Line 31: Line 31:
 
|  <math>\frac{d^n}{dx^n}\left( u v \right)= u \frac{d^n v }{dx^n} + \left( \begin{array}{cc}n \\ 1 \end{array}\right) \frac{du }{dx}\frac{d^{n-1}v }{dx^{n-1}} + \left( \begin{array}{cc}n \\ 2 \end{array}\right) \frac{d^2u}{dx^2}\frac{d^{n-2}v }{dx^{n-2}}+ \ldots + v \frac{d^n u }{dx^n}</math>
 
|  <math>\frac{d^n}{dx^n}\left( u v \right)= u \frac{d^n v }{dx^n} + \left( \begin{array}{cc}n \\ 1 \end{array}\right) \frac{du }{dx}\frac{d^{n-1}v }{dx^{n-1}} + \left( \begin{array}{cc}n \\ 2 \end{array}\right) \frac{d^2u}{dx^2}\frac{d^{n-2}v }{dx^{n-2}}+ \ldots + v \frac{d^n u }{dx^n}</math>
 
|}
 
|}
 +
{|
 +
|-
 +
! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="3" | More rules for derivatives
 +
|-
 +
|<math> \frac{d}{dx} ( \frac{u}{v} ) = \frac{v ( \frac{du}{dx} ) - u ( \frac{dv}{dx} )}{v^2}</math>
 +
|-
 +
|<math> \frac{d}{dx} ( u^n ) = n u^{n-1} \frac{du}{dx}</math>
 +
|-
 +
|<math> \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}</math>
 +
|-
 +
|<math> \frac{du}{dx} = \frac{1}{\frac{dx}{du}}</math>
 +
|-
 +
|<math> \frac{dy}{dx} = \frac{ \frac{dy}{du}}{\frac{dx}{du}}</math>
 +
|-
 +
 
{|
 
{|
 
|-
 
|-

Revision as of 11:57, 14 November 2010

Table of Derivatives
General Rules
Derivative of a constant $ \frac{d}{dx}\left( c \right) = 0, \ \text{ for any constant }c $
$ \frac{d}{dx}\left( c x \right) = c, \ \text{ for any constant }c $
Linearity $ \frac{d}{dx}\left( c_1 u_1+c_2 u_2 \right) = c_1 \frac{d}{dx}\left( u_1 \right)+c_2 \frac{d}{dx}\left( u_2 \right), \ \text{ for any constants }c_1, c_2 $
Please continue write a rule here
Leibnitz Rule for Successive Derivatives of a Product
first order $ \frac{d}{dx}\left( u v \right)= u \frac{dv }{dx} + v \frac{du }{dx} $
second order $ \frac{d^2}{dx^2}\left( u v \right)= u \frac{d^2v }{dx^2} + 2\frac{du }{dx}\frac{dv }{dx}+ v \frac{d^2u }{dx^2} $
third order $ \frac{d^3}{dx^3}\left( u v \right)= u \frac{d^3v }{dx^3} + 3 \frac{du }{dx}\frac{d^2v }{dx^2}+ 3 \frac{du^2 }{dx^2}\frac{d v }{dx}+ v \frac{d^3u }{dx^3} $ credit
n-th order $ \frac{d^n}{dx^n}\left( u v \right)= u \frac{d^n v }{dx^n} + \left( \begin{array}{cc}n \\ 1 \end{array}\right) \frac{du }{dx}\frac{d^{n-1}v }{dx^{n-1}} + \left( \begin{array}{cc}n \\ 2 \end{array}\right) \frac{d^2u}{dx^2}\frac{d^{n-2}v }{dx^{n-2}}+ \ldots + v \frac{d^n u }{dx^n} $
More rules for derivatives
$ \frac{d}{dx} ( \frac{u}{v} ) = \frac{v ( \frac{du}{dx} ) - u ( \frac{dv}{dx} )}{v^2} $
$ \frac{d}{dx} ( u^n ) = n u^{n-1} \frac{du}{dx} $
$ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} $
$ \frac{du}{dx} = \frac{1}{\frac{dx}{du}} $
$ \frac{dy}{dx} = \frac{ \frac{dy}{du}}{\frac{dx}{du}} $
Derivatives of trigonometric functions
$ \frac {d}{dx} \sin u = \cos u \frac{du}{dx} $
$ \frac {d}{dx} \cos u = - \sin u \frac{du}{dx} $
$ \frac {d}{dx} \tan u = \frac{1}{\cos^2 u} \frac{du}{dx} $
$ \frac {d}{dx} \cot u = - \frac{1}{\sin^2 u} \frac{du}{dx} $
$ \frac {d}{dx} \frac{1}{\cos u} = \frac{\tan u}{\cos u} \frac{du}{dx} $
$ \frac {d}{dx} \frac{1}{\sin u} = - \frac{\cot u}{\sin u} \frac{du}{dx} $
$ \frac {d}{dx} \arcsin u = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx} \qquad ( - \frac{\pi}{2} < \arcsin u < \frac{\pi}{2} ) $
$ \frac {d}{dx} \arccos u = - \frac{1}{\sqrt{1-u^2}} \frac{du}{dx} \qquad ( 0 < \arccos u < \pi ) $
$ \frac {d}{dx} \arctan u = \frac{1}{1+u^2} \frac{du}{dx} \qquad ( - \frac{\pi}{2} < \arctan u < \frac{\pi}{2} ) $
$ \frac {d}{dx} \arccot u = - \frac{1}{1+u^2} \frac{du}{dx} \qquad ( 0 < \arccot u < \pi ) $
add function here derivative here
Derivatives of exponential and logarithm functions
$ \frac{d}{dx} \log_a u = \frac{log_a e}{u} \frac{du}{dx} \qquad a \neq 0,1 $
$ \frac{d}{dx} \ln u = \frac{d}{dx} log_e u = \frac{1}{u} \frac{du}{dx} $
$ \frac{d}{dx} a^u = a^u \ln a \frac{du}{dx} $
$ \frac{d}{dx} e^u = e^u \frac{du}{dx} $
$ \frac{d}{dx} u^v = \frac{d}{dx} e^{v ln u} = e^{v ln u} \frac {d}{dx} [ v ln u ] = v u^{v-1} \frac{du}{dx} + u^v ln u \frac{dv}{dx} $
eu $ e^u \frac{du}{dx} $
add function here derivative here
Derivatives of hyperbolic functions
$ \frac{d}{dx} \operatorname{sh}\ u = \operatorname{ch}\ u \frac{du}{dx} $
$ \frac{d}{dx} \operatorname{ch}\ u = \operatorname{sh}\ u \frac{du}{dx} $
$ \frac{d}{dx} \operatorname{th}\ u = \frac{1}{\operatorname{ch^2}\ u} \frac{du}{dx} $
$ \frac{d}{dx} \operatorname{coth}\ u = - \frac{1}{\operatorname{sh^2}\ u} \frac{du}{dx} $
$ \frac{d}{dx} \frac{1}{\operatorname{ch}\ u} = - \frac{\operatorname{th}\ u}{\operatorname{ch}\ u} \frac{du}{dx} $
$ \frac{d}{dx} \frac{1}{\operatorname{sh}\ u} = - \frac{\operatorname{coth}\ u}{\operatorname{sh}\ u} \frac{du}{dx} $
$ \frac{d}{dx} \arg \operatorname{sh}\ u = \frac{1}{\sqrt{u^2+1}} \frac{du}{dx} $
$ \frac{d}{dx} \arg \operatorname{ch}\ u = \frac{\pm 1}{\sqrt{u^2-1}} \frac{du}{dx} \qquad ( + si \ arg \ \operatorname{ch}\ u > 0, \ u > 1 \ ; \ - si \ arg \ \operatorname{ch}\ u < 0, \ u > 1 ) $
$ \frac{d}{dx} \arg \operatorname{th}\ u = \frac{1}{1-u^2} \frac{du}{dx} \qquad ( \ -1 < u < 1 \ ) $
$ \frac{d}{dx} \arg \operatorname{coth}\ u = \frac{1}{1-u^2} \frac{du}{dx} \qquad ( \ u > 1 \ or \ u < -1 \ ) $
$ \text{sh } u $ $ \text{ch } u \frac{du}{dx} $
add function here derivative here


Back to Collective Table of Formulas

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang