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Lösung 1.2:1f

Aus Online Mathematik Brückenkurs 2

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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
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<math>\sin x</math>
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". As a first step, we therefore use the quotient rule,
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<math>\left( \frac{x\ln x}{\sin x} \right)^{\prime }=\frac{\left( x\ln x \right)^{\prime }\centerdot \sin x-x\ln x\centerdot \left( \sin x \right)^{\prime }}{\left( \sin x \right)^{2}}</math>
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We can, in turn, differentiate the expression
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<math>x\ln x</math>
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by using the product rule:
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<math>\begin{align}
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& \left( x\ln x \right)^{\prime }=\left( x \right)^{\prime }\ln x+x\left( \ln x \right)^{\prime } \\
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& \\
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& =1\centerdot \ln x+x\centerdot \frac{1}{x}=\ln x+1 \\
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\end{align}</math>
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All in all, we thus obtain
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<math>\begin{align}
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& \left( \frac{x\ln x}{\sin x} \right)^{\prime }=\frac{\left( \ln x+1 \right)\centerdot \sin x-x\ln x\centerdot \cos x}{\left( \sin x \right)^{2}} \\
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& \\
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& =\frac{\ln x+1}{\sin x}-\frac{x\ln x\cos x}{\sin ^{2}x} \\
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\end{align}</math>

Version vom 15:53, 12. Sep. 2008

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 sinx ". As a first step, we therefore use the quotient rule,


sinxxlnx=sinx2xlnxsinx−xlnxsinx 


We can, in turn, differentiate the expression xlnx by using the product rule:


xlnx=xlnx+xlnx=1lnx+xx1=lnx+1


All in all, we thus obtain


sinxxlnx=sinx2lnx+1sinx−xlnxcosx=sinxlnx+1−sin2xxlnxcosx

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