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 1.2:3a

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 1.2:3a moved to Solution 1.2:3a: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
There is a "ln of something", so a first step in the differentiation is to take the derivative of the logarithm:
-
<center> [[Image:1_2_3a-1(2).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
-
{{NAVCONTENT_START}}
+
<math>\frac{d}{dx}\ln \left( \left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right. \right)=\frac{1}{\left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right.}\centerdot \left( \left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right. \right)^{\prime }</math>
-
<center> [[Image:1_2_3a-2(2).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
 +
We can carry out the differentiation of
 +
<math>\sqrt{x}+\sqrt{x+1}</math>
 +
on the right-hand side term by term to obtain
 +
 
 +
 
 +
<math>\quad =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \left( \sqrt{x} \right)^{\prime }+\left( \sqrt{x+1} \right)^{\prime } \right]</math>
 +
 
 +
 
 +
and it remains then only to differentiate
 +
<math>\sqrt{x}</math>,which we do directly, and
 +
<math>\sqrt{x+1}</math>
 +
which a simple inner derivative)
 +
 
 +
 
 +
<math></math>
 +
 
 +
<math>\begin{align}
 +
& =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\centerdot \left( x+1 \right)^{\prime } \right] \\
 +
& =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\centerdot 1 \right] \\
 +
\end{align}</math>
 +
 
 +
 
 +
 
 +
If we rewrite the expression inside the square brackets using a common denominator, we get
 +
 
 +
 
 +
<math>=\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{\sqrt{x+1}+\sqrt{x}}{2\sqrt{x}\sqrt{x+1}} \right]</math>,
 +
 
 +
and we can then eliminate the factor
 +
<math>\sqrt{x+1}+\sqrt{x}</math>
 +
from the numerator and denominator to get
 +
 
 +
 
 +
<math>=\frac{1}{2\sqrt{x}\sqrt{x+1}}</math>

Version vom 14:17, 11. Okt. 2008

There is a "ln of something", so a first step in the differentiation is to take the derivative of the logarithm:


ddxlnx+x+1=1x+x+1x+x+1 


We can carry out the differentiation of x+x+1  on the right-hand side term by term to obtain


=1x+x+1x+x+1 


and it remains then only to differentiate x ,which we do directly, and x+1  which a simple inner derivative)


=1x+x+112x+12x+1x+1=1x+x+112x+12x+11


If we rewrite the expression inside the square brackets using a common denominator, we get


=1x+x+12xx+1x+1+x ,

and we can then eliminate the factor x+1+x  from the numerator and denominator to get


=12xx+1

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:3a“