Lösung 1.2:4b
Aus Online Mathematik Brückenkurs 2
K |
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
||
| Zeile 1: | Zeile 1: | ||
To start with, we determine the first derivative and begin by using the product rule, | To start with, we determine the first derivative and begin by using the product rule, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr] | \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] | &= (x)'\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\\[5pt] | ||
| Zeile 9: | Zeile 9: | ||
We divide up the differentiation of the second term in sections and use the chain rule, | We divide up the differentiation of the second term in sections and use the chain rule, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
(\sin\ln x + \cos\ln x)' | (\sin\ln x + \cos\ln x)' | ||
&= (\sin\ln x)' + (\cos\ln x)'\\[5pt] | &= (\sin\ln x)' + (\cos\ln x)'\\[5pt] | ||
| Zeile 18: | Zeile 18: | ||
This means that | This means that | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr] | \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] | &= \sin \ln x + \cos \ln x + \cos \ln x - \sin \ln x\\[5pt] | ||
| Zeile 26: | Zeile 26: | ||
The second derivative is | The second derivative is | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{d}{dx}\,2\cos\ln x | \frac{d}{dx}\,2\cos\ln x | ||
&= -2\sin\ln x\cdot (\ln x)'\\[5pt] | &= -2\sin\ln x\cdot (\ln x)'\\[5pt] | ||
Version vom 12:55, 10. Mär. 2009
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} |
