Lösung 1.2:1b

Aus Online Mathematik Brückenkurs 2

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We use the product rule with
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We use the product rule with <math>x^2</math> and <math>\ln x</math> as factors,
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<math>x^{2}</math>
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and
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<math>\ln x</math>
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as factors:
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\bigl(x^{2}\ln x\bigr)'
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& \left( x^{2}\ln x \right)^{\prime }=\left( x^{2} \right)^{\prime }\ln x+x^{2}\left( \ln x \right)^{\prime } \\
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&= \bigl(x^2\bigr)'\,\ln x + x^2\,(\ln x)'\\[5pt]
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& \\
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&= 2x\cdot\ln x + x^2\cdot\frac{1}{x}\\[5pt]
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& =2x\centerdot \ln x+x^{2}\centerdot \frac{1}{x}=2x\ln x+x \\
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&= 2x\ln x + x\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}

Version vom 13:44, 14. Okt. 2008

We use the product rule with \displaystyle x^2 and \displaystyle \ln x as factors,

\displaystyle \begin{align}

\bigl(x^{2}\ln x\bigr)' &= \bigl(x^2\bigr)'\,\ln x + x^2\,(\ln x)'\\[5pt] &= 2x\cdot\ln x + x^2\cdot\frac{1}{x}\\[5pt] &= 2x\ln x + x\,\textrm{.} \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:1b“