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Lösung 1.2:2e

Aus Online Mathematik Brückenkurs 2

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One way to differentiate the expression could be to expand
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<center> [[Image:1_2_2e-1(2).gif]] </center>
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<math>\left( 2x+1 \right)^{4}</math>
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multiply by
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<math>x</math>
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<center> [[Image:1_2_2e-2(2).gif]] </center>
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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.
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To begin with, we have a product of
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<math>x</math>
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and
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<math>\left( 2x+1 \right)^{4}</math>
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so the product rule gives that
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<math>\begin{align}
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& \frac{d}{dx}x\left( 2x+1 \right)^{4}=\left( x \right)^{\prime }\centerdot \left( 2x+1 \right)^{4}+x\centerdot \left( \left( 2x+1 \right)^{4} \right)^{\prime } \\
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& =1\centerdot \left( 2x+1 \right)^{4}+x\centerdot \left( \left( 2x+1 \right)^{4} \right)^{\prime } \\
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\end{align}</math>
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We can differentiate the expression
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<math>\left( 2x+1 \right)^{4}</math>
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by viewing it as "something raised to the
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<math>\text{4}</math>",
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<math>\left\{ \left. {} \right\} \right.^{4}</math>.
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The chain rule then gives
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<math>\begin{align}
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& \frac{d}{dx}\left\{ \left. {} \right\} \right.^{4}=4\centerdot \left\{ \left. {} \right\} \right.^{3}\centerdot \left( \left\{ \left. {} \right\} \right. \right)^{\prime } \\
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& \frac{d}{dx}\left( 2x+1 \right)^{4}=4\centerdot \left( 2x+1 \right)^{3}\centerdot \left( 2x+1 \right)^{\prime } \\
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\end{align}</math>
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We carry out the last differentiation directly, and obtain
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<math>\left( 2x+1 \right)^{\prime }=2</math>
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If we go through the whole calculation from the beginning, it is
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<math>\begin{align}
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& \frac{d}{dx}x\left( 2x+1 \right)^{4}=\left( x \right)^{\prime }\centerdot \left( 2x+1 \right)^{4}+x\centerdot \left( \left( 2x+1 \right)^{4} \right)^{\prime } \\
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& =1\centerdot \left( 2x+1 \right)^{4}+x\centerdot 4\left( 2x+1 \right)^{3}\centerdot \left( 2x+1 \right)^{\prime } \\
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& =\left( 2x+1 \right)^{4}+x\centerdot 4\left( 2x+1 \right)^{3}\centerdot 2 \\
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& =\left( 2x+1 \right)^{4}+8x\left( 2x+1 \right)^{3} \\
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\end{align}</math>
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Both terms contain a common factor
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<math>\left( 2x+1 \right)^{3}</math>
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which we can take out to get an answer in factorized form:
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<math>\begin{align}
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& \frac{d}{dx}x\left( 2x+1 \right)^{4}=\left( 2x+1 \right)^{3}\left( \left( 2x+1 \right)+8x \right) \\
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& =\left( 2x+1 \right)^{3}\left( 10x+1 \right) \\
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\end{align}</math>

Version vom 13:39, 11. Okt. 2008

One way to differentiate the expression could be to expand 2x+14  multiply by 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 x and 2x+14  so the product rule gives that


ddxx2x+14=x2x+14+x2x+14=12x+14+x2x+14


We can differentiate the expression 2x+14  by viewing it as "something raised to the 4",


4.

The chain rule then gives


ddx4=43ddx2x+14=42x+132x+1


We carry out the last differentiation directly, and obtain


2x+1=2 


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


ddxx2x+14=x2x+14+x2x+14=12x+14+x42x+132x+1=2x+14+x42x+132=2x+14+8x2x+13


Both terms contain a common factor 2x+13  which we can take out to get an answer in factorized form:


ddxx2x+14=2x+132x+1+8x=2x+1310x+1

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:2e“