Inverse function theorem: Difference between revisions

Revision as of 14:04, 1 March 2006

Revision as of 06:54, 7 March 2006

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If the derivative of ''f'' is an isomorphism at all points ''p'' in ''M'' then the map ''f'' is a [[local diffeomorphism]].

This can be expressed more clearly as <math>(f^{-1})'(f(a)) = {{1} \over {(f^{-1})'(f(a))}}</math>. Where ' indicates the derivative of the function.

===Examples===

Several functions exist for which differentiating the inverse is much easier than differentiating the function itself. Using the inverse function theorem, a derivative of a function's inverse indicates the derivative of the original function. Perhaps the most well-known example is the method used to compute the derivative of the [[natural logarithm]], whose inverse is the [[exponential function]]. Let <math>u = \ln x</math> and restrict the ___domain to <math>x > 0</math>.

<math>\frac{d}{dx}\ln x = {{1} \over {\frac{d}{du}e^u}} = {{1} \over {e^u}} = {{1} \over {e^{\ln x}}} = {{1} \over {x}}</math>

A similar approach can be used to differentiate an inverse [[trigonometric function]]. Let <math>u = \tan x</math>.

<math>\frac{d}{dx}\arctan x = {{1} \over {\frac{d}{du}\tan u}} = \cos^2{u} = \cos^2{\arctan x} = \left({{1} \over {\sqrt{1+x^2}}}\right)^2 = {{1} \over {1+x^2}}</math>

[[Category:Multivariate calculus]]