Lösung 1.2:1f

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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In this case, we have a slightly more complicated expression, but if we focus on the expression's outer form, we have essentially "something divided by <math>\sin x</math>". As a first step, we therefore use the quotient rule,
In this case, we have a slightly more complicated expression, but if we focus on the expression's outer form, we have essentially "something divided by <math>\sin x</math>". As a first step, we therefore use the quotient rule,
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{{Displayed math||<math>\Bigl(\frac{x\ln x}{\sin x}\Bigr)' = \frac{(x\ln x)'\cdot \sin x - x\ln x\cdot (\sin x)'}{(\sin x)^2}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\Bigl(\frac{x\ln x}{\sin x}\Bigr)' = \frac{(x\ln x)'\cdot \sin x - x\ln x\cdot (\sin x)'}{(\sin x)^2}\,\textrm{.}</math>}}
We can, in turn, differentiate the expression <math>x\ln x</math> by using the product rule,
We can, in turn, differentiate the expression <math>x\ln x</math> by using the product rule,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
(x\ln x)'
(x\ln x)'
&= (x)'\ln x + x\,(\ln x)'\\[5pt]
&= (x)'\ln x + x\,(\ln x)'\\[5pt]
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All in all, we thus obtain
All in all, we thus obtain
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\Bigl(\frac{x\ln x}{\sin x}\Bigr)'
\Bigl(\frac{x\ln x}{\sin x}\Bigr)'
&= \frac{(\ln x+1)\cdot\sin x - x\ln x\cdot \cos x}{(\sin x)^2}\\[5pt]
&= \frac{(\ln x+1)\cdot\sin x - x\ln x\cdot \cos x}{(\sin x)^2}\\[5pt]
&= \frac{\ln x+1}{\sin x}-\frac{x\ln x\cos x}{\sin^2\!x}\,\textrm{.}
&= \frac{\ln x+1}{\sin x}-\frac{x\ln x\cos x}{\sin^2\!x}\,\textrm{.}
\end{align}</math>}}
\end{align}</math>}}

Version vom 12:52, 10. Mär. 2009

In this case, we have a slightly more complicated expression, but if we focus on the expression's outer form, we have essentially "something divided by \displaystyle \sin x". As a first step, we therefore use the quotient rule,

\displaystyle \Bigl(\frac{x\ln x}{\sin x}\Bigr)' = \frac{(x\ln x)'\cdot \sin x - x\ln x\cdot (\sin x)'}{(\sin x)^2}\,\textrm{.}

We can, in turn, differentiate the expression \displaystyle x\ln x by using the product rule,

\displaystyle \begin{align}

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

All in all, we thus obtain

\displaystyle \begin{align}

\Bigl(\frac{x\ln x}{\sin x}\Bigr)' &= \frac{(\ln x+1)\cdot\sin x - x\ln x\cdot \cos x}{(\sin x)^2}\\[5pt] &= \frac{\ln x+1}{\sin x}-\frac{x\ln x\cos x}{\sin^2\!x}\,\textrm{.} \end{align}

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