Lösung 1.2:1d

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We have a quotient between <math>\sin x</math> and <math>x</math>, and therefore one way to differentiate the expression is to use the quotient rule,
We have a quotient between <math>\sin x</math> and <math>x</math>, and therefore one way to differentiate the expression is to use the quotient rule,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\Bigl(\frac{\sin x}{x}\Bigr)'
\Bigl(\frac{\sin x}{x}\Bigr)'
&= \frac{(\sin x)'\cdot x - \sin x\cdot (x)'}{x^2}\\[5pt]
&= \frac{(\sin x)'\cdot x - \sin x\cdot (x)'}{x^2}\\[5pt]
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<math>1/x</math>, and to use the product rule,
<math>1/x</math>, and to use the product rule,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\Bigl(\sin x\cdot\frac{1}{x}\Bigr)'
\Bigl(\sin x\cdot\frac{1}{x}\Bigr)'
&= (\sin x)'\cdot\frac{1}{x} + \sin x\cdot\Bigl(\frac{1}{x}\Bigr)'\\[5pt]
&= (\sin x)'\cdot\frac{1}{x} + \sin x\cdot\Bigl(\frac{1}{x}\Bigr)'\\[5pt]
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where we have used
where we have used
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{{Displayed math||<math>\Bigl(\frac{1}{x}\Bigr)' = \bigl(x^{-1}\bigr)' = (-1)x^{-1-1} = -1\cdot x^{-2} = -\frac{1}{x^2}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\Bigl(\frac{1}{x}\Bigr)' = \bigl(x^{-1}\bigr)' = (-1)x^{-1-1} = -1\cdot x^{-2} = -\frac{1}{x^2}\,\textrm{.}</math>}}

Version vom 12:52, 10. Mär. 2009

We have a quotient between \displaystyle \sin x and \displaystyle x, and therefore one way to differentiate the expression is to use the quotient rule,

\displaystyle \begin{align}

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

It is also possible to see the expression as a product of \displaystyle \sin x and \displaystyle 1/x, and to use the product rule,

\displaystyle \begin{align}

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

where we have used

\displaystyle \Bigl(\frac{1}{x}\Bigr)' = \bigl(x^{-1}\bigr)' = (-1)x^{-1-1} = -1\cdot x^{-2} = -\frac{1}{x^2}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:1d“