# Proving That Two Different Variance Formulas Are Equal To One Another

Prove that:

$\frac { \Sigma { \left( x-\overline { x } \right) }^{ 2 } }{ n } =\frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 }$

Well you first identify these truths:

$x=\left\{ a,\quad b,\quad c \right\} \\ \\ \therefore \quad \overline { x } =\frac { \Sigma x }{ n } =\frac { a+b+c }{ n } ,\quad \\ \\ \therefore \quad n=3\\ \\ \therefore \quad n\overline { x } =a+b+c\\ \\ \therefore \quad \frac { \Sigma { x }^{ 2 } }{ n } =\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ n } .\\ \\ \therefore \quad \Sigma { x }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }$

And next:

The standard deviation formula looks like this:

$\sigma =\sqrt { \frac { \Sigma { \left( x-\overline { x } \right) }^{ 2 } }{ n } } =\sqrt { \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } }$

# Integration Proof (1)

Prove that:

$\int { { x }^{ n }dx=\frac { { x }^{ n+1 } }{ n+1 } } +C$

Proof:

# Implicit Differentiation Rules

Rule 1:

$\frac { dy }{ dy } \cdot \frac { dy }{ dx } =\frac { dy }{ dx }$

Rule 2: