一元函数微分学的计算

2022.09.25

基本求导公式

(一些太简单的就不写了) $(ln|x|{)}^{{}^{\prime }}=\frac{1}{x},\text{视}\text{绝}\text{对}\text{值}\text{而}\text{不}\text{见}$$(ln|x|)^{'}=\frac{1}{x},视绝对值而不见$ $(\tan x{)}^{{}^{\prime }}={\sec }^{2}x$$(\tan x)^{'}=\sec^2x$ $(\cot x{)}^{{}^{\prime }}=-{\csc }^{2}x$$(\cot x)^{'}=-\csc^2x$ $(\sec x{)}^{{}^{\prime }}=\sec x\tan x$$(\sec x)^{'}=\sec x\tan x$ $(\csc x{)}^{{}^{\prime }}=-\csc x\cot x$$(\csc x)^{'}=-\csc x\cot x$ $(\ln |\cos x|{)}^{{}^{\prime }}=-\tan x$$(\ln|\cos x|)^{'}=-\tan x$ $(\ln |\sin x|{)}^{{}^{\prime }}=\cot x$$(\ln|\sin x|)^{'}=\cot x$ $(\ln |\sec x+\tan x|{)}^{{}^{\prime }}=\sec x$$(\ln|\sec x+\tan x|)^{'}=\sec x$ $(\ln |\csc x-\cot x|{)}^{{}^{\prime }}=\csc x$$(\ln|\csc x-\cot x|)^{'}=\csc x$ $(\arctan x{)}^{{}^{\prime }}=\frac{1}{1+x^2}$$(\arctan x)^{'}=\frac{1}{1+x^2}$ $(\arctan \frac{x-1}{x+1}{)}^{{}^{\prime }}=\frac{1}{1+x^2}$$(\arctan {\frac{x-1}{x+1}})^{'}=\frac{1}{1+x^2}$ $(arccotx{)}^{{}^{\prime }}=-\frac{1}{1+x^2}$$(arccot\ x)^{'}=-\frac{1}{1+x^2}$ $(\arcsin x{)}^{{}^{\prime }}=\frac{1}{\sqrt{1-x^2}}$$(\arcsin x)^{'}=\frac{1}{\sqrt{1-x^2}}$ $(\arccos x{)}^{{}^{\prime }}=-\frac{1}{\sqrt{1-x^2}}$$(\arccos x)^{'}=-\frac{1}{\sqrt{1-x^2}}$ $[\ln (x+\sqrt{x^2+a^2}){]}^{{}^{\prime }}=\frac{1}{\sqrt{a^2+x^2}}$$[\ln(x+\sqrt{x^2+a^2})]^{'}=\frac{1}{\sqrt{a^2+x^2}}$ $[\ln (x+\sqrt{x^2-a^2}){]}^{{}^{\prime }}=\frac{1}{\sqrt{-a^2+x^2}}$$[\ln(x+\sqrt{x^2-a^2})]^{'}=\frac{1}{\sqrt{-a^2+x^2}}$

高阶导数

$(\frac{1}{1-x}{)}^{(n)}=\frac{n!}{(1-x{)}^{n+1}}$$(\frac{1}{1-x})^{(n)}=\frac{n!}{(1-x)^{n+1}}$ $(\frac{1}{1+x}{)}^{(n)}=\frac{(-1{)}^{n}n!}{(1-x{)}^{n+1}}$$(\frac{1}{1+x})^{(n)}=\frac{(-1)^{n}n!}{(1-x)^{n+1}}$ $(\sin kx{)}^{(n)}=k^n\sin (kx+\frac{n}{2}\pi )$$(\sin kx)^{(n)} = k^n \sin(kx + \frac{n}{2}\pi)$ $(\cos kx{)}^{(n)}=k^n\cos (kx+\frac{n}{2}\pi )$$(\cos kx)^{(n)} = k^n \cos(kx + \frac{n}{2}\pi)$