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Lösung 1.2:1d

Aus Online Mathematik Brückenkurs 2

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We have a quotient between
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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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<math>\begin{align}
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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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& \\
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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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where we have used
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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>

Version vom 15:33, 12. Sep. 2008

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


xsinx=x2sinxx−sinxx=x2cosxx−sinx1=xcosx−x2sinx

It is also possible to see the expression as a product of

sinx and x1, and to use the product rule,


sinxx1=sinxx1+sinxx1=cosxx1+sinx−1x2=xcosx−x2sinx


where we have used


x1=x−1=−1x−1−1=−1x−2=−1x2 

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