补充说明：
抽象函数使用洛必达法则最多可用到

基本初等函数的导数公式

$$\begin{array}{r}\begin{array}{rl}& (1)(C{)}^{\prime }＝0\\ & (3)(\sin x{)}^{\prime }=\cos x\\ & (5)(\tan x{)}^{\prime }={\sec }^{2}x(\sec x=\frac{1}{\cos x})\\ & (7)(\sec x{)}^{\prime }=\sec x\tan x\\ & (9)(a^x{)}^{\prime }=a^x\ln a\\ & (11)({\log }_{a}x{)}^{\prime }=\frac{1}{x\ln a}\\ & (13)(\arcsin x{)}^{\prime }=\frac{1}{\sqrt{1-x^2}}\\ & (15)(\arctan x{)}^{\prime }=\frac{1}{\sqrt{1+x^2}}\\ & (17)[\ln (x+\sqrt{x^2+1}){]}^{\prime }=\frac{1}{\sqrt{x^2+1}}\\ & \text{注：(17)(18)是推导}\end{array}\begin{array}{rl}& (2)(x^{\mu }{)}^{\prime }=\mu x^{\mu -1}\\ & (4)(\cos x{)}^{\prime }=-\sin x\\ & (6)(\cot x{)}^{\prime }=-{\csc }^{2}x(\csc x=\frac{1}{\sin x})\\ & (8)(\csc x{)}^{\prime }=-\csc x\cot x\\ & (10)(e^x{)}^{\prime }=e^x\\ & (12)(\ln x{)}^{\prime }=\frac{1}{x},(\ln |x|{)}^{\prime }=\frac{1}{x}\\ & (14)(\arccos x{)}^{\prime }=-\frac{1}{\sqrt{1-x^2}}\\ & (16)(\text{arccot}x{)}^{\prime }=-\frac{1}{1+x^2}\\ & (18)[\ln (x+\sqrt{x^2-1}){]}^{\prime }=\frac{1}{\sqrt{x^2-1}}\end{array}\end{array}$$

函数的和、差、积、商的求导法则

设$u=u(x),v=v(x)$都可导，则
(1) $(u+v{)}^{\prime }=u^{\prime }\pm v^{\prime }$
(2) $(Cu{)}^{\prime }=C{u}^{\prime }(C是常数)$
(3) $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$
(4) $(uv{)}^{\prime }=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}(v\ne 0)$

复合函数的求导法则

设$u=\phi (x)$在$x$处可导，$y=f(u)$在对应点处可导，则复合函数$y=f[\phi (x)]$在$x$处的导数为
$$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}=f^{\prime }(u)\cdot {\phi }^{\prime }(x)$$

隐函数的求导法则

设$y=y(x)$是由方程$F(x,y)=0$所确定的可导函数，为求得$y^{\prime }$，可在方程$F(x,y)=0$两边对$x$求导，得到一个含有$y^{\prime }$的方程，从中解出$y^{\prime }$即可。
另有隐函数求导公式：
$$\frac{dy}{dx}=-\frac{F_x^{\prime }}{F_y^{\prime }}$$

反函数的求导法则

若$y=f(x)$在某区间内单调可导，且$f^{\prime }(x)\ne 0$，则其反函数$x=\phi (y)$在对应区间内也可导，且
$${\phi }^{\prime }(y)=\frac{1}{f^{\prime }(x)}或\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$$

参数方程的求导法则