Not just the formula. The condition that earns the point.

Continuity at a point$$\lim_{x\to a}f(x)=f(a)$$三件事：函数值存在、极限存在、二者相等。

Intermediate Value Theorem$$f\text{ continuous on }[a,b],\;N\text{ strictly between }f(a),f(b)\Rightarrow\exists c\in(a,b):f(c)=N$$

L'Hospital's Rule$$\lim\frac{f}{g}=\lim\frac{f'}{g'}\quad\text{only after verifying }0/0\text{ or }\infty/\infty$$先写未定式，再分别求导；不是商法则。

Definition$$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}h$$

Product / quotient / chain$$ (uv)'=u'v+uv',\qquad\left(\frac uv\right)'=\frac{u'v-uv'}{v^2},\qquad(f\circ g)'=f'(g)g'$$

Inverse derivative$$\left(f^{-1}\right)'(a)=\frac1{f'(f^{-1}(a))}$$

Linearization$$L(x)=f(a)+f'(a)(x-a)$$写清中心 $a$；误差方向由凹凸性判断。

Trigonometric derivatives$$\frac{d}{dx}\sin x=\cos x,\quad\frac{d}{dx}\cos x=-\sin x,\quad\frac{d}{dx}\tan x=\sec^2x$$$$\frac{d}{dx}\sec x=\sec x\tan x,\quad\frac{d}{dx}\csc x=-\csc x\cot x,\quad\frac{d}{dx}\cot x=-\csc^2x$$以 co- 开头的三个（cos、csc、cot）导数带负号。

Inverse trig derivatives$$\frac{d}{dx}\arcsin x=\frac1{\sqrt{1-x^2}},\quad\frac{d}{dx}\arccos x=-\frac1{\sqrt{1-x^2}},\quad\frac{d}{dx}\arctan x=\frac1{1+x^2}$$

Exponential / logarithmic derivatives$$\frac{d}{dx}e^x=e^x,\quad\frac{d}{dx}a^x=a^x\ln a,\quad\frac{d}{dx}\ln|x|=\frac1x,\quad\frac{d}{dx}\log_ax=\frac1{x\ln a}$$

Basic antiderivatives$$\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\;(n\ne-1),\quad\int\frac{dx}{x}=\ln|x|+C,\quad\int e^x\,dx=e^x+C,\quad\int a^x\,dx=\frac{a^x}{\ln a}+C$$$$\int\sin x\,dx=-\cos x+C,\quad\int\cos x\,dx=\sin x+C,\quad\int\sec^2x\,dx=\tan x+C,\quad\int\sec x\tan x\,dx=\sec x+C$$$$\int\tan x\,dx=-\ln|\cos x|+C,\quad\int\frac{dx}{\sqrt{1-x^2}}=\arcsin x+C,\quad\int\frac{dx}{a^2+x^2}=\frac1a\arctan\frac xa+C$$不定积分必须写 $+C$；定积分才不用。

Mean Value Theorem$$f\text{ continuous on }[a,b],\;f\text{ differentiable on }(a,b)\Rightarrow f'(c)=\frac{f(b)-f(a)}{b-a}$$

First / second derivative tests$$f':+\to-\Rightarrow\text{local max},\quad f':-\to+\Rightarrow\text{local min}$$$$f'(c)=0,\;f''(c)>0\Rightarrow\text{local min}$$

Inflection-point check$$f''\text{ must change sign}$$仅有 $f''(c)=0$ 不够。

FTC$$\frac d{dx}\int_a^{g(x)}f(t)\,dt=f(g(x))g'(x),\qquad\int_a^bf'(x)\,dx=f(b)-f(a)$$

Substitution / integration by parts$$\int f(g(x))g'(x)\,dx=\int f(u)\,du,\qquad\int u\,dv=uv-\int v\,du$$

Partial fractions (non-repeated linear factors)$$\frac{P(x)}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}\;(\deg P<2),\qquad\int\frac{dx}{(x-a)(x-b)}=\frac1{a-b}\ln\left|\frac{x-a}{x-b}\right|+C$$分子次数不低于分母时先做多项式除法；BC 只考不重复的一次因式。

Net change / accumulation$$f(b)=f(a)+\int_a^bf'(x)\,dx,\qquad\text{displacement}=\int_a^bv(t)\,dt,\qquad\text{distance}=\int_a^b|v(t)|\,dt$$"变化量 = 变化率的积分"：先写初值，再加累积；路程要对速度取绝对值。

Improper integral$$\int_a^\infty f(x)\,dx=\lim_{b\to\infty}\int_a^bf(x)\,dx$$必须写极限并判断有限或发散。

Euler's method$$y_{n+1}=y_n+f(x_n,y_n)\Delta x$$

Exponential / logistic$$\frac{dy}{dt}=ky\Rightarrow y=Ce^{kt},\qquad\frac{dP}{dt}=kP\left(1-\frac PL\right)$$

Logistic facts (0 < P(0) < L)$$\lim_{t\to\infty}P(t)=L,\qquad\frac{dP}{dt}\text{ is greatest when }P=\frac L2\;\left(\text{where }\frac{d^2P}{dt^2}=0\right)$$承载量 $L$ 直接从 $dP/dt=0$ 读出；无需解出 $P(t)$ 就能回答这两个高频问题。

Separation$$\frac{1}{g(y)}\,dy=f(x)\,dx$$积分后再用初值；若除以含 $y$ 的因子，检查是否丢失平衡解。

Average value / area$$f_{avg}=\frac1{b-a}\int_a^bf(x)\,dx,\qquad A=\int_a^b(\text{top}-\text{bottom})\,dx$$

Disk / washer / cross sections$$V=\pi\int_a^b(R^2-r^2)\,dx,\qquad V=\int_a^bA(x)\,dx$$

Arc length$$L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx$$

Parametric derivatives$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt},\qquad\frac{d^2y}{dx^2}=\frac{d}{dt}(dy/dx)\bigg/\frac{dx}{dt}$$要求 $dx/dt\ne0$。

Vector motion$$\vec v(t)=\vec{r}'(t),\quad\vec a(t)=\vec{v}'(t),\quad\text{speed}=\|\vec v(t)\|=\sqrt{x'(t)^2+y'(t)^2},\quad\text{distance}=\int_a^b\|\vec v(t)\|\,dt$$

Parametric arc length$$L=\int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$$与向量运动的路程是同一个积分；被积函数就是速率。

Polar slope / area / length$$\frac{dy}{dx}=\frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta},\quad A=\frac12\int_\alpha^\beta r^2\,d\theta$$$$L=\int_\alpha^\beta\sqrt{r^2+(dr/d\theta)^2}\,d\theta$$

Geometric / p-series$$\sum_{n=0}^\infty ar^n=\frac a{1-r}\;( |r|<1),\qquad\sum\frac1{n^p}\text{ converges iff }p>1$$

Alternating / ratio tests$$b_n>0,\;b_{n+1}\le b_n,\;\lim_{n\to\infty}b_n=0\Rightarrow\sum(-1)^nb_n\text{ converges},\qquad|S-S_n|\le b_{n+1}$$$$L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1\Rightarrow\text{absolute convergence};\;L>1\Rightarrow\text{diverges};\;L=1\text{ inconclusive}$$交错级数三个条件缺一不可；误差界 $b_{n+1}$ 是下一项的绝对值。

nth-term / integral / comparison tests$$\lim_{n\to\infty}a_n\ne0\Rightarrow\sum a_n\text{ diverges}\quad(\text{converse is false})$$$$f\text{ positive, continuous, decreasing on }[k,\infty),\;a_n=f(n)\Rightarrow\sum_{n=k}^\infty a_n\text{ and }\int_k^\infty f(x)\,dx\text{ both converge or both diverge}$$$$0\le a_n\le b_n:\;\sum b_n\text{ converges}\Rightarrow\sum a_n\text{ converges};\quad a_n,b_n>0,\;\lim_{n\to\infty}\frac{a_n}{b_n}=c\in(0,\infty)\Rightarrow\text{same behavior}$$$\sum|a_n|$ 收敛则 $\sum a_n$ 收敛（绝对收敛）；写检验名称并核对其假设才能拿到理由分。

Taylor series & Lagrange error$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n,\qquad|R_n(x)|\le\frac{M|x-a|^{n+1}}{(n+1)!},\;M\ge\max\left|f^{(n+1)}(z)\right|\text{ for }z\text{ between }a\text{ and }x$$必须写明 $M$ 是 $(n+1)$ 阶导数在 $a$ 与 $x$ 之间的最大值界；交错级数则改用 $b_{n+1}$。

Exponential and trig (all real x)$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots,\quad(-\infty,\infty)$$$$\sin x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots,\quad(-\infty,\infty)$$$$\cos x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots,\quad(-\infty,\infty)$$

Geometric, log and arctan (finite radius)$$\frac1{1-x}=\sum_{n=0}^\infty x^n=1+x+x^2+\cdots,\quad(-1,1)$$$$\ln(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}{n}=x-\frac{x^2}2+\frac{x^3}3-\cdots,\quad(-1,1]$$$$\arctan x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}=x-\frac{x^3}3+\frac{x^5}5-\cdots,\quad[-1,1]$$代换、逐项求导/积分不改变收敛半径，但端点必须重新检验。