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Lösung 1.2:1d

Aus Online Mathematik Brückenkurs 2

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We have a quotient between sinx and x, and therefore one way to differentiate the expression is to use the quotient rule,

xsinx=x2(sinx)x−sinx(x)=x2cosxx−sinx1=xcosx−x2sinx.

It is also possible to see the expression as a product of sinx and 1x, and to use the product rule,

sinxx1=(sinx)x1+sinxx1=cosxx1+sinx−1x2=xcosx−x2sinx

where we have used

x1=x−1=(−1)x−1−1=−1x−2=−1x2. 
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:1d“