Inverse function theorem: Difference between revisions

Revision as of 06:54, 7 March 2006

Revision as of 07:18, 7 March 2006

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<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>

For more general [[logarithms]], we see that <math>\frac{d}{dx} \log_b(x) = \frac{1}{x \ln(b)} = \frac{\log_b(e)}{x}</math>.

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