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Lösung 1.2:3e

Aus Online Mathematik Brückenkurs 2

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Version vom 10:06, 11. Mär. 2009

At first sight, the expression looks like “e raised to something” and therefore we differentiate using the chain rule,

ddxesinx2=esinx2sinx2. 

Then, we differentiate “sine of something”,

esinx2sinx2=esinx2cosx2x2=esinx2cosx22x.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:3e“