Lösung 1.2:1f
Aus Online Mathematik Brückenkurs 2
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Version vom 10:03, 11. 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} |
