Solution 1.2:2e - 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 1.2:2e
			
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- {{NAVCONTENT_START}} + One way to differentiate the expression could be to expand <math>(2x+1)^4</math> multiply by <math>x</math> 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.
- <center> [[Image:1_2_2e-1(2).gif]] </center> +  
- {{NAVCONTENT_STOP}} + To begin with, we have a product of <math>x</math> and <math>(2x+1)^4</math> so the product rule gives that 
- {{NAVCONTENT_START}} +  
- <center> [[Image:1_2_2e-2(2).gif]] </center> + {{Displayed math||<math>\begin{align}
- {{NAVCONTENT_STOP}} + \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}</math>}}
  +  
  + 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>}}
  +  
  + The chain rule then gives 
  +  
  + {{Displayed math||<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] 
  + \frac{d}{dx}\,\bigl[(2x+1)^4\bigr] &= 4\cdot (2x+1)^3\cdot (2x+1)'\,\textrm{.} 
  + \end{align}</math>}}
  +  
  + We carry out the last differentiation directly,  and obtain
  +  
  + {{Displayed math||<math>(2x+1)' = 2\,\textrm{.}</math>}}
  +  
  + If we go through the whole calculation from the beginning, it is 
  +  
  + {{Displayed math||<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] 
  + &= 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}</math>}}
  +  
  + 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}
  + \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>}}
Current revision
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}





Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_1.2:2e"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+One way to differentiate the expression could be to expand <math>(2x+1)^4</math> multiply by <math>x</math> 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.
-<center> [[Image:1_2_2e-1(2).gif]] </center>
-{{NAVCONTENT_STOP}}
+To begin with, we have a product of <math>x</math> and <math>(2x+1)^4</math> so the product rule gives that
-{{NAVCONTENT_START}}
-<center> [[Image:1_2_2e-2(2).gif]] </center>
+{{Displayed math||<math>\begin{align}
-{{NAVCONTENT_STOP}}
+\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}</math>}}
+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>}}
+The chain rule then gives
+{{Displayed math||<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]
+\frac{d}{dx}\,\bigl[(2x+1)^4\bigr] &= 4\cdot (2x+1)^3\cdot (2x+1)'\,\textrm{.}
+\end{align}</math>}}
+We carry out the last differentiation directly,  and obtain
+{{Displayed math||<math>(2x+1)' = 2\,\textrm{.}</math>}}
+If we go through the whole calculation from the beginning, it is
+{{Displayed math||<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]
+&= 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}</math>}}
+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}
+\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}