Lösung 1.2:4a - Online Mathematik Brückenkurs 2

                       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 1.2:4a
			
			  Aus Online Mathematik Brückenkurs 2
			  (Unterschied zwischen Versionen)
			  			                            Wechseln zu: Navigation, Suche
                          
		          
			
			
			
			
			
			
				Version vom 13:47, 15. Okt. 2008 (bearbeiten)
Tek  (Diskussion | Beiträge)
K 
← Zum vorherigen Versionsunterschied
				Version vom 12:54, 10. Mär. 2009 (bearbeiten) (rückgängig)
Tekbot  (Diskussion | Beiträge) 
K  (Robot: Automated text replacement  (-{{Displayed math +{{Abgesetzte Formel))
Zum nächsten Versionsunterschied →
			
		Zeile 1:
Zeile 1:
 We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives  We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}  \frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}
 &= {}\rlap{\frac{(x)'\sqrt{1-x^2}-x\bigl(\sqrt{1-x^2}\bigr)'}{\bigl(\sqrt{1-x^2}\bigr)^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]  &= {}\rlap{\frac{(x)'\sqrt{1-x^2}-x\bigl(\sqrt{1-x^2}\bigr)'}{\bigl(\sqrt{1-x^2}\bigr)^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]
Zeile 9:
Zeile 9:
 We determine the derivative <math>\bigl(\sqrt{1-x^2}\bigr)'</math> by using the chain rule  We determine the derivative <math>\bigl(\sqrt{1-x^2}\bigr)'</math> by using the chain rule
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}  \phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
 &= \frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}\\[5pt]  &= \frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}\\[5pt]
Zeile 17:
Zeile 17:
 We simplify the result as far as possible, so as to make the second differentiation easier,  We simplify the result as far as possible, so as to make the second differentiation easier,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}  \phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
 &= {}\rlap{\frac{\sqrt{1-x^2} + \dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]  &= {}\rlap{\frac{\sqrt{1-x^2} + \dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]
Zeile 27:
Zeile 27:
 The second derivative is  The second derivative is
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d^2}{dx^2}\,\frac{x}{\sqrt{1-x^2}}  \frac{d^2}{dx^2}\,\frac{x}{\sqrt{1-x^2}}
 &= \frac{d}{dx}\,\frac{1}{(1-x^2)^{3/2}}\\[5pt]  &= \frac{d}{dx}\,\frac{1}{(1-x^2)^{3/2}}\\[5pt]
Version vom 12:54, 10. Mär. 2009
We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives





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



We determine the derivative \displaystyle \bigl(\sqrt{1-x^2}\bigr)' by using the chain rule





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



We simplify the result as far as possible, so as to make the second differentiation easier,





\displaystyle \begin{align}
\phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
&= {}\rlap{\frac{\sqrt{1-x^2} + \dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]
&= \frac{\dfrac{\bigl(\sqrt{1-x^2}\bigr)^2}{\sqrt{1-x^2}}+\dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}\\[5pt]
&= \frac{\dfrac{1-x^2+x^2}{\sqrt{1-x^2}}}{1-x^2}\\[5pt]
&= \frac{1}{(1-x^2)^{3/2}}\,\textrm{.} 
\end{align}



The second derivative is





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




Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:4a“
                       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 1.2:4a
			
			  Aus Online Mathematik Brückenkurs 2
			  (Unterschied zwischen Versionen)
			  			                            Wechseln zu: Navigation, Suche
                          
		          
			
			
			
			
			
			
				Version vom 13:47, 15. Okt. 2008 (bearbeiten)
Tek  (Diskussion | Beiträge)
K 
← Zum vorherigen Versionsunterschied
				Version vom 12:54, 10. Mär. 2009 (bearbeiten) (rückgängig)
Tekbot  (Diskussion | Beiträge) 
K  (Robot: Automated text replacement  (-{{Displayed math +{{Abgesetzte Formel))
Zum nächsten Versionsunterschied →
			
		Zeile 1:
Zeile 1:
 We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives  We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}  \frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}
 &= {}\rlap{\frac{(x)'\sqrt{1-x^2}-x\bigl(\sqrt{1-x^2}\bigr)'}{\bigl(\sqrt{1-x^2}\bigr)^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]  &= {}\rlap{\frac{(x)'\sqrt{1-x^2}-x\bigl(\sqrt{1-x^2}\bigr)'}{\bigl(\sqrt{1-x^2}\bigr)^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]
Zeile 9:
Zeile 9:
 We determine the derivative <math>\bigl(\sqrt{1-x^2}\bigr)'</math> by using the chain rule  We determine the derivative <math>\bigl(\sqrt{1-x^2}\bigr)'</math> by using the chain rule
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}  \phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
 &= \frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}\\[5pt]  &= \frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}\\[5pt]
Zeile 17:
Zeile 17:
 We simplify the result as far as possible, so as to make the second differentiation easier,  We simplify the result as far as possible, so as to make the second differentiation easier,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}  \phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
 &= {}\rlap{\frac{\sqrt{1-x^2} + \dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]  &= {}\rlap{\frac{\sqrt{1-x^2} + \dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]
Zeile 27:
Zeile 27:
 The second derivative is  The second derivative is
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d^2}{dx^2}\,\frac{x}{\sqrt{1-x^2}}  \frac{d^2}{dx^2}\,\frac{x}{\sqrt{1-x^2}}
 &= \frac{d}{dx}\,\frac{1}{(1-x^2)^{3/2}}\\[5pt]  &= \frac{d}{dx}\,\frac{1}{(1-x^2)^{3/2}}\\[5pt]
Version vom 12:54, 10. Mär. 2009
We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives





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



We determine the derivative \displaystyle \bigl(\sqrt{1-x^2}\bigr)' by using the chain rule





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



We simplify the result as far as possible, so as to make the second differentiation easier,





\displaystyle \begin{align}
\phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
&= {}\rlap{\frac{\sqrt{1-x^2} + \dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]
&= \frac{\dfrac{\bigl(\sqrt{1-x^2}\bigr)^2}{\sqrt{1-x^2}}+\dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}\\[5pt]
&= \frac{\dfrac{1-x^2+x^2}{\sqrt{1-x^2}}}{1-x^2}\\[5pt]
&= \frac{1}{(1-x^2)^{3/2}}\,\textrm{.} 
\end{align}



The second derivative is





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




Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:4a“
-                      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 1.2:4a
			
			  Aus Online Mathematik Brückenkurs 2
			  (Unterschied zwischen Versionen)
			  			                            Wechseln zu: Navigation, Suche
                          
		          
			
			
			
			
			
			
				Version vom 13:47, 15. Okt. 2008 (bearbeiten)
Tek  (Diskussion | Beiträge)
K 
← Zum vorherigen Versionsunterschied
				Version vom 12:54, 10. Mär. 2009 (bearbeiten) (rückgängig)
Tekbot  (Diskussion | Beiträge) 
K  (Robot: Automated text replacement  (-{{Displayed math +{{Abgesetzte Formel))
Zum nächsten Versionsunterschied →
			
		Zeile 1:
Zeile 1:
 We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives  We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}  \frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}
 &= {}\rlap{\frac{(x)'\sqrt{1-x^2}-x\bigl(\sqrt{1-x^2}\bigr)'}{\bigl(\sqrt{1-x^2}\bigr)^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]  &= {}\rlap{\frac{(x)'\sqrt{1-x^2}-x\bigl(\sqrt{1-x^2}\bigr)'}{\bigl(\sqrt{1-x^2}\bigr)^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]
Zeile 9:
Zeile 9:
 We determine the derivative <math>\bigl(\sqrt{1-x^2}\bigr)'</math> by using the chain rule  We determine the derivative <math>\bigl(\sqrt{1-x^2}\bigr)'</math> by using the chain rule
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}  \phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
 &= \frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}\\[5pt]  &= \frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}\\[5pt]
Zeile 17:
Zeile 17:
 We simplify the result as far as possible, so as to make the second differentiation easier,  We simplify the result as far as possible, so as to make the second differentiation easier,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}  \phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
 &= {}\rlap{\frac{\sqrt{1-x^2} + \dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]  &= {}\rlap{\frac{\sqrt{1-x^2} + \dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]
Zeile 27:
Zeile 27:
 The second derivative is  The second derivative is
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{d^2}{dx^2}\,\frac{x}{\sqrt{1-x^2}}  \frac{d^2}{dx^2}\,\frac{x}{\sqrt{1-x^2}}
 &= \frac{d}{dx}\,\frac{1}{(1-x^2)^{3/2}}\\[5pt]  &= \frac{d}{dx}\,\frac{1}{(1-x^2)^{3/2}}\\[5pt]
Version vom 12:54, 10. Mär. 2009
We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives





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



We determine the derivative \displaystyle \bigl(\sqrt{1-x^2}\bigr)' by using the chain rule





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



We simplify the result as far as possible, so as to make the second differentiation easier,





\displaystyle \begin{align}
\phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
&= {}\rlap{\frac{\sqrt{1-x^2} + \dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]
&= \frac{\dfrac{\bigl(\sqrt{1-x^2}\bigr)^2}{\sqrt{1-x^2}}+\dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}\\[5pt]
&= \frac{\dfrac{1-x^2+x^2}{\sqrt{1-x^2}}}{1-x^2}\\[5pt]
&= \frac{1}{(1-x^2)^{3/2}}\,\textrm{.} 
\end{align}



The second derivative is





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




Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:4a“
@@ Zeile 1: / Zeile 1: @@
 We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives
-{{Displayed math||<math>\begin{align}
+{{Abgesetzte Formel||<math>\begin{align}
 \frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}
 &= {}\rlap{\frac{(x)'\sqrt{1-x^2}-x\bigl(\sqrt{1-x^2}\bigr)'}{\bigl(\sqrt{1-x^2}\bigr)^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]
@@ Zeile 9: / Zeile 9: @@
 We determine the derivative <math>\bigl(\sqrt{1-x^2}\bigr)'</math> by using the chain rule
-{{Displayed math||<math>\begin{align}
+{{Abgesetzte Formel||<math>\begin{align}
 \phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
 &= \frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}\\[5pt]
@@ Zeile 17: / Zeile 17: @@
 We simplify the result as far as possible, so as to make the second differentiation easier,
-{{Displayed math||<math>\begin{align}
+{{Abgesetzte Formel||<math>\begin{align}
 \phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
 &= {}\rlap{\frac{\sqrt{1-x^2} + \dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]
@@ Zeile 27: / Zeile 27: @@
 The second derivative is
-{{Displayed math||<math>\begin{align}
+{{Abgesetzte Formel||<math>\begin{align}
 \frac{d^2}{dx^2}\,\frac{x}{\sqrt{1-x^2}}
 &= \frac{d}{dx}\,\frac{1}{(1-x^2)^{3/2}}\\[5pt]
 \displaystyle \begin{align}
\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}
&= {}\rlap{\frac{(x)'\sqrt{1-x^2}-x\bigl(\sqrt{1-x^2}\bigr)'}{\bigl(\sqrt{1-x^2}\bigr)^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]
&= \frac{1\cdot\sqrt{1-x^2}-x\bigl(\sqrt{1-x^2}\bigr)'}{1-x^2}\,\textrm{.} 
\end{align}
 \displaystyle \begin{align}
\phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
&= \frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}\\[5pt]
&= \frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot (-2x)}{1-x^2}\,\textrm{.} 
\end{align}
 \displaystyle \begin{align}
\phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
&= {}\rlap{\frac{\sqrt{1-x^2} + \dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt]
&= \frac{\dfrac{\bigl(\sqrt{1-x^2}\bigr)^2}{\sqrt{1-x^2}}+\dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}\\[5pt]
&= \frac{\dfrac{1-x^2+x^2}{\sqrt{1-x^2}}}{1-x^2}\\[5pt]
&= \frac{1}{(1-x^2)^{3/2}}\,\textrm{.} 
\end{align}
 \displaystyle \begin{align}
\frac{d^2}{dx^2}\,\frac{x}{\sqrt{1-x^2}}
&= \frac{d}{dx}\,\frac{1}{(1-x^2)^{3/2}}\\[5pt]
&= \frac{d}{dx}\,\bigl(1-x^2\bigr)^{-3/2}\\[5pt]
&= -\tfrac{3}{2}\bigl(1-x^2\bigr)^{-3/2-1}\cdot\bigl(1-x^2\bigr)'\\[5pt] 
&= -\tfrac{3}{2}\bigl(1-x^2\bigr)^{-5/2}\cdot (-2x)\\[5pt]
&= 3x\bigl(1-x^2\bigr)^{-5/2}\\[5pt]
&= \frac{3x}{\bigl(1-x^2\bigr)^{5/2}}\,\textrm{.} 
\end{align}