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Lösung 1.2:4b

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
To start with, we determine the first derivative and begin by using the product rule,
To start with, we determine the first derivative and begin by using the product rule,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr]
\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr]
&= (x)'\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\\[5pt]
&= (x)'\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\\[5pt]
Zeile 9: Zeile 9:
We divide up the differentiation of the second term in sections and use the chain rule,
We divide up the differentiation of the second term in sections and use the chain rule,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
(\sin\ln x + \cos\ln x)'
(\sin\ln x + \cos\ln x)'
&= (\sin\ln x)' + (\cos\ln x)'\\[5pt]
&= (\sin\ln x)' + (\cos\ln x)'\\[5pt]
Zeile 18: Zeile 18:
This means that
This means that
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr]
\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr]
&= \sin \ln x + \cos \ln x + \cos \ln x - \sin \ln x\\[5pt]
&= \sin \ln x + \cos \ln x + \cos \ln x - \sin \ln x\\[5pt]
Zeile 26: Zeile 26:
The second derivative is
The second derivative is
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{d}{dx}\,2\cos\ln x
\frac{d}{dx}\,2\cos\ln x
&= -2\sin\ln x\cdot (\ln x)'\\[5pt]
&= -2\sin\ln x\cdot (\ln x)'\\[5pt]

Version vom 12:55, 10. 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)=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“