# Even And Odd Functions

An even function exists when $f\left( x \right) =f\left( -x \right)$ for all values of x.

The graph of an even function must be symmetrical about the y-axis.

Examples of even functions have been listed below:

$f\left( x \right) ={ x }^{ 2 }$

$f\left( x \right) ={ x }^{ 4 }$

$f\left( x \right) =cosx$

———————————————————————–

An odd function exists when $f\left( x \right) =-f\left( -x \right)$ for all values of x.

Graphs of odd functions should have 180 degrees rotational symmetry about the origin (0,0).

Examples of odd functions are listed below:

$f\left( x \right) =x$

$f\left( x \right) ={ x }^{ 3 }$

$f\left( x \right) =sinx$

——————————————————————–

Most functions are neither even or odd.

An example of a function that is neither even or odd would be:

$f\left( x \right) ={ x }^{ 2 }+\frac { 1 }{ x }$

# Mathematical Programming (Manipulation) Tips

Commutative means that the order of elements contained within an expression do not alter a  certain mathematical result.

For example:

$2\cdot 3=3\cdot 2=6\\ \\ 1\cdot 2\cdot 3=2\cdot 1\cdot 3=3\cdot 1\cdot 2=6\\ \\ Or:\\ \\ 1+2=2+1=3\\ \\ 1+2+3=3+2+1=2+1+3=6$

Knowing this, you can say that:

$a\cdot b\cdot c=c\cdot a\cdot b=b\cdot a\cdot c=c\cdot b\cdot a\\ \\ Or:\\ \\ a+b+c=c+a+b=b+a+c=c+b+a$

—————————————————————–

Associative means that you can alter the grouping of certain elements in an expression without changing its result.

$\left( 2\cdot 3 \right) \cdot 4=2\cdot \left( 3\cdot 4 \right) =24\\ \\ Or:\\ \\ \left( 2+3 \right) +4=2+\left( 3+4 \right) =9$

Knowing this you could say that:

$\left( a\cdot b \right) \cdot c=a\cdot \left( b\cdot c \right) \\ \\ Or:\\ \\ \left( a+b \right) +c=a+\left( b+c \right)$

——————————————————————

Warning:

Commutative and Associative manipulation should not be used when subtracting  and dividing.

# How To Multiply Surds Contained within Fractions

First Scenario:

$\frac { 1 }{ a+\sqrt { b } } =\frac { 1 }{ \left( a+\sqrt { b } \right) } \cdot \frac { \left( a-\sqrt { b } \right) }{ \left( a-\sqrt { b } \right) } \\ \\ =\frac { a-\sqrt { b } }{ { a }^{ 2 }-a\sqrt { b } +a\sqrt { b } -b } =\frac { a-\sqrt { b } }{ { a }^{ 2 }-b }$

Second Scenario:

$\frac { 1 }{ a-\sqrt { b } } =\frac { 1 }{ \left( a-\sqrt { b } \right) } \cdot \frac { \left( a+\sqrt { b } \right) }{ \left( a+\sqrt { b } \right) } \\ \\ =\frac { a+\sqrt { b } }{ { a }^{ 2 }+a\sqrt { b } -a\sqrt { b } -b } =\frac { a+\sqrt { b } }{ { a }^{ 2 }-b }$

Notice that what we’re ultimately doing in both cases is multiplying the surd within a fraction by 1. When the value of the numerator is exactly the same as the value of the denominator in a fraction, what you have is 1.

You should know that 1/1, 2/2, 3/3, (a+b)/(a+b) are all equal to 1.

# How To Multiply Surds

In order to multiply surds, you should first know these rules:

${ a }^{ m }\cdot { a }^{ n }={ a }^{ m+n }\\ \\ { a }^{ m }\div { a }^{ n }={ a }^{ m-n }$

You should also know that:

${ a }^{ \frac { 1 }{ 2 } }=\sqrt [ 2 ]{ { a }^{ 1 } } =\sqrt { a }$

So, knowing these rules, what would you get if you multiplied: $\sqrt { 3 } \cdot \sqrt { 3 }$?

Well, $\sqrt { 3 } \cdot \sqrt { 3 } ={ 3 }^{ \frac { 1 }{ 2 } }\cdot { 3 }^{ \frac { 1 }{ 2 } }={ 3 }^{ \frac { 1 }{ 2 } +\frac { 1 }{ 2 } }={ 3 }^{ 1 }=3$.

——————————————————————————————–

Now how about $\sqrt { 3 } \cdot \left( -\sqrt { 3 } \right)$?

$\sqrt { 3 } \cdot \left( -\sqrt { 3 } \right) =\sqrt { 3 } \cdot \left( -1 \right) \cdot \sqrt { 3 } \\ \\ =\sqrt { 3 } \cdot \sqrt { 3 } \cdot \left( -1 \right) =3\cdot \left( -1 \right) =-3$

——————————————————————————————–

What about $\left( -\sqrt { 3 } \right) \left( -\sqrt { 3 } \right)$?

$\left( -\sqrt { 3 } \right) \left( -\sqrt { 3 } \right) =\left( -1 \right) \cdot \sqrt { 3 } \cdot \left( -1 \right) \cdot \sqrt { 3 } \\ \\ =\left( -1 \right) \left( -1 \right) \sqrt { 3 } \sqrt { 3 } =1\cdot 3=3$

# Derivative of y=cotx

$y=cotx=\frac { cosx }{ sinx } =\frac { u }{ v } ,\\ \\ If\quad y=\frac { u }{ v } ,\quad \frac { dy }{ dx } =\frac { v\frac { du }{ dx } -u\frac { dv }{ dx } }{ { v }^{ 2 } } .\\ \\ u=cosx,\quad \therefore \quad \frac { du }{ dx } =-sinx\\ v=sinx,\quad \therefore \quad \frac { dv }{ dx } =cosx,\quad { v }^{ 2 }=\sin ^{ 2 }{ x } \\ \\ So:\\ \\ \frac { dy }{ dx } =\frac { sinx\cdot \left( -sinx \right) -cosxcosx }{ \sin ^{ 2 }{ x } } \\ \\ =\frac { -\sin ^{ 2 }{ x } -\cos ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } =\frac { -\sin ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } -\frac { \cos ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } \\ \\ =-1-\cot ^{ 2 }{ x } =-{ cosec }^{ 2 }x\\ \\ \therefore \quad If\quad y=cotx,\quad \frac { dy }{ dx } =-{ cosec }^{ 2 }x\\ \\ \\$

# tan(A-B)=(tanA-tanB)/(1+tanAtanB)

Prove that:

$tan\left( A-B \right) =\frac { tanA-tanB }{ 1+tanAtanB }$

Firstly:

$tan\left( \alpha +\beta \right) =\frac { tan\alpha +tan\beta }{ 1-tan\alpha tan\beta }$

Also, remember that:

$tan\left( -\theta \right) =-tan\theta$

So:

$\beta =-\phi \\ \\ tan\left( \alpha +\left( -\phi \right) \right) =\frac { tan\alpha +tan\left( -\phi \right) }{ 1-tan\alpha tan\left( -\phi \right) } \\ \\ =\frac { tan\alpha -tan\phi }{ 1+tan\alpha tan\phi } =tan\left( \alpha -\phi \right) \\ \\ \therefore \quad tan\left( A-B \right) =\frac { tanA-tanB }{ 1+tanAtanB }$

# How To Derive The Quadratic Formula

$a{ x }^{ 2 }+bx+c=0\\ \\ a{ x }^{ 2 }+bx=-c\\ \\ { x }^{ 2 }+\frac { b }{ a } x=-\frac { c }{ a } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }-{ \left( \frac { b }{ 2a } \right) }^{ 2 }=-\frac { c }{ a } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }-\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } =-\frac { c }{ a } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }=\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } -\frac { c }{ a } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }=\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } -\frac { 4ac }{ 4{ a }^{ 2 } } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }=\frac { { b }^{ 2 }-4ac }{ 4{ a }^{ 2 } } \\ \\ x+\frac { b }{ 2a } =\pm \frac { \sqrt { { b }^{ 2 }-4ac } }{ \sqrt { 4 } \sqrt { { a }^{ 2 } } } =\pm \frac { \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \\ x=-\frac { b }{ 2a } \pm \frac { \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \\ x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

— USEFUL FORMULAS:

$\frac { a }{ c } \pm \frac { b }{ c } =\frac { a\pm b }{ c } \\ \\ \frac { a }{ b } \pm \frac { c }{ d } =\frac { ad }{ bd } \pm \frac { bc }{ bd } =\frac { ad\pm bc }{ bd } \\ \\ \sqrt { { a }^{ 2 } } ={ a }^{ \frac { 2 }{ 2 } }=a\\ \\ \sqrt { a } \sqrt { b } =\sqrt { ab } \\ \\ \frac { \sqrt { a } }{ \sqrt { b } } =\sqrt { \frac { a }{ b } }$

# Quick Way To Derive The Integration By Parts Formula

Firstly, you need to know what the product rule is:

$If\quad y=u\cdot v,\quad \frac { dy }{ dx } =u\frac { dv }{ dx } +v\frac { du }{ dx } .$

Then…

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

Now integrate each term with respect to x:

$\int { u\frac { dv }{ dx } } dx=\int { \frac { dy }{ dx } } dx-\int { v\frac { du }{ dx } } dx$

Leaving:

$\int { u\frac { dv }{ dx } } dx=y-\int { v\frac { du }{ dx } } dx\\ \\ As\quad y=u\cdot v,\\ \\ \int { u\frac { dv }{ dx } } dx=uv-\int { v\frac { du }{ dx } } dx$

# Derivate Of y=cosx Proof

Prove that:

$If\quad f(x)=cosx,\quad f$

$\lim _{ \delta x\rightarrow 0 }{ \frac { f\left( x+\delta x \right) -f\left( x \right) }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cos\left( x+\delta x \right) -cosx }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cosxcos\delta x-sinxsin\delta x-cosx }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cosx\left( cos\delta x-1 \right) -sinxsin\delta x }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cosx\left( cos\delta x-1 \right) }{ \delta x } } -\frac { sinxsin\delta x }{ \delta x } \\ \\ =0-sinx=-sinx\\ \\ As:\\ \\ \lim _{ \delta x\rightarrow 0 }{ \frac { \left( cos\delta x-1 \right) }{ \delta x } } =0\\ \\ \lim _{ \delta x\rightarrow 0 }{ \frac { sin\delta x }{ \delta x } } =1$

# sin(A-B)=sinAcosB-cosAsinB

Prove that:

$sin(A-B)=sinAcosB-cosAsinB$

$sin\left( \alpha +\beta \right) =sin\alpha cos\beta +cos\alpha sin\beta \\ \\ But:\quad \\ \\ sin(-\theta )=-sin\theta \\ cos(-\theta )=cos\theta \\ tan(-\theta )=-tan\theta \\ \\ Say:\quad \beta =-\phi \\ \\ \therefore \quad sin\left( \alpha +\left( -\phi \right) \right) =sin\alpha cos\left( -\phi \right) +cos\alpha sin\left( -\phi \right) \\ \\ =sin\alpha cos\phi -cos\alpha sin\phi =sin\left( \alpha -\phi \right) \\ \\ If\quad \alpha =A\quad and\quad \phi =B\\ \\ sin\left( A-B \right) =sinAcosB-cosAsinB.$