Lösung 1.2:4b

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 1.2:4b moved to Solution 1.2:4b: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
To start with, we determine the first derivative and begin by using the product rule:
-
<center> [[Image:1_2_4b-1(2).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
-
{{NAVCONTENT_START}}
+
<math>\begin{align}
-
<center> [[Image:1_2_4b-2(2).gif]] </center>
+
& \frac{d}{dx}x\left( \sin \ln x+\cos \ln x \right) \\
-
{{NAVCONTENT_STOP}}
+
& =\left( x \right)^{\prime }\centerdot \left( \sin \ln x+\cos \ln x \right)+x\centerdot \left( \sin \ln x+\cos \ln x \right)^{\prime } \\
 +
& =1\centerdot \left( \sin \ln x+\cos \ln x \right)+x\centerdot \left( \sin \ln x+\cos \ln x \right)^{\prime } \\
 +
\end{align}</math>
 +
 
 +
 
 +
We divide up the differentiation of the second term in sections and use the chain rule:
 +
 
 +
 
 +
 
 +
<math>\begin{align}
 +
& \left( \sin \ln x+\cos \ln x \right)^{\prime }=\left( \sin \ln x \right)^{\prime }+\left( \cos \ln x \right)^{\prime } \\
 +
& =\cos \ln x\centerdot \left( \ln x \right)^{\prime }-\sin \ln x\centerdot \left( \ln x \right)^{\prime } \\
 +
& =\cos \ln x\centerdot \frac{1}{x}-\sin \ln x\centerdot \frac{1}{x} \\
 +
\end{align}</math>
 +
 
 +
 
 +
This means that
 +
 
 +
 
 +
<math>\begin{align}
 +
& \frac{d}{dx}x\left( \sin \ln x+\cos \ln x \right) \\
 +
& =\sin \ln x+\cos \ln x+\cos \ln x-\sin \ln x \\
 +
& =2\cos \ln x \\
 +
\end{align}</math>
 +
 
 +
 
 +
The second derivative is
 +
 
 +
 
 +
<math>\begin{align}
 +
& \frac{d}{dx}2\cos \ln x=-2\sin \ln x\centerdot \left( \ln x \right)^{\prime } \\
 +
& =-2\sin \ln x\centerdot \frac{1}{x}=-\frac{2\sin \ln x}{x} \\
 +
\end{align}</math>

Version vom 14:15, 12. Okt. 2008

To start with, we determine the first derivative and begin by using the product rule:


\displaystyle \begin{align} & \frac{d}{dx}x\left( \sin \ln x+\cos \ln x \right) \\ & =\left( x \right)^{\prime }\centerdot \left( \sin \ln x+\cos \ln x \right)+x\centerdot \left( \sin \ln x+\cos \ln x \right)^{\prime } \\ & =1\centerdot \left( \sin \ln x+\cos \ln x \right)+x\centerdot \left( \sin \ln x+\cos \ln x \right)^{\prime } \\ \end{align}


We divide up the differentiation of the second term in sections and use the chain rule:


\displaystyle \begin{align} & \left( \sin \ln x+\cos \ln x \right)^{\prime }=\left( \sin \ln x \right)^{\prime }+\left( \cos \ln x \right)^{\prime } \\ & =\cos \ln x\centerdot \left( \ln x \right)^{\prime }-\sin \ln x\centerdot \left( \ln x \right)^{\prime } \\ & =\cos \ln x\centerdot \frac{1}{x}-\sin \ln x\centerdot \frac{1}{x} \\ \end{align}


This means that


\displaystyle \begin{align} & \frac{d}{dx}x\left( \sin \ln x+\cos \ln x \right) \\ & =\sin \ln x+\cos \ln x+\cos \ln x-\sin \ln x \\ & =2\cos \ln x \\ \end{align}


The second derivative is


\displaystyle \begin{align} & \frac{d}{dx}2\cos \ln x=-2\sin \ln x\centerdot \left( \ln x \right)^{\prime } \\ & =-2\sin \ln x\centerdot \frac{1}{x}=-\frac{2\sin \ln x}{x} \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:4b“