Lösung 1.2:4b

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To start with, we determine the first derivative and begin by using the product rule:
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To start with, we determine the first derivative and begin by using the product rule,
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{{Displayed math||<math>\begin{align}
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\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr]
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&= (x)'\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\\[5pt]
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&= 1\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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We divide up the differentiation of the second term in sections and use the chain rule,
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& \frac{d}{dx}x\left( \sin \ln x+\cos \ln x \right) \\
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& =\left( x \right)^{\prime }\centerdot \left( \sin \ln x+\cos \ln x \right)+x\centerdot \left( \sin \ln x+\cos \ln x \right)^{\prime } \\
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& =1\centerdot \left( \sin \ln x+\cos \ln x \right)+x\centerdot \left( \sin \ln x+\cos \ln x \right)^{\prime } \\
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\end{align}</math>
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We divide up the differentiation of the second term in sections and use the chain rule:
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<math>\begin{align}
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& \left( \sin \ln x+\cos \ln x \right)^{\prime }=\left( \sin \ln x \right)^{\prime }+\left( \cos \ln x \right)^{\prime } \\
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& =\cos \ln x\centerdot \left( \ln x \right)^{\prime }-\sin \ln x\centerdot \left( \ln x \right)^{\prime } \\
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& =\cos \ln x\centerdot \frac{1}{x}-\sin \ln x\centerdot \frac{1}{x} \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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(\sin\ln x + \cos\ln x)'
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&= (\sin\ln x)' + (\cos\ln x)'\\[5pt]
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&= \cos\ln x\cdot (\ln x)' - \sin\ln x\cdot (\ln x)'\\[5pt]
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&= \cos\ln x\cdot\frac{1}{x} - \sin\ln x\cdot\frac{1}{x}\,\textrm{.}
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\end{align}</math>}}
This means that
This means that
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr]
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& \frac{d}{dx}x\left( \sin \ln x+\cos \ln x \right) \\
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&= \sin \ln x + \cos \ln x + \cos \ln x - \sin \ln x\\[5pt]
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& =\sin \ln x+\cos \ln x+\cos \ln x-\sin \ln x \\
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&= 2\cos \ln x\,\textrm{.}
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& =2\cos \ln x \\
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\end{align}</math>}}
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\end{align}</math>
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The second derivative is
The second derivative is
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\frac{d}{dx}\,2\cos\ln x
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& \frac{d}{dx}2\cos \ln x=-2\sin \ln x\centerdot \left( \ln x \right)^{\prime } \\
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&= -2\sin\ln x\cdot (\ln x)'\\[5pt]
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& =-2\sin \ln x\centerdot \frac{1}{x}=-\frac{2\sin \ln x}{x} \\
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&= -2\sin\ln x\cdot \frac{1}{x}\\[5pt]
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\end{align}</math>
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&= -\frac{2\sin\ln x}{x}\,\textrm{.}
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\end{align}</math>}}

Version vom 13:58, 15. Okt. 2008

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

\displaystyle \begin{align}

\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] &= 1\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\,\textrm{.} \end{align}

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

\displaystyle \begin{align}

(\sin\ln x + \cos\ln x)' &= (\sin\ln x)' + (\cos\ln x)'\\[5pt] &= \cos\ln x\cdot (\ln x)' - \sin\ln x\cdot (\ln x)'\\[5pt] &= \cos\ln x\cdot\frac{1}{x} - \sin\ln x\cdot\frac{1}{x}\,\textrm{.} \end{align}

This means that

\displaystyle \begin{align}

\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] &= 2\cos \ln x\,\textrm{.} \end{align}

The second derivative is

\displaystyle \begin{align}

\frac{d}{dx}\,2\cos\ln x &= -2\sin\ln x\cdot (\ln x)'\\[5pt] &= -2\sin\ln x\cdot \frac{1}{x}\\[5pt] &= -\frac{2\sin\ln x}{x}\,\textrm{.} \end{align}

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