mentioned the derivative of a function at only one point. Meanwhile, we can find the derivative of the whole function - since every $a$ corresponds to a new $f’(a)$, the function that inputs $a$ and returns $f’(a)$ (for all $a$ in the domain of definition of $f(x)$) is the derivative of $f(x)$. We can notate it as $f’(x)$, $y’$, $\dfrac{dy}{dx}$, $\dfrac{d}{dx}f(x)$. Note that the correct code for ${f}'(x)$ is {f}'(x) while this note uses f’ as I am using an iPad which does not allow directly typing ': see '‘.
Actually, $f’(a)$ is the slope of the tangent line to curve $f(x)$ at $x=a$.
Note that $dy$ represents the change in $y$ here: dx & dy for point A $\color{#D61F06} f’(x)=\frac{dy}{dx}=\lim _{\Delta x\rightarrow 0} \frac{\Delta y}{\Delta x}= \lim _{\Delta x\rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}.$ P.S. If we differentiate a function $f(x)$ twice(sometimes this isn’t possible), we get $f’’(x)$. Similarly, we have $f’’’(x),f^4(x)$ etc.

Here is how we find derivatives: $\displaystyle f’(x)=\lim _{\Delta x\rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$. (For simplicity:$f’(x)=\dfrac{f(x+dx)-f(x)}{dx}$).
For example, let $f(x)=2x$.Then $f’(x)=\lim _{\Delta x\rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{(2x+2\Delta x)-2x}{\Delta x} =\lim _{\Delta x\rightarrow 0} \frac{2\Delta x}{\Delta x}=2.$ Here is the derivative table for common derivatives:

$f(x)$	$f’(x)$
$k$ (a constant)	0
$x^a$	$ax^{a-1}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec ^2 x$
$\cot x$	$-\csc ^2 x$
$\sec x$	$\sec x \tan x$
$\csc x$	$-\csc x \cot x$
$\arcsin x$ (or $\sin ^{-1} x$)	$\dfrac{1}{\sqrt{1-x^2}}$
$\arccos x$	$-\dfrac{1}{\sqrt{1-x^2}}$
$\arctan x$	$\dfrac{1}{1+x^2}$
$\cot ^{-1} x$	$-\dfrac{1}{1+x^2}$
$\sec ^{-1} x$	$\dfrac{1}{x\sqrt{x^2-1}}$
$\csc ^{-1} x$	$-\dfrac{1}{x\sqrt{x^2-1}}$
$e^x$ ($e$ is Euler’s number)	$e^x$
$a^x$	$a^x\ln a$ ($\ln =\log _e$)
$\ln x$	$\dfrac{1}{x}$
$\log _a x$	$\dfrac{1}{x\ln a}$

This chapter will be separated into two sections.

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1. Differentiating simple functions
There are a few formulae you need to know to differentiate more complicated functions(note that they only work when used in the region where both $f(x)$ & $g(x)$ are ‘differentiable’):
$(1)~~~[f(x)\pm g(x)]’=f’(x)\pm g’(x)$;
$(2)~~~[kf(x)]’=kf’(x)$;
$(3)~~~[f(x)g(x)]’=f’(x)g(x)+f(x)g’(x)$;
$(4)~~~[\dfrac{1}{g(x)} ]’=\dfrac{-g’(x)}{[g(x)]^2}$;
$(5)~~~[\dfrac{f(x)}{g(x)}]’=\dfrac{f’(x)g(x)-f(x)g’(x)}{[g(x)]^2}$.

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2. Differentiating other complicated functions
What do I do when I need to differentiate functions like $\sqrt{x^2+x}$?
Here’s what you can do: let $a=g(x)=x^2+x$, $y=f(a)=\sqrt a$. Then we can differentiate $f[g(x)]=\sqrt{x^2+x}$: $\frac{d}{dx} f[g(x)]=\frac{dy}{da} \times \frac{da}{dx}=\frac{1}{2\sqrt a}\times (2x+1)=\frac{2x+1}{2\sqrt a}=\frac{2x+1}{2\sqrt{x^2+x}}.$ So $\left \{ f[g(x)] \right \} ’=f’[g(x)]g’(x)$, otherwise known as the chain rule.

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Questions:
1. Find the derivative of:
a) $\color{#3D99F6} f=\dfrac{1}{x}$,
b) $g=2^x \cdot x^3$,
c) $\color{#3D99F6} p=6\log _3 x$,
d) $q=\displaystyle \sqrt{\sin x+e^{\sin x}}$.
2. Given properties of $f(x)$, find $f(x)$:
a) $\color{#20A900} f(x)=ax^2+2x$ and $\color{#20A900}f’(1)=8$;
b) $\color{#20A900}f(x)=a^x$ and $\color{#20A900} f’(1)=e$.

Answers:

here you go

1.
a) $\color{#3D99F6} f’=\{ x^{-1} \}’ =-x^{-1-1}=-x^{-2}=-\dfrac{1}{x^2}$.
b) $g’=2^x\cdot \{ x^3 \}’ +\{ 2^x\}’x^3 =2^x 3x^2 +(2^x\ln 2)x^3 =2^x x^2 (x\ln 2+3)$.
c) $\color{#3D99F6} p’=6\cdot \{\log _3 x\} ’=6\cdot \dfrac{1}{x\ln 3}=\dfrac{6}{x\ln 3}$.
d) $\text{Let} ~y=\sqrt a, ~a=b+e^b,~ b=\sin x.$
Then, by the chain rule, $q’=\dfrac{dy}{dx}=\dfrac{dy}{da}\dfrac{da}{db}\dfrac{db}{dx}=\dfrac{1}{2\sqrt{a}}\cdot (1+e^b)\cdot \cos x=\dfrac{\cos x +e^{\sin x}\cos x}{2\sqrt{\sin x+e^{\sin x}}}$.
2.
a) Since $\color{#20A900}f’(x)=2ax+2,~f’(1)=8$, $\color{#20A900} a$ solves to $\color{#20A900}3$. Therefore $\color{#20A900}f(x)=3x^2+2x$.
b) Since $\color{#20A900}f’(1)=a^1\ln a=a\ln a=e$, the function solves to $\color{#20A900}a=e$. Therefore $\color{#20A900}f(x)=e^x$.

There are three types of functions that only involve $x$ and $y$ that are not defined as $y=f(x)$.

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1.$x=g(y)$
As for a function with $y$ as an independent variable, we can differentiate it with respect to $y$: $\dfrac{dx}{dy} =\dfrac{g(y+dy)-g(y)}{dy}$. If the inverse function is differentiable in a region, then in this region $\dfrac{dy}{dx}=\cfrac{1}{\frac{dx}{dy}}$ always holds. This may seem useless but comes handy sometimes.

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2.$P=(f(t),g(t))$
In this situation, $x=f(t),y=g(t)$, making $P$ an equation in its parametric form. Therefore it would be reasonable to say:$\dfrac{dy}{dx}=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$.
e.g. To prove this works, say for example $P=(2t,4t^2+1)$ is equivalent to $y=x^2+1$. Differentiate both:
$(1) \frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{8t}{2}=4t=2x.$ $(2) y’=\left \{ x^2\right \} ’+\{ 1\} ’=2x+0=2x.$

Brilliant wiki:
By the chain rule, $\frac{dy}{dx}\frac{dx}{dt}=\frac{dy}{dt}.$ The first term on the right side is the first derivative of the parametric function. In order to isolate it, divide both sides by $\frac{dx}{dt}$.

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3.$f(x,y)=0$ When $f(x,y)=0$, we can: 1.either rewrite this into $y=f(x)$ (or $\color{#3D99F6}y_1=f(x) ~\text{and}~ y_2=g(x)$ etc.); 2. or differentiate it with respect to $x$ first, and then to $y$.

Examples of 1:
If $x^2+y^2-1=0$, then $y^2=1-x^2$. Therefore $y=\sqrt{1-x^2} ~\text{or}~ y=-\sqrt{1-x^2}$. Differentiate both:
$\dfrac{dy}{dx}=\pm \frac{-2x}{2\sqrt{1-x^2}}= \pm \frac{x}{\sqrt{1-x^2}},$ And when $xy\gt 0,~f’(x)=\dfrac{x}{\sqrt{1-x^2}}$; when $xy=0,~f’(x)=0$ ; when $xy\lt 0,~f’(x)=-\dfrac{x}{\sqrt{1-x^2}}.$

Examples of 2:
e.g.1.
Let $x^2+y^2-1=0$ still. Treat $y^2$ as a nonzero function of $x$ while differentiating, that is to say, instead of $\dfrac{d}{dx} y^2=0$, let $\dfrac{d}{dx} y^2=\dfrac{d}{dy}y^2 \times \dfrac{dy}{dx}$. Then we can solve for $\dfrac{dy}{dx}$ like solving an equation.
1. Differentiate with respect to $x$: $2x+\dfrac{d}{dy}y^2 \frac{dy}{dx}=0.$ 2. Differentiate with respect to $y$:
$2x+2y\dfrac{dy}{dx}=0$. So $\dfrac{dy}{dx}=-\dfrac{x}{y}.$

e.g.2.
Let $xy-1=0$. Obviously, $y=\frac{1}{x}$, implying $\frac{dy}{dx}=-\dfrac{1}{x^2}$.
1. Differentiate with respect to $x$:
The only term including $x$ is $xy$. How do I work on that? Well, since $[f(x)g(x)]’=f’(x)g(x)+f(x)g’(x)$, ${xy}’=\frac{d}{dx}x\cdot y+xy’=y+xy’.$ So $y+xy’=0$, which solves to $y’=-\dfrac{y}{x}=-\dfrac{\frac{1}{x}}{x}=-\dfrac{1}{x^2}$.

The big question: what is an extremum?

Brilliant wiki:
An extremum (or extreme value) of a function is a point at which a maximum or minimum value of the function is obtained in some interval. A local extremum (or relative extremum) of a function is the point at which a maximum or minimum value of the function in some open interval containing the point is obtained.

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Let’s start with local extrema first.
Mathematically, if $f(x)$ is the maximum or minimum of a function in range $(x-c,x+c)$ for sufficiently small $c$, then it is a local extremum of the function.
Logically speaking, if a function decreases on one side of $f(x)$ and increases on the other, then $f(x)$ is a local extremum of the function.

Given a function $f$ and interval $[a,b]$, the local extrema may be points of discontinuity, points of non-differentiability, or points at which the derivative has value $0$. However, none of these points are necessarily local extrema, so the local behavior of the function must be examined for each point.
So here are two simple ways of finding the extrema after examining endpoints and un-differentiable points:
1. Maximum occurs when for all $a$ in range $(x-c,x+c)$ for sufficiently small $c$, $f(a)\lt f(x)$. Minimum occurs when similarly when $f(a)\gt f(x)$.
2.We can also test using derivatives: see First Derivative Test and Second Derivative Test.

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Next we have global extrema.

Global extrema is rather like extrema when we are to find it. Unlike local extrema though, it takes only one value. Why?
This is because global extrema is defined as the absolute maximum or minimum in range $[a,b]$, while local extrema is defined for $x-c,x+c)\in [a,b]$ for sufficiently small $c$. Take $f(x)=(x+1)^2(x+2)(x+3)$ and $[a,b]=[-3,0]$ for example: Credits to GeoGebra This function in range $[-3,0]$ has five local extrema: $(-3,0);~~(-2.640388...,-0.61968...);~~(-1.60961...,0.20171...);~~(-1,0);(0,6)$.
In the same range, there exists only one absolute maximum $(0,6)$ and one absolute minimum $(-2.64038...,-0.61968...)$.
The way to find global extrema is to test for endpoints, points where $f’(x)=0$ and points where $f’(x)$ is undefined and compare to find absolute maximum and minimum.

3. Integrals

Gabriel’s horn(trumpet): TOOT!

4. Infinite sums

Whooo!