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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:


ddxxsinlnx+coslnx=xsinlnx+coslnx+xsinlnx+coslnx=1sinlnx+coslnx+xsinlnx+coslnx


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


sinlnx+coslnx=sinlnx+coslnx=coslnxlnx−sinlnxlnx=coslnxx1−sinlnxx1


This means that


ddxxsinlnx+coslnx=sinlnx+coslnx+coslnx−sinlnx=2coslnx


The second derivative is


ddx2coslnx=−2sinlnxlnx=−2sinlnxx1=−x2sinlnx

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:4b“