Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
No jsMath TeX fonts found -- using image fonts instead.
These may be slow and might not print well.
Use the jsMath control panel to get additional information.
jsMath Control PanelHide this Message

jsMath

Lösung 1.2:1d

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:
We have a quotient between <math>\sin x</math> and <math>x</math>, and therefore one way to differentiate the expression is to use the quotient rule,
We have a quotient between <math>\sin x</math> and <math>x</math>, and therefore one way to differentiate the expression is to use the quotient rule,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\Bigl(\frac{\sin x}{x}\Bigr)'
\Bigl(\frac{\sin x}{x}\Bigr)'
&= \frac{(\sin x)'\cdot x - \sin x\cdot (x)'}{x^2}\\[5pt]
&= \frac{(\sin x)'\cdot x - \sin x\cdot (x)'}{x^2}\\[5pt]
Zeile 11: Zeile 11:
<math>1/x</math>, and to use the product rule,
<math>1/x</math>, and to use the product rule,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\Bigl(\sin x\cdot\frac{1}{x}\Bigr)'
\Bigl(\sin x\cdot\frac{1}{x}\Bigr)'
&= (\sin x)'\cdot\frac{1}{x} + \sin x\cdot\Bigl(\frac{1}{x}\Bigr)'\\[5pt]
&= (\sin x)'\cdot\frac{1}{x} + \sin x\cdot\Bigl(\frac{1}{x}\Bigr)'\\[5pt]
Zeile 20: Zeile 20:
where we have used
where we have used
-
{{Displayed math||<math>\Bigl(\frac{1}{x}\Bigr)' = \bigl(x^{-1}\bigr)' = (-1)x^{-1-1} = -1\cdot x^{-2} = -\frac{1}{x^2}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\Bigl(\frac{1}{x}\Bigr)' = \bigl(x^{-1}\bigr)' = (-1)x^{-1-1} = -1\cdot x^{-2} = -\frac{1}{x^2}\,\textrm{.}</math>}}

Version vom 12:52, 10. Mär. 2009

We have a quotient between sinx and x, and therefore one way to differentiate the expression is to use the quotient rule,

xsinx=x2(sinx)x−sinx(x)=x2cosxx−sinx1=xcosx−x2sinx.

It is also possible to see the expression as a product of sinx and 1x, and to use the product rule,

sinxx1=(sinx)x1+sinxx1=cosxx1+sinx−1x2=xcosx−x2sinx

where we have used

x1=x−1=(−1)x−1−1=−1x−2=−1x2. 
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:1d“