Aus Online Mathematik Brückenkurs 2

To start with, we determine the first derivative and begin by using the product rule:

\displaystyle \begin{align} & \frac{d}{dx}x\left( \sin \ln x+\cos \ln x \right) \\ & =\left( x \right)^{\prime }\centerdot \left( \sin \ln x+\cos \ln x \right)+x\centerdot \left( \sin \ln x+\cos \ln x \right)^{\prime } \\ & =1\centerdot \left( \sin \ln x+\cos \ln x \right)+x\centerdot \left( \sin \ln x+\cos \ln x \right)^{\prime } \\ \end{align}

We divide up the differentiation of the second term in sections and use the chain rule:

\displaystyle \begin{align} & \left( \sin \ln x+\cos \ln x \right)^{\prime }=\left( \sin \ln x \right)^{\prime }+\left( \cos \ln x \right)^{\prime } \\ & =\cos \ln x\centerdot \left( \ln x \right)^{\prime }-\sin \ln x\centerdot \left( \ln x \right)^{\prime } \\ & =\cos \ln x\centerdot \frac{1}{x}-\sin \ln x\centerdot \frac{1}{x} \\ \end{align}

This means that

\displaystyle \begin{align} & \frac{d}{dx}x\left( \sin \ln x+\cos \ln x \right) \\ & =\sin \ln x+\cos \ln x+\cos \ln x-\sin \ln x \\ & =2\cos \ln x \\ \end{align}

The second derivative is

\displaystyle \begin{align} & \frac{d}{dx}2\cos \ln x=-2\sin \ln x\centerdot \left( \ln x \right)^{\prime } \\ & =-2\sin \ln x\centerdot \frac{1}{x}=-\frac{2\sin \ln x}{x} \\ \end{align}