Inverse function theorem: Difference between revisions

Revision as of 05:56, 5 May 2006

Revision as of 02:16, 9 June 2006

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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 = \tanarctan x.</math> Then

:<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>