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Lösung 1.2:4b

Aus Online Mathematik Brückenkurs 2

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Version vom 10:07, 11. Mär. 2009

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)=coslnxx1sinlnxx1.

This means that

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

The second derivative is

ddx2coslnx=2sinlnx(lnx)=2sinlnxx1=x2sinlnx.