Math

Basics

To render any math equation, the math equation must be between <math></math> i.e., <math>f(x)=x^2</math> gives $f(x)=x^{2}$.

Common math commands

Superscript & Subscript

Superscript: <math>x^{5+y}</math> gives $x^{5+y}$
Subscript: <math>x_{5+t}</math> gives $x_{5+t}$
Together: <math>x_{5+t}^{5+y}</math> gives $x_{5+t}^{5+y}$

Fractions, radicals and brackets

Fractions: <math>\frac{1}{x}</math> gives ${\frac {1}{x}}$
Normal brackets, parentheses, etc.: <math>(\frac{1}{x})^3</math> gives $({\frac {1}{x}})^{3}$
Large brackets, parentheses, etc.: <math>\left(\frac{1}{x}\right)^3</math> gives $\left({\frac {1}{x}}\right)^{3}$
Square root: <math>\sqrt{x+1}</math> gives ${\sqrt {x+1}}$
General radical: <math>\sqrt[3]{64}=4</math> gives ${\sqrt[{3}]{64}}=4$

Trig. & Log Functions

Sin, cos, tan, etc.: <math>\sin{(\theta)}</math> gives $\sin {(\theta )}$
Arcsin, arccos, arctan, etc.: <math>\arcsin{(\theta)}</math> gives $\arcsin {(\theta )}$
Log: <math>\log_{5}{5^2}=2</math> gives $\log _{5}{5^{2}}=2$
Ln: <math>\ln{e^3}=3</math> gives $\ln {e^{3}}=3$

Calculus

Sum: <math>\sum_{i=1}^{n}i=\frac{n(n+1)}{2}</math> gives $\sum _{i=1}^{n}i={\frac {n(n+1)}{2}}$

Limit: <math>\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}</math> gives $\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Derivative: <math>\frac{d}{dx}\left[\frac{1}{x}\right]=-\frac{1}{x^2}</math> gives ${\frac {d}{dx}}\left[{\frac {1}{x}}\right]=-{\frac {1}{x^{2}}}$

Integral: <math>\int_{1}^{x+1}\frac{1}{r}dr</math> gives $\int _{1}^{x+1}{\frac {1}{r}}dr$

Limit bar: <math>\bigg|_{0}^{1}</math> gives ${\bigg |}_{0}^{1}$

Advanced

Sometimes it might be necessary to break up and align a long equation:

${\begin{aligned}\int _{0}^{1}\left(3+x{\sqrt {x}}\right)dx&=\int _{0}^{1}\left(3+x^{1}{x}^{\frac {1}{2}}\right)dx=\int _{0}^{1}\left(3+x^{1+{\frac {1}{2}}}\right)dx=\int _{0}^{1}\left(3+x^{\frac {3}{2}}\right)dx\\[2ex]&=3x+{\frac {x^{{\frac {3}{2}}+1}}{{\frac {3}{2}}+1}}{\bigg |}_{0}^{1}=3x+{\frac {x^{\tfrac {5}{2}}}{\frac {5}{2}}}{\bigg |}_{0}^{1}=3x+{\frac {2x^{\frac {5}{2}}}{5}}{\bigg |}_{0}^{1}\\[2ex]&=\left[3(1)+{\frac {2(1)^{5/2}}{5}}\right]-\left[3(0)+{\frac {2(0)^{5/2}}{5}}\right]\\[2ex]&=3+{\frac {2}{5}}={\frac {15}{5}}+{\frac {2}{5}}={\frac {17}{5}}\end{aligned}}$