In fact, I don't wanna write this. But I'm in china school, they don't care what you have learned, they only care about what you can do on a test paper.

How to read

$\underline{sine} = \sin$

$\underline{co} \underline{sine} = \cos$

$\underline{tan} \underline{gent} = \tan$

$\underline{co} \underline{tangent} = \cot$

$\underline{se} \underline{cant} = \sec$

$\underline{cose} \underline{cant} = \csc$

Basic Derivative Formula

$$\begin{align*} \\ (C)^\prime &= 0 &\text{(C is a constant)} \\ \\ (x^n)^\prime &= nx^{n-1} \\ \\ (a^x)^\prime &= a^x\ln{a} &(e^x)^\prime &= e^x \\ \\ (\log_a{x})^\prime &= \frac{1}{x\ln{a}} &(\ln{x})^\prime &= \frac{1}{x} \\ \\ \\ (\sin{x})^\prime &= \cos{x} &(\cos{x})^\prime &= -\sin{x} \\ \\ (\tan{x})^\prime &= \sec^2{x} &(\cot{x})^\prime &= -\csc^2{x} \\ \\ (\sec{x})^\prime &= \sec{x}\tan{x} &(\csc{x})^\prime &= -\csc{x}\cot{x} \\ \\ \\ (\arcsin{x})^\prime &= \frac{1}{\sqrt{1 - x^2}} &(\arccos{x})^\prime &= -\frac{1}{\sqrt{1 - x^2}} \\ \\ (\arctan{x})^\prime &= \frac{1}{1 + x^2} &(arccot{x})^\prime &= -\frac{1}{1 + x^2} \\ \\ \end{align*}$$

Additional

$$\begin{align*} &\csc x = \frac{1}{\sin x} \\ \\ &\sec x = \frac{1}{\cos x} \\ \\ &\cot x = \frac{1}{\tan x} \\ \\ \\ &\sin ^2{x} + \cos ^2{x} = 1 \\ \\ &1 + \tan ^2{x} = \sec ^2{x} \\ \\ &1 + \cot ^2{x} = \csc ^2{x} \\ \\ \\ &\tan{x} = \frac{\sin{x}}{\cos{x}} \\ \\ &\cot{x} = \frac{\cos{x}}{\sin{x}} \end{align*}$$