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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)xsinx(x)=x2cosxxsinx1=xcosxx2sinx.

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+sinx1x2=xcosxx2sinx

where we have used

x1=x1=(1)x11=1x2=1x2.