数学公式

$$\sin x = x - \frac{x^3}{3!} + o(x^3)$$

$$\arcsin x = x + \frac{x^3}{3!} + o(x^3)$$

$$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + o(x^4)$$

$$\tan x = x + \frac{x^3}{3} + o(x^3)$$

$$\arctan x = x - \frac{x^3}{3} + o(x^3)$$

$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + o(x^3)$$

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + o(x^3)$$

$$(1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!} x^2 + o(x^2)$$

$$(log_{a}x)' = \frac{1}{x \ln{a}} (a>0, a \ne 1)$$

$$(\ln{\left|x\right|})' = \frac{1}{x}$$

$$(arcsin x)' = \frac{1}{\sqrt{1-x^2}}$$

$$(arccos x)' = - \frac{1}{\sqrt{1-x^2}}$$

$$(tan x)' = sec^2 x$$

$$(cot x)' = -csc^2 x$$

$$(arctan x)' =\frac{1}{1 + x^2}$$

$$(arccot x)' = - \frac{1}{1 + x^2}$$

$$(sec x)' = sec x tan x$$

$$(csc x)' = -csc x cot x$$

$$[\ln(x + \sqrt{x^2 \pm 1})]' = \frac{1}{\sqrt{x^2 \pm 1}}$$

$\int \frac{sin x}{x}\, {\rm d}x$$\int \frac{cos x}{x}\, {\rm d}x$$\int \frac{tan x}{x}\, {\rm d}x$$\int \frac{e^x}{x}\, {\rm d}x$

$\int sin x^2\, {\rm d}x$$\int cos x^2\, {\rm d}x$$\int tan x^2\, {\rm d}x$$\int \frac{e^{a x^2 + b x + c}}{x}\, {\rm d}x (a \ne 0)$

$\int \frac{1}{sin x}\, {\rm d}x$$\int \frac{1}{cos x}\, {\rm d}x$

$\int \frac{1}{\ln x}\, {\rm d}x$$\int \frac{\ln(1+x)}{x}\, {\rm d}x$

$$y=f(x) = \sum_{n=0}^\infty \frac{f^{(n)} (x_0)}{n!} (x-x_0)^n$$

$$(uv)^{(n)} = \sum_{k=0}^n C^k_n u^{(n-k)} v^{(k)} = \sum_{k=0}^n C^k_n u^{(k)} v^{(n-k)}$$

举个例子，因为 $C^2_6$ 和 $C^4_6$ 结果是一样的，所以可以互换。