Lösung 1.2:2e - Online Mathematik Brückenkurs 2

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                      1. Differentialrechnung
                       
                        1.1 Einführung 
                        1.2 Ableitungsregeln 
                        1.3 Max/Min probleme
                      
                    
                      2. Integralrechnung
                       
                        2.1 Einführung 
                        2.2 Substitution 
                        2.3 Partielle Integration
                      
                    
                      3. Komplexe Zahlen
                       
                        3.1 Rechnungen 
                        3.2 Polarform 
                        3.3 Potenzen und Wurzeln 
                        3.4 Komplexe Polynome
                      
                    
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			Lösung 1.2:2e
			
			  Aus Online Mathematik Brückenkurs 2
			  (Unterschied zwischen Versionen)
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				Version vom 08:56, 15. Okt. 2008 (bearbeiten)
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← Zum vorherigen Versionsunterschied
				Version vom 12:53, 10. Mär. 2009 (bearbeiten) (rückgängig)
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Zum nächsten Versionsunterschied →
			
		Zeile 3:
Zeile 3:
 To begin with, we have a product of <math>x</math> and <math>(2x+1)^4</math> so the product rule gives that  To begin with, we have a product of <math>x</math> and <math>(2x+1)^4</math> so the product rule gives that 
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]  \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
 &= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]  &= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]
Zeile 11:
Zeile 11:
 We can differentiate the expression <math>(2x+1)^4</math> by viewing it as "something raised to the 4",  We can differentiate the expression <math>(2x+1)^4</math> by viewing it as "something raised to the 4",
  
- {{Displayed math||<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}} + {{Abgesetzte Formel||<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}}
  
 The chain rule then gives  The chain rule then gives 
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]  \frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]
 &= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt]  &= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt] 
Zeile 23:
Zeile 23:
 We carry out the last differentiation directly,  and obtain  We carry out the last differentiation directly,  and obtain
  
- {{Displayed math||<math>(2x+1)' = 2\,\textrm{.}</math>}} + {{Abgesetzte Formel||<math>(2x+1)' = 2\,\textrm{.}</math>}}
  
 If we go through the whole calculation from the beginning, it is  If we go through the whole calculation from the beginning, it is 
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]  \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
 &= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt]  &= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt] 
Zeile 37:
Zeile 37:
 Both terms contain a common factor <math>(2x+1)^3</math> which we can take out to get an answer in factorized form,  Both terms contain a common factor <math>(2x+1)^3</math> which we can take out to get an answer in factorized form,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]  \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
 &= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt]  &= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt] 
 &= (2x+1)^3(10x+1)\,\textrm{.}  &= (2x+1)^3(10x+1)\,\textrm{.} 
 \end{align}</math>}}  \end{align}</math>}}
Version vom 12:53, 10. Mär. 2009
One way to differentiate the expression could be to expand \displaystyle (2x+1)^4 multiply by \displaystyle x and differentiate term by term, but it is simpler instead to use the structure of the expression and differentiate step by step using the differentiation rules.
To begin with, we have a product of \displaystyle x and \displaystyle (2x+1)^4 so the product rule gives that 





\displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]
&= 1\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\,\textrm{.} 
\end{align}



We can differentiate the expression \displaystyle (2x+1)^4 by viewing it as "something raised to the 4",





\displaystyle \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}


The chain rule then gives 





\displaystyle \begin{align}
\frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]
&= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt] 
\frac{d}{dx}\,\bigl[(2x+1)^4\bigr] &= 4\cdot (2x+1)^3\cdot (2x+1)'\,\textrm{.} 
\end{align}



We carry out the last differentiation directly,  and obtain





\displaystyle (2x+1)' = 2\,\textrm{.}


If we go through the whole calculation from the beginning, it is 





\displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt] 
&= 1\cdot (2x+1)^4 + x\cdot 4(2x+1)^3\cdot (2x+1)'\\[5pt] 
&= (2x+1)^4 + x\cdot 4(2x+1)^3\cdot 2\\[5pt] 
&= (2x+1)^4 + 8x(2x+1)^3\,\textrm{.} 
\end{align}



Both terms contain a common factor \displaystyle (2x+1)^3 which we can take out to get an answer in factorized form,





\displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
&= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt] 
&= (2x+1)^3(10x+1)\,\textrm{.} 
\end{align}





Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:2e“
                       Willkommen zum Kurs
                       
                        Information über den Kurs 
                        Information über Prüfungen
                      
                    
                      1. Differentialrechnung
                       
                        1.1 Einführung 
                        1.2 Ableitungsregeln 
                        1.3 Max/Min probleme
                      
                    
                      2. Integralrechnung
                       
                        2.1 Einführung 
                        2.2 Substitution 
                        2.3 Partielle Integration
                      
                    
                      3. Komplexe Zahlen
                       
                        3.1 Rechnungen 
                        3.2 Polarform 
                        3.3 Potenzen und Wurzeln 
                        3.4 Komplexe Polynome
                      
                    
                      Kurs als PDF
                                            
                    
                    
		       Suche
		       
		         

                           
                            
			 
		       
		    
		    
		  
		  
		  
		    
		      
		        
			
			Lösung 1.2:2e
			
			  Aus Online Mathematik Brückenkurs 2
			  (Unterschied zwischen Versionen)
			  			                            Wechseln zu: Navigation, Suche
                          
		          
			
			
			
			
			
			
				Version vom 08:56, 15. Okt. 2008 (bearbeiten)
Tek  (Diskussion | Beiträge)
K 
← Zum vorherigen Versionsunterschied
				Version vom 12:53, 10. Mär. 2009 (bearbeiten) (rückgängig)
Tekbot  (Diskussion | Beiträge) 
K  (Robot: Automated text replacement  (-{{Displayed math +{{Abgesetzte Formel))
Zum nächsten Versionsunterschied →
			
		Zeile 3:
Zeile 3:
 To begin with, we have a product of <math>x</math> and <math>(2x+1)^4</math> so the product rule gives that  To begin with, we have a product of <math>x</math> and <math>(2x+1)^4</math> so the product rule gives that 
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]  \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
 &= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]  &= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]
Zeile 11:
Zeile 11:
 We can differentiate the expression <math>(2x+1)^4</math> by viewing it as "something raised to the 4",  We can differentiate the expression <math>(2x+1)^4</math> by viewing it as "something raised to the 4",
  
- {{Displayed math||<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}} + {{Abgesetzte Formel||<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}}
  
 The chain rule then gives  The chain rule then gives 
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]  \frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]
 &= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt]  &= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt] 
Zeile 23:
Zeile 23:
 We carry out the last differentiation directly,  and obtain  We carry out the last differentiation directly,  and obtain
  
- {{Displayed math||<math>(2x+1)' = 2\,\textrm{.}</math>}} + {{Abgesetzte Formel||<math>(2x+1)' = 2\,\textrm{.}</math>}}
  
 If we go through the whole calculation from the beginning, it is  If we go through the whole calculation from the beginning, it is 
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]  \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
 &= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt]  &= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt] 
Zeile 37:
Zeile 37:
 Both terms contain a common factor <math>(2x+1)^3</math> which we can take out to get an answer in factorized form,  Both terms contain a common factor <math>(2x+1)^3</math> which we can take out to get an answer in factorized form,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]  \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
 &= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt]  &= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt] 
 &= (2x+1)^3(10x+1)\,\textrm{.}  &= (2x+1)^3(10x+1)\,\textrm{.} 
 \end{align}</math>}}  \end{align}</math>}}
Version vom 12:53, 10. Mär. 2009
One way to differentiate the expression could be to expand \displaystyle (2x+1)^4 multiply by \displaystyle x and differentiate term by term, but it is simpler instead to use the structure of the expression and differentiate step by step using the differentiation rules.
To begin with, we have a product of \displaystyle x and \displaystyle (2x+1)^4 so the product rule gives that 





\displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]
&= 1\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\,\textrm{.} 
\end{align}



We can differentiate the expression \displaystyle (2x+1)^4 by viewing it as "something raised to the 4",





\displaystyle \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}


The chain rule then gives 





\displaystyle \begin{align}
\frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]
&= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt] 
\frac{d}{dx}\,\bigl[(2x+1)^4\bigr] &= 4\cdot (2x+1)^3\cdot (2x+1)'\,\textrm{.} 
\end{align}



We carry out the last differentiation directly,  and obtain





\displaystyle (2x+1)' = 2\,\textrm{.}


If we go through the whole calculation from the beginning, it is 





\displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt] 
&= 1\cdot (2x+1)^4 + x\cdot 4(2x+1)^3\cdot (2x+1)'\\[5pt] 
&= (2x+1)^4 + x\cdot 4(2x+1)^3\cdot 2\\[5pt] 
&= (2x+1)^4 + 8x(2x+1)^3\,\textrm{.} 
\end{align}



Both terms contain a common factor \displaystyle (2x+1)^3 which we can take out to get an answer in factorized form,





\displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
&= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt] 
&= (2x+1)^3(10x+1)\,\textrm{.} 
\end{align}





Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:2e“
-                      Willkommen zum Kurs
                       
                        Information über den Kurs 
                        Information über Prüfungen
                      
                    
                      1. Differentialrechnung
                       
                        1.1 Einführung 
                        1.2 Ableitungsregeln 
                        1.3 Max/Min probleme
                      
                    
                      2. Integralrechnung
                       
                        2.1 Einführung 
                        2.2 Substitution 
                        2.3 Partielle Integration
                      
                    
                      3. Komplexe Zahlen
                       
                        3.1 Rechnungen 
                        3.2 Polarform 
                        3.3 Potenzen und Wurzeln 
                        3.4 Komplexe Polynome
                      
                    
                      Kurs als PDF
                                            
                    
                    
		       Suche
+			Lösung 1.2:2e
			
			  Aus Online Mathematik Brückenkurs 2
			  (Unterschied zwischen Versionen)
			  			                            Wechseln zu: Navigation, Suche
                          
		          
			
			
			
			
			
			
				Version vom 08:56, 15. Okt. 2008 (bearbeiten)
Tek  (Diskussion | Beiträge)
K 
← Zum vorherigen Versionsunterschied
				Version vom 12:53, 10. Mär. 2009 (bearbeiten) (rückgängig)
Tekbot  (Diskussion | Beiträge) 
K  (Robot: Automated text replacement  (-{{Displayed math +{{Abgesetzte Formel))
Zum nächsten Versionsunterschied →
			
		Zeile 3:
Zeile 3:
 To begin with, we have a product of <math>x</math> and <math>(2x+1)^4</math> so the product rule gives that  To begin with, we have a product of <math>x</math> and <math>(2x+1)^4</math> so the product rule gives that 
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]  \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
 &= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]  &= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]
Zeile 11:
Zeile 11:
 We can differentiate the expression <math>(2x+1)^4</math> by viewing it as "something raised to the 4",  We can differentiate the expression <math>(2x+1)^4</math> by viewing it as "something raised to the 4",
  
- {{Displayed math||<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}} + {{Abgesetzte Formel||<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}}
  
 The chain rule then gives  The chain rule then gives 
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]  \frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]
 &= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt]  &= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt] 
Zeile 23:
Zeile 23:
 We carry out the last differentiation directly,  and obtain  We carry out the last differentiation directly,  and obtain
  
- {{Displayed math||<math>(2x+1)' = 2\,\textrm{.}</math>}} + {{Abgesetzte Formel||<math>(2x+1)' = 2\,\textrm{.}</math>}}
  
 If we go through the whole calculation from the beginning, it is  If we go through the whole calculation from the beginning, it is 
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]  \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
 &= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt]  &= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt] 
Zeile 37:
Zeile 37:
 Both terms contain a common factor <math>(2x+1)^3</math> which we can take out to get an answer in factorized form,  Both terms contain a common factor <math>(2x+1)^3</math> which we can take out to get an answer in factorized form,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]  \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
 &= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt]  &= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt] 
 &= (2x+1)^3(10x+1)\,\textrm{.}  &= (2x+1)^3(10x+1)\,\textrm{.} 
 \end{align}</math>}}  \end{align}</math>}}
Version vom 12:53, 10. Mär. 2009
One way to differentiate the expression could be to expand \displaystyle (2x+1)^4 multiply by \displaystyle x and differentiate term by term, but it is simpler instead to use the structure of the expression and differentiate step by step using the differentiation rules.
To begin with, we have a product of \displaystyle x and \displaystyle (2x+1)^4 so the product rule gives that 





\displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]
&= 1\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\,\textrm{.} 
\end{align}



We can differentiate the expression \displaystyle (2x+1)^4 by viewing it as "something raised to the 4",





\displaystyle \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}


The chain rule then gives 





\displaystyle \begin{align}
\frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]
&= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt] 
\frac{d}{dx}\,\bigl[(2x+1)^4\bigr] &= 4\cdot (2x+1)^3\cdot (2x+1)'\,\textrm{.} 
\end{align}



We carry out the last differentiation directly,  and obtain





\displaystyle (2x+1)' = 2\,\textrm{.}


If we go through the whole calculation from the beginning, it is 





\displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt] 
&= 1\cdot (2x+1)^4 + x\cdot 4(2x+1)^3\cdot (2x+1)'\\[5pt] 
&= (2x+1)^4 + x\cdot 4(2x+1)^3\cdot 2\\[5pt] 
&= (2x+1)^4 + 8x(2x+1)^3\,\textrm{.} 
\end{align}



Both terms contain a common factor \displaystyle (2x+1)^3 which we can take out to get an answer in factorized form,





\displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
&= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt] 
&= (2x+1)^3(10x+1)\,\textrm{.} 
\end{align}





Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:2e“
@@ Zeile 3: / Zeile 3: @@
 To begin with, we have a product of <math>x</math> and <math>(2x+1)^4</math> so the product rule gives that
-{{Displayed math||<math>\begin{align}
+{{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
 &= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]
@@ Zeile 11: / Zeile 11: @@
 We can differentiate the expression <math>(2x+1)^4</math> by viewing it as "something raised to the 4",
-{{Displayed math||<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}}
+{{Abgesetzte Formel||<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}}
 The chain rule then gives
-{{Displayed math||<math>\begin{align}
+{{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]
 &= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt]
@@ Zeile 23: / Zeile 23: @@
 We carry out the last differentiation directly,  and obtain
-{{Displayed math||<math>(2x+1)' = 2\,\textrm{.}</math>}}
+{{Abgesetzte Formel||<math>(2x+1)' = 2\,\textrm{.}</math>}}
 If we go through the whole calculation from the beginning, it is
-{{Displayed math||<math>\begin{align}
+{{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
 &= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt]
@@ Zeile 37: / Zeile 37: @@
 Both terms contain a common factor <math>(2x+1)^3</math> which we can take out to get an answer in factorized form,
-{{Displayed math||<math>\begin{align}
+{{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
 &= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt]
 &= (2x+1)^3(10x+1)\,\textrm{.}
 \end{align}</math>}}
 \displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]
&= 1\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\,\textrm{.} 
\end{align}
 \displaystyle \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}
 \displaystyle \begin{align}
\frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]
&= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt] 
\frac{d}{dx}\,\bigl[(2x+1)^4\bigr] &= 4\cdot (2x+1)^3\cdot (2x+1)'\,\textrm{.} 
\end{align}
 \displaystyle (2x+1)' = 2\,\textrm{.}
 \displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt] 
&= 1\cdot (2x+1)^4 + x\cdot 4(2x+1)^3\cdot (2x+1)'\\[5pt] 
&= (2x+1)^4 + x\cdot 4(2x+1)^3\cdot 2\\[5pt] 
&= (2x+1)^4 + 8x(2x+1)^3\,\textrm{.} 
\end{align}
 \displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
&= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt] 
&= (2x+1)^3(10x+1)\,\textrm{.} 
\end{align}