Revision as of 06:21, 5 April 2012 by Lrprice (Talk | contribs)

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!

                                         HKNlogo.jpg


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 \ ) $


Back to Collective Table of Formulas

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett