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Lösung 1.2:4b

Aus Online Mathematik Brückenkurs 2

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To start with, we determine the first derivative and begin by using the product rule,

ddxx(sinlnx+coslnx)=(x)(sinlnx+coslnx)+x(sinlnx+coslnx)=1(sinlnx+coslnx)+x(sinlnx+coslnx).

We divide up the differentiation of the second term in sections and use the chain rule,

(sinlnx+coslnx)=(sinlnx)+(coslnx)=coslnx(lnx)−sinlnx(lnx)=coslnxx1−sinlnxx1.

This means that

ddxx(sinlnx+coslnx)=sinlnx+coslnx+coslnx−sinlnx=2coslnx.

The second derivative is

ddx2coslnx=−2sinlnx(lnx)=−2sinlnxx1=−x2sinlnx.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:4b“