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| <math> \frac {d}{dx} \tan u = \frac{1}{\cos^2 u} \frac{du}{dx} </math>
 
| <math> \frac {d}{dx} \tan u = \frac{1}{\cos^2 u} \frac{du}{dx} </math>
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| <math> \frac {d}{dx} \cot u = - \frac{1}{\sin^2 u} \frac{du}{dx} </math>
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| <math> \frac {d}{dx} \frac{1}{\cos u} = \frac{\tan u}{\cos u} \frac{du}{dx} </math>
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| <math> \frac {d}{dx} \frac{1}{\sin u} = - \frac{\cot u}{\sin u} \frac{du}{dx} </math>
 
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| <math> \frac {d}{dx} \arcsin u = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx} \qquad ( - \frac{\pi}{2} < \arcsin u < \frac{\pi}{2} )</math>
 
| <math> \frac {d}{dx} \arcsin u = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx} \qquad ( - \frac{\pi}{2} < \arcsin u < \frac{\pi}{2} )</math>
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| <math> \frac {d}{dx} \arccos u = - \frac{1}{\sqrt{1-u^2}} \frac{du}{dx} \qquad ( 0 < \arccos u < \pi ) </math>
 
| <math> \frac {d}{dx} \arccos u = - \frac{1}{\sqrt{1-u^2}} \frac{du}{dx} \qquad ( 0 < \arccos u < \pi ) </math>
 
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| <math> \frac {d}{dx} \arctan u = \frac{1}{1+u^2} \frac{du}{dx} \qquad ( - \frac{\pi}{2} < \arctan u < \frac{\pi}{2} )</math>
 
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| <span class="texhtml">sin'' u''</span>
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| align="left" | <math>\cos u \frac{du}{dx}</math>
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| <math> \frac {d}{dx} \arccot u = - \frac{1}{1+u^2} \frac{du}{dx} \qquad ( 0 < \arccot u < \pi ) </math>
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| add function here  
 
| add function here  
 
|  derivative here
 
|  derivative here

Revision as of 21:21, 11 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} $
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
exponential eu $ e^u \frac{du}{dx} $
add function here derivative here
Derivatives of hyperbolic functions
hyperbolic sine $ \text{sh } u $ $ \text{ch } u \frac{du}{dx} $
add function here derivative here


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Ruth Enoch, PhD Mathematics