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Lösung 1.2:1f

Aus Online Mathematik Brückenkurs 2

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Version vom 10:03, 11. Mär. 2009

In this case, we have a slightly more complicated expression, but if we focus on the expression's outer form, we have essentially "something divided by sinx". As a first step, we therefore use the quotient rule,

sinxxlnx=(sinx)2(xlnx)sinx−xlnx(sinx). 

We can, in turn, differentiate the expression xlnx by using the product rule,

(xlnx)=(x)lnx+x(lnx)=1lnx+xx1=lnx+1.

All in all, we thus obtain

sinxxlnx=(sinx)2(lnx+1)sinx−xlnxcosx=sinxlnx+1−sin2xxlnxcosx.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:1f“