Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Lösung 1.2:1f

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
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 <math>\sin x</math>". As a first step, we therefore use the quotient rule,
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 <math>\sin x</math>". As a first step, we therefore use the quotient rule,
-
{{Displayed math||<math>\Bigl(\frac{x\ln x}{\sin x}\Bigr)' = \frac{(x\ln x)'\cdot \sin x - x\ln x\cdot (\sin x)'}{(\sin x)^2}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\Bigl(\frac{x\ln x}{\sin x}\Bigr)' = \frac{(x\ln x)'\cdot \sin x - x\ln x\cdot (\sin x)'}{(\sin x)^2}\,\textrm{.}</math>}}
We can, in turn, differentiate the expression <math>x\ln x</math> by using the product rule,
We can, in turn, differentiate the expression <math>x\ln x</math> by using the product rule,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
(x\ln x)'
(x\ln x)'
&= (x)'\ln x + x\,(\ln x)'\\[5pt]
&= (x)'\ln x + x\,(\ln x)'\\[5pt]
Zeile 14: Zeile 14:
All in all, we thus obtain
All in all, we thus obtain
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\Bigl(\frac{x\ln x}{\sin x}\Bigr)'
\Bigl(\frac{x\ln x}{\sin x}\Bigr)'
&= \frac{(\ln x+1)\cdot\sin x - x\ln x\cdot \cos x}{(\sin x)^2}\\[5pt]
&= \frac{(\ln x+1)\cdot\sin x - x\ln x\cdot \cos x}{(\sin x)^2}\\[5pt]
&= \frac{\ln x+1}{\sin x}-\frac{x\ln x\cos x}{\sin^2\!x}\,\textrm{.}
&= \frac{\ln x+1}{\sin x}-\frac{x\ln x\cos x}{\sin^2\!x}\,\textrm{.}
\end{align}</math>}}
\end{align}</math>}}

Version vom 12:52, 10. Mär. 2009

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 sinx". As a first step, we therefore use the quotient rule,

sinxxlnx=(sinx)2(xlnx)sinx−xlnx(sinx). 

We can, in turn, differentiate the expression xlnx by using the product rule,

(xlnx)=(x)lnx+x(lnx)=1lnx+xx1=lnx+1.

All in all, we thus obtain

sinxxlnx=(sinx)2(lnx+1)sinx−xlnxcosx=sinxlnx+1−sin2xxlnxcosx.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:1f“