Solution 1.2:3b - Förberedande kurs i matematik 2

-                      Welcome to the course
                       
                        About the course 
                        Examinations
                      
                    
                      1. Derivatives
                       
                        1.1 Introduction 
                        1.2 Rules of differentiation 
                        1.3 Max and min problems
                      
                    
                      2. Integrals
                       
                        2.1 Introduction 
                        2.2 Substitution 
                        2.3 Integration by parts
                      
                    
                      3. Complex numbers
                       
                        3.1 Calculations 
                        3.2 Polar form 
                        3.3 Powers and roots 
                        3.4 Complex polynomials
                      
                    
                    
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+			Solution 1.2:3b
			
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 (Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_3b-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:1_2_3b-2(2).gif </center> {{NAVCONTENT_STOP}})
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- {{NAVCONTENT_START}} + The outer function in the expression is "the square root of something",
- <center> [[Bild:1_2_3b-1(2).gif]] </center> +  
- {{NAVCONTENT_STOP}} + {{Displayed math||<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } }</math>}}
- {{NAVCONTENT_START}} +  
- <center> [[Bild:1_2_3b-2(2).gif]] </center> + and differentiating with the chain rule gives
- {{NAVCONTENT_STOP}} +  
  + {{Displayed math||<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\dfrac{x+1}{x-1} } } }\cdot \Bigl( \frac{x+1}{x-1}\Bigr)'\,\textrm{.}</math>}}
  +  
  + We establish the inner derivative by using the quotient rule,
  +  
  + {{Displayed math||<math>\begin{align}
  + \frac{d}{dx}\,\sqrt{\frac{x+1}{x-1}}
  + &= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot\frac{(x+1)'\cdot (x-1) - (x+1)\cdot (x-1)'}{(x-1)^2}\\[5pt] 
  + &= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot \frac{1\cdot (x-1) - (x+1)\cdot 1}{(x-1)^2}\\[5pt] 
  + &= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot \frac{-2}{(x-1)^2}\\[5pt] 
  + &= -\sqrt{\frac{x-1}{x+1}}\cdot\frac{1}{(x-1)^2}\\[5pt] 
  + &= -\frac{1}{(x-1)^{3/2}\sqrt{x+1}}\,, 
  + \end{align}</math>}}
  +  
  + where we have used the simplification
  +  
  + {{Displayed math||<math>\frac{\sqrt{x-1}}{(x-1)^2}
  + = \frac{(x-1)^{1/2}}{(x-1)^2}
  + = (x-1)^{1/2-2}
  + = (x-1)^{-3/2}
  + = \frac{1}{(x-1)^{3/2}}\,\textrm{.}</math>}}
Current revision
The outer function in the expression is "the square root of something",





\displaystyle \sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } }


and differentiating with the chain rule gives





\displaystyle \frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\dfrac{x+1}{x-1} } } }\cdot \Bigl( \frac{x+1}{x-1}\Bigr)'\,\textrm{.}


We establish the inner derivative by using the quotient rule,





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



where we have used the simplification





\displaystyle \frac{\sqrt{x-1}}{(x-1)^2}
= \frac{(x-1)^{1/2}}{(x-1)^2}
= (x-1)^{1/2-2}
= (x-1)^{-3/2}
= \frac{1}{(x-1)^{3/2}}\,\textrm{.}




Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-ImperialCollege/index.php/Solution_1.2:3b"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+The outer function in the expression is "the square root of something",
-<center> [[Bild:1_2_3b-1(2).gif]] </center>
-{{NAVCONTENT_STOP}}
+{{Displayed math||<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } }</math>}}
-{{NAVCONTENT_START}}
-<center> [[Bild:1_2_3b-2(2).gif]] </center>
+and differentiating with the chain rule gives
-{{NAVCONTENT_STOP}}
+{{Displayed math||<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\dfrac{x+1}{x-1} } } }\cdot \Bigl( \frac{x+1}{x-1}\Bigr)'\,\textrm{.}</math>}}
+We establish the inner derivative by using the quotient rule,
+{{Displayed math||<math>\begin{align}
+\frac{d}{dx}\,\sqrt{\frac{x+1}{x-1}}
+&= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot\frac{(x+1)'\cdot (x-1) - (x+1)\cdot (x-1)'}{(x-1)^2}\\[5pt]
+&= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot \frac{1\cdot (x-1) - (x+1)\cdot 1}{(x-1)^2}\\[5pt]
+&= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot \frac{-2}{(x-1)^2}\\[5pt]
+&= -\sqrt{\frac{x-1}{x+1}}\cdot\frac{1}{(x-1)^2}\\[5pt]
+&= -\frac{1}{(x-1)^{3/2}\sqrt{x+1}}\,,
+\end{align}</math>}}
+where we have used the simplification
+{{Displayed math||<math>\frac{\sqrt{x-1}}{(x-1)^2}
+= \frac{(x-1)^{1/2}}{(x-1)^2}
+= (x-1)^{1/2-2}
+= (x-1)^{-3/2}
+= \frac{1}{(x-1)^{3/2}}\,\textrm{.}</math>}}
 \displaystyle \sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } }
 \displaystyle \frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\dfrac{x+1}{x-1} } } }\cdot \Bigl( \frac{x+1}{x-1}\Bigr)'\,\textrm{.}
 \displaystyle \begin{align}
\frac{d}{dx}\,\sqrt{\frac{x+1}{x-1}}
&= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot\frac{(x+1)'\cdot (x-1) - (x+1)\cdot (x-1)'}{(x-1)^2}\\[5pt] 
&= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot \frac{1\cdot (x-1) - (x+1)\cdot 1}{(x-1)^2}\\[5pt] 
&= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot \frac{-2}{(x-1)^2}\\[5pt] 
&= -\sqrt{\frac{x-1}{x+1}}\cdot\frac{1}{(x-1)^2}\\[5pt] 
&= -\frac{1}{(x-1)^{3/2}\sqrt{x+1}}\,, 
\end{align}
 \displaystyle \frac{\sqrt{x-1}}{(x-1)^2}
= \frac{(x-1)^{1/2}}{(x-1)^2}
= (x-1)^{1/2-2}
= (x-1)^{-3/2}
= \frac{1}{(x-1)^{3/2}}\,\textrm{.}