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Lösung 1.2:1d

Aus Online Mathematik Brückenkurs 2

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K (Lösning 1.2:1d moved to Solution 1.2:1d: Robot: moved page)
K
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We have a quotient between
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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,
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<math>\sin x</math>
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and
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<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}
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\Bigl(\frac{\sin x}{x}\Bigr)'
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&= \frac{(\sin x)'\cdot x - \sin x\cdot (x)'}{x^2}\\[5pt]
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&= \frac{\cos x\cdot x - \sin x\cdot 1}{x^2}\\[5pt]
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&= \frac{\cos x}{x} - \frac{\sin x}{x^2}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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It is also possible to see the expression as a product of <math>\sin x</math> and
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& \left( \frac{\sin x}{x} \right)^{\prime }=\frac{\left( \sin x \right)^{\prime }\centerdot x-\sin x\centerdot \left( x \right)^{\prime }}{x^{2}} \\
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<math>1/x</math>, and to use the product rule,
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& \\
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& =\frac{\cos x\centerdot x-\sin x\centerdot 1}{x^{2}}=\frac{\cos x}{x}-\frac{\sin x}{x^{2}} \\
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\end{align}</math>
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It is also possible to see the expression as a product of
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<math>\sin x</math>
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and
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<math>\frac{1}{x}</math>, and to use the product rule,
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<math>\begin{align}
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& \left( \sin x\centerdot \frac{1}{x} \right)^{\prime }=\left( \sin x \right)^{\prime }\centerdot \frac{1}{x}+\sin x\centerdot \left( \frac{1}{x} \right)^{\prime } \\
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& \\
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& =\cos x\centerdot \frac{1}{x}+\sin x\centerdot \left( -\frac{1}{x^{2}} \right)=\frac{\cos x}{x}-\frac{\sin x}{x^{2}} \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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\Bigl(\sin x\cdot\frac{1}{x}\Bigr)'
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&= (\sin x)'\cdot\frac{1}{x} + \sin x\cdot\Bigl(\frac{1}{x}\Bigr)'\\[5pt]
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&= \cos x\cdot\frac{1}{x} + \sin x\cdot\Bigl(-\frac{1}{x^2}\Bigr)\\[5pt]
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&= \frac{\cos x}{x} - \frac{\sin x}{x^2}\,,
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\end{align}</math>}}
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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<math>\left( \frac{1}{x} \right)^{\prime }=\left( x^{-1} \right)^{\prime }=\left( -1 \right)x^{-1-1}=-1\centerdot x^{-2}=-\frac{1}{x^{2}}</math>
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Version vom 14:10, 14. Okt. 2008

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

xsinx=x2(sinx)x−sinx(x)=x2cosxx−sinx1=xcosx−x2sinx.

It is also possible to see the expression as a product of sinx and 1x, and to use the product rule,

sinxx1=(sinx)x1+sinxx1=cosxx1+sinx−1x2=xcosx−x2sinx

where we have used

x1=x−1=(−1)x−1−1=−1x−2=−1x2. 
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:1d“