Lösung 1.2:4b
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}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr] &= (x)'\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\\[5pt] &= 1\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\,\textrm{.} \end{align} |
We divide up the differentiation of the second term in sections and use the chain rule,
| \displaystyle \begin{align}
(\sin\ln x + \cos\ln x)' &= (\sin\ln x)' + (\cos\ln x)'\\[5pt] &= \cos\ln x\cdot (\ln x)' - \sin\ln x\cdot (\ln x)'\\[5pt] &= \cos\ln x\cdot\frac{1}{x} - \sin\ln x\cdot\frac{1}{x}\,\textrm{.} \end{align} |
This means that
| \displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr] &= \sin \ln x + \cos \ln x + \cos \ln x - \sin \ln x\\[5pt] &= 2\cos \ln x\,\textrm{.} \end{align} |
The second derivative is
| \displaystyle \begin{align}
\frac{d}{dx}\,2\cos\ln x &= -2\sin\ln x\cdot (\ln x)'\\[5pt] &= -2\sin\ln x\cdot \frac{1}{x}\\[5pt] &= -\frac{2\sin\ln x}{x}\,\textrm{.} \end{align} |
