數學要背的內容
常用泰勒公式:
\(\sin x = x - {\Large\frac{x^3}{3!}} + o(x^3)\)
\(\cos x = 1 - {\Large \frac{x^2}{2!} + \frac{x^4}{4!}} + o(x^4)\)
\(\tan x = x + {\Large\frac{x^3}{3!}} + o(x^3)\)
\(\arcsin x = x + {\Large\frac{x^3}{3!}} + o(x^3)\)
\(\arctan x = x - {\Large\frac{x^3}{3!}} + o(x^3)\)
\(\ln (1 + x) = x - {\Large \frac{x^2}{2!} + \frac{x^3}{3!}} + o(x^3)\)
\(e^x = 1 + x + {\Large \frac{x^2}{2!} + \frac{x^3}{3!}} + o(x^3)\)
\((1 + x)^a = 1 + ax + {\Large \frac{a(a - 1)}{2!}}x^2 + o(x^2)\)
常用等價替換:
當 x->0 時，有：
\(\sin x \sim x，\tan x \sim x，\arcsin x \sim x，\arctan x \sim x\)
\(1-\cos x \sim {\Large \frac{x^2}{2}} ， a^x-1 \sim x\ln a ，e^x-1 \sim x\)
\(\ln(1+x) \sim x ，(1+x)^a-1 \sim ax\)
當 \(x\rightarrow\infty\) 時，有：
重要極限:
\(\underset{x\rightarrow \infty}{\lim}(1 + \frac{1}{x})^x = e\)
\(\underset{x\rightarrow \infty}{\lim}{\Large \frac{\sin x}{x}} = 1\)
三角函式公式：
基本求導公式
\((x^a)' = ax^{a - 1}, a為常數\)
\((a^x)' = a^x\ln a,(a>0, a\neq 1)\)
\((e^x)' = e^x\)
\((\log _ax)' = {\Large \frac{1}{x\ln a}},(a>0, a\neq 1）\)
\((\ln|x|)' = \Large \frac{1}{x}\)
\((\sin x)' = \cos x\)
\((\cos x)' = -\sin x\)
\((\arcsin x)'=\Large \frac{1}{\sqrt{1 - x^2}}\)
\((\arccos x)' = -\Large \frac{1}{\sqrt{1 - x^2}}\)
\((\tan x)' = \sec ^2x\)
\((\cot x)' = -\csc ^2x\)
\((\arctan x)' = \Large \frac{1}{1 + x^2}\)
\((arccot x)' = -\Large \frac{1}{1 + x^2}\)
\((\sec x)'= \sec x \tan x\)
\((\csc x)' = -\csc x \cot x\)
\((\ln(x + \sqrt{x^2 + 1}))' = \Large \frac{1}{\sqrt{x^2 + 1}}\)
\((\ln(x + \sqrt{x^2 - 1}))' = \Large \frac{1}{\sqrt{x^2 - 1}}\)