Lösung 1.2:1f
Aus Online Mathematik Brückenkurs 2
K (Lösning 1.2:1f moved to Solution 1.2:1f: Robot: moved page) |
Version vom 14:23, 16. 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 \displaystyle \sin x ". As a first step, we therefore use the quotient rule,
\displaystyle \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}}
We can, in turn, differentiate the expression
\displaystyle x\ln x
by using the product rule:
\displaystyle \begin{align}
& \left( x\ln x \right)^{\prime }=\left( x \right)^{\prime }\ln x+x\left( \ln x \right)^{\prime } \\
& \\
& =1\centerdot \ln x+x\centerdot \frac{1}{x}=\ln x+1 \\
\end{align}
All in all, we thus obtain
\displaystyle \begin{align}
& \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}} \\
& \\
& =\frac{\ln x+1}{\sin x}-\frac{x\ln x\cos x}{\sin ^{2}x} \\
\end{align}
