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Revision as of 06:57, 24 April 2012

This Collective table of formulas is proudly sponsored
by the Nice Guys of Eta Kappa Nu.

Visit us at the HKN Lounge in EE24 for hot coffee and fresh bagels only $1 each!

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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 $
Quotient rule $ \frac{d}{dx} ( \frac{u}{v} ) = \frac{v ( \frac{du}{dx} ) - u ( \frac{dv}{dx} )}{v^2} $
Exponent rule $ \frac{d}{dx} ( u^n ) = n u^{n-1} \frac{du}{dx} $
Chain rule $ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} $
$ \frac{du}{dx} = \frac{1}{\frac{dx}{du}} $
$ \frac{dy}{dx} = \frac{dy}{du}/\frac{dx}{du} $
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 ) $
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} $
Derivatives of hyperbolic functions
$ \frac{d}{dx} \sinh u = \cosh u \frac{du}{dx} $
$ \frac{d}{dx} \cosh u = \sinh u \frac{du}{dx} $
$ \frac{d}{dx} \tanh u = \frac{1}{\cosh^2 u} \frac{du}{dx} $
$ \frac{d}{dx} \coth u = - \frac{1}{\sinh^2 u} \frac{du}{dx} $
$ \frac{d}{dx} \frac{1}{\cosh u} = - \frac{\tanh u}{\cosh u} \frac{du}{dx} $
$ \frac{d}{dx} \frac{1}{\sinh u} = - \frac{\coth u}{\sinh u} \frac{du}{dx} $
$ \frac{d}{dx}\ \operatorname{arsinh}\ u = \frac{1}{\sqrt{u^2+1}} \frac{du}{dx} $
$ \frac{d}{dx}\ \operatorname{arcosh}\ u = \frac{1}{\sqrt{u^2-1}} \frac{du}{dx} $
$ \frac{d}{dx}\ \operatorname{artanh}\ u = \frac{1}{1-u^2} \frac{du}{dx} \qquad ( \ -1 < u < 1 \ ) $
$ \frac{d}{dx}\ \operatorname{arcoth}\ u = \frac{1}{1-u^2} \frac{du}{dx} \qquad ( \ u > 1 \ or \ u < -1 \ ) $


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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

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