Aus Online Mathematik Brückenkurs 2

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}