# Function Rules

$fg\left( x \right) =f\left( g\left( x \right) \right) \\ \\ ff\left( x \right) =f\left( f\left( x \right) \right) ={ f }^{ 2 }\left( x \right) \\ \\ gf\left( x \right) =g\left( f\left( x \right) \right) \\ \\ gg\left( x \right) =g\left( g\left( x \right) \right) ={ g }^{ 2 }\left( x \right)$

# Completing The Square (Why It Works)

Prove that:

${ \left( x-\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ x }^{ 2 }-bx$

Proof:

${ \left( x-\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ =\left( x-\frac { b }{ 2 } \right) \left( x-\frac { b }{ 2 } \right) -\frac { { b }^{ 2 } }{ 4 } \\ \\ ={ x }^{ 2 }-2x\frac { b }{ 2 } +\frac { { b }^{ 2 } }{ 4 } -\frac { { b }^{ 2 } }{ 4 } \\ \\ ={ x }^{ 2 }-bx$

Second Proof:

$As:\\ \\ { p }^{ 2 }-{ q }^{ 2 }=\left( p+q \right) \left( p-q \right) \\ \\ And:\\ \\ p=\left( x-\frac { b }{ 2 } \right) \\ \\ And:\\ \\ q=\frac { b }{ 2 } ,\\ \\ { \left( x-\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ =\left( x-\frac { b }{ 2 } +\frac { b }{ 2 } \right) \left( x-\frac { b }{ 2 } -\frac { b }{ 2 } \right) \\ \\ =x\left( x-2\frac { b }{ 2 } \right) \\ \\ =x\left( x-b \right) \\ \\ ={ x }^{ 2 }-bx$

Prove that:

${ \left( x+\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ x }^{ 2 }+bx$

Proof:

${ \left( x+\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ =\left( x+\frac { b }{ 2 } \right) \left( x+\frac { b }{ 2 } \right) -\frac { { b }^{ 2 } }{ 4 } \\ \\ ={ x }^{ 2 }+2x\frac { b }{ 2 } +\frac { { b }^{ 2 } }{ 4 } -\frac { { b }^{ 2 } }{ 4 } \\ \\ ={ x }^{ 2 }+bx$

Second Proof:

$As:\\ \\ { p }^{ 2 }-{ q }^{ 2 }=\left( p+q \right) \left( p-q \right) \\ \\ And:\\ \\ p=\left( x+\frac { b }{ 2 } \right) \\ \\ And:\\ \\ q=\frac { b }{ 2 } ,\\ \\ { \left( x+\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ =\left( x+\frac { b }{ 2 } +\frac { b }{ 2 } \right) \left( x+\frac { b }{ 2 } -\frac { b }{ 2 } \right) \\ \\ =\left( x+2\frac { b }{ 2 } \right) x\\ \\ =x\left( x+b \right) \\ \\ ={ x }^{ 2 }+bx$

# Obtaining The Mean From Two Sets Of Data

Prove that if set A, of size ${ n }_{ 1 }$, has mean $\overline { { x }_{ 1 } }$ and set B, of size ${ n }_{ 2 }$, has a mean $\overline { { x }_{ 2} }$ then the mean of the combined set of A and B is $\overline { x } =\frac { { n }_{ 1 }\overline { { x }_{ 1 } } +{ n }_{ 2 }\overline { { x }_{ 2 } } }{ { n }_{ 1 }+{ n }_{ 2 } }$.

——————

Set A:

${ q }_{ 1 },{ q }_{ 2 },{ q }_{ 3 },{ q }_{ 4 }$

${ n }_{ 1 }=4$

Therefore:

$\overline { { x }_{ 1 } } =\frac { { q }_{ 1 }+{ q }_{ 2 }+{ q }_{ 3 }+{ q }_{ 4 } }{ 4 } =\frac { { q }_{ 1 }+{ q }_{ 2 }+{ q }_{ 3 }+{ q }_{ 4 } }{ { n }_{ 1 } }$

Therefore:

${ n }_{ 1 }\overline { { x }_{ 1 } } ={ q }_{ 1 }+{ q }_{ 2 }+{ q }_{ 3 }+{ q }_{ 4 }$

——————

Set B:

${ q }_{ 5 },{ q }_{ 6 },{ q }_{ 7 },{ q }_{ 8 },{ q }_{ 9 }$

${ n }_{ 2 }=5$

Therefore:

$\overline { { x }_{ 2 } } =\frac { { q }_{ 5 }+{ q }_{ 6 }+{ q }_{ 7 }+{ q }_{ 8 }+{ q }_{ 9 } }{ 5 } =\frac { { q }_{ 5 }+{ q }_{ 6 }+{ q }_{ 7 }+{ q }_{ 8 }+{ q }_{ 9 } }{ { n }_{ 2 } }$

Therefore:

${ n }_{ 2 }\overline { { x }_{ 2 } } ={ q }_{ 5 }+{ q }_{ 6 }+{ q }_{ 7 }+{ q }_{ 8 }+{ q }_{ 9 }$

So:

$\frac { \left( { q }_{ 1 }+{ q }_{ 2 }+{ q }_{ 3 }+{ q }_{ 4 } \right) +\left( { q }_{ 5 }+{ q }_{ 6 }+{ q }_{ 7 }+{ q }_{ 8 }+{ q }_{ 9 } \right) }{ 9 } \\ \\ =\frac { { n }_{ 1 }\overline { { x }_{ 1 } } +{ n }_{ 2 }\overline { { x }_{ 2 } } }{ 4+5 } \\ \\ =\frac { { n }_{ 1 }\overline { { x }_{ 1 } } +{ n }_{ 2 }\overline { { x }_{ 2 } } }{ { n }_{ 1 }+{ n }_{ 2 } }$

# If y=e^(f(x)), dy/dx=f'(x)e^(f(x))

Proof:

$y={ e }^{ n }\\ \\ But:\quad n=f\left( x \right) \\ \\ \frac { dy }{ dn } ={ e }^{ n }\\ \\ n=f\left( x \right) \\ \\ \frac { dn }{ dx } =f$

# If y=e^x, dy/dx=e^x

$\underset { \delta x\rightarrow 0 }{ lim } \frac { f\left( x+\delta x \right) -f\left( x \right) }{ \delta x } \\ \\ =\underset { \delta x\rightarrow 0 }{ lim } \frac { { e }^{ x+\delta x }-{ e }^{ x } }{ \delta x } \\ \\ =\underset { \delta x\rightarrow 0 }{ lim } \frac { { e }^{ x }{ e }^{ \delta x }-{ e }^{ x } }{ \delta x } \\ \\ =\underset { \delta x\rightarrow 0 }{ lim } \frac { { e }^{ x }\left( { e }^{ \delta x }-1 \right) }{ \delta x } \\ \\ ={ e }^{ x }\underset { \delta x\rightarrow 0 }{ lim } \frac { \left( { e }^{ \delta x }-1 \right) }{ \delta x } \\ \\ ={ e }^{ x }\cdot 1\\ \\ ={ e }^{ x }\\ \\ Therefore,\quad if:\\ \\ y={ e }^{ x },\\ \\ \frac { dy }{ dx } ={ e }^{ x }$

Prove that:

$\log _{ a }{ x } =\frac { \log _{ b }{ x } }{ \log _{ b }{ a } }$

Say that:

$\log _{ b }{ x } =p\\ \\ \therefore \quad { b }^{ p }=x$

And that:

$\log _{ b }{ a } =q\\ \\ \therefore \quad { b }^{ q }=a$

Therefore:

$\log _{ a }{ \left( x \right) } \\ \\ =\log _{ a }{ \left( { b }^{ p } \right) } \\ \\ =p\log _{ a }{ \left( b \right) } \\ \\ =p\cdot \frac { 1 }{ \log _{ b }{ a } } \\ \\ =\frac { \log _{ b }{ x } }{ \log _{ b }{ a } }$

# If y=a^x, dy/dx=a^xlna proof

${ a }^{ x }=y\\ \\ \log _{ a }{ y=x } \\ \\ \frac { \ln { y } }{ \ln { a } } =x\\ \\ x=\frac { 1 }{ \ln { a } } \cdot \ln { y } \\ \\ x=n\cdot u\\ \\ \therefore \quad \frac { dx }{ du } =n\\ \\ u=\ln { y,\quad \frac { du }{ dy } } =\frac { 1 }{ y } \\ \\ \frac { dx }{ dy } =\frac { dx }{ du } \cdot \frac { du }{ dy } =\frac { n }{ y } \\ \\ \frac { dy }{ dx } =\frac { 1 }{ \frac { dx }{ dy } } \\ \frac { dy }{ dx } =\frac { 1 }{ \frac { n }{ y } } \\ \frac { dy }{ dx } =\frac { y }{ n } \\ \\ but,\quad y={ a }^{ x }\quad and\quad n=\frac { 1 }{ \ln { a } } \\ \\ \therefore \quad \frac { dy }{ dx } =\frac { { a }^{ x } }{ \frac { 1 }{ \ln { a } } } \\ \\ \frac { dy }{ dx } ={ a }^{ x }\cdot \ln { a } \\ \\ \\$

# If y=lnx, dy/dx=1/x proof

$\log _{ e }{ x } =y\\ \\ \therefore \quad x={ e }^{ y }\\ \\ \frac { dx }{ dy } ={ e }^{ y }\\ \\$

Now:

$\frac { dy }{ dx } =\frac { 1 }{ \frac { dx }{ dy } } \\ \\ \frac { dy }{ dx } =\frac { 1 }{ { e }^{ y } } \\ \\ \frac { dy }{ dx } =\frac { 1 }{ x } \\ \\$