Solution 1.2:4b - 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 1.2:4b
			
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 (Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_4b-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:1_2_4b-2(2).gif </center> {{NAVCONTENT_STOP}})
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- {{NAVCONTENT_START}} + To start with, we determine the first derivative and begin by using the product rule,
- <center> [[Bild:1_2_4b-1(2).gif]] </center> +  
- {{NAVCONTENT_STOP}} + {{Displayed math||<math>\begin{align}
- {{NAVCONTENT_START}} + \frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr] 
- <center> [[Bild:1_2_4b-2(2).gif]] </center> + &= (x)'\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\\[5pt] 
- {{NAVCONTENT_STOP}} + &= 1\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\,\textrm{.} 
  + \end{align}</math>}}
  +  
  + We divide up the differentiation of the second term in sections and use the chain rule,
  +  
  + {{Displayed math||<math>\begin{align}
  + (\sin\ln x + \cos\ln x)'
  + &= (\sin\ln x)' + (\cos\ln x)'\\[5pt] 
  + &= \cos\ln x\cdot (\ln x)' - \sin\ln x\cdot (\ln x)'\\[5pt] 
  + &= \cos\ln x\cdot\frac{1}{x} - \sin\ln x\cdot\frac{1}{x}\,\textrm{.} 
  + \end{align}</math>}}
  +  
  + This means that
  +  
  + {{Displayed math||<math>\begin{align}
  + \frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr] 
  + &= \sin \ln x + \cos \ln x + \cos \ln x - \sin \ln x\\[5pt] 
  + &= 2\cos \ln x\,\textrm{.}
  + \end{align}</math>}}
  +  
  + The second derivative is
  +  
  + {{Displayed math||<math>\begin{align}
  + \frac{d}{dx}\,2\cos\ln x
  + &= -2\sin\ln x\cdot (\ln x)'\\[5pt] 
  + &= -2\sin\ln x\cdot \frac{1}{x}\\[5pt]
  + &= -\frac{2\sin\ln x}{x}\,\textrm{.} 
  + \end{align}</math>}}
Current revision
To start with, we determine the first derivative and begin by using the product rule,





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



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





\displaystyle \begin{align}
(\sin\ln x + \cos\ln x)'
&= (\sin\ln x)' + (\cos\ln x)'\\[5pt] 
&= \cos\ln x\cdot (\ln x)' - \sin\ln x\cdot (\ln x)'\\[5pt] 
&= \cos\ln x\cdot\frac{1}{x} - \sin\ln x\cdot\frac{1}{x}\,\textrm{.} 
\end{align}



This means that





\displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr] 
&= \sin \ln x + \cos \ln x + \cos \ln x - \sin \ln x\\[5pt] 
&= 2\cos \ln x\,\textrm{.}
\end{align}



The second derivative is





\displaystyle \begin{align}
\frac{d}{dx}\,2\cos\ln x
&= -2\sin\ln x\cdot (\ln x)'\\[5pt] 
&= -2\sin\ln x\cdot \frac{1}{x}\\[5pt]
&= -\frac{2\sin\ln x}{x}\,\textrm{.} 
\end{align}




Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_1.2:4b"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+To start with, we determine the first derivative and begin by using the product rule,
-<center> [[Bild:1_2_4b-1(2).gif]] </center>
-{{NAVCONTENT_STOP}}
+{{Displayed math||<math>\begin{align}
-{{NAVCONTENT_START}}
+\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr]
-<center> [[Bild:1_2_4b-2(2).gif]] </center>
+&= (x)'\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\\[5pt]
-{{NAVCONTENT_STOP}}
+&= 1\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\,\textrm{.}
+\end{align}</math>}}
+We divide up the differentiation of the second term in sections and use the chain rule,
+{{Displayed math||<math>\begin{align}
+(\sin\ln x + \cos\ln x)'
+&= (\sin\ln x)' + (\cos\ln x)'\\[5pt]
+&= \cos\ln x\cdot (\ln x)' - \sin\ln x\cdot (\ln x)'\\[5pt]
+&= \cos\ln x\cdot\frac{1}{x} - \sin\ln x\cdot\frac{1}{x}\,\textrm{.}
+\end{align}</math>}}
+This means that
+{{Displayed math||<math>\begin{align}
+\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr]
+&= \sin \ln x + \cos \ln x + \cos \ln x - \sin \ln x\\[5pt]
+&= 2\cos \ln x\,\textrm{.}
+\end{align}</math>}}
+The second derivative is
+{{Displayed math||<math>\begin{align}
+\frac{d}{dx}\,2\cos\ln x
+&= -2\sin\ln x\cdot (\ln x)'\\[5pt]
+&= -2\sin\ln x\cdot \frac{1}{x}\\[5pt]
+&= -\frac{2\sin\ln x}{x}\,\textrm{.}
+\end{align}</math>}}
 \displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr] 
&= (x)'\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\\[5pt] 
&= 1\cdot (\sin\ln x + \cos\ln x) + x\cdot (\sin\ln x + \cos\ln x)'\,\textrm{.} 
\end{align}
 \displaystyle \begin{align}
(\sin\ln x + \cos\ln x)'
&= (\sin\ln x)' + (\cos\ln x)'\\[5pt] 
&= \cos\ln x\cdot (\ln x)' - \sin\ln x\cdot (\ln x)'\\[5pt] 
&= \cos\ln x\cdot\frac{1}{x} - \sin\ln x\cdot\frac{1}{x}\,\textrm{.} 
\end{align}
 \displaystyle \begin{align}
\frac{d}{dx}\,\bigl[x(\sin\ln x + \cos\ln x)\bigr] 
&= \sin \ln x + \cos \ln x + \cos \ln x - \sin \ln x\\[5pt] 
&= 2\cos \ln x\,\textrm{.}
\end{align}
 \displaystyle \begin{align}
\frac{d}{dx}\,2\cos\ln x
&= -2\sin\ln x\cdot (\ln x)'\\[5pt] 
&= -2\sin\ln x\cdot \frac{1}{x}\\[5pt]
&= -\frac{2\sin\ln x}{x}\,\textrm{.} 
\end{align}