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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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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.
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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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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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To begin with, we have a product of <math>x</math> and <math>(2x+1)^4</math> so the product rule gives that
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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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{{Displayed math||<math>\begin{align}
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\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
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&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]
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&= 1\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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We can differentiate the expression <math>(2x+1)^4</math> by viewing it as "something raised to the 4",
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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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{{Displayed math||<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</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
The chain rule then gives
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]
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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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&= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt]
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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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\frac{d}{dx}\,\bigl[(2x+1)^4\bigr] &= 4\cdot (2x+1)^3\cdot (2x+1)'\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}
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We carry out the last differentiation directly, and obtain
We carry out the last differentiation directly, and obtain
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{{Displayed math||<math>(2x+1)' = 2\,\textrm{.}</math>}}
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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
If we go through the whole calculation from the beginning, it is
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{{Displayed math||<math>\begin{align}
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\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
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&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt]
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&= 1\cdot (2x+1)^4 + x\cdot 4(2x+1)^3\cdot (2x+1)'\\[5pt]
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&= (2x+1)^4 + x\cdot 4(2x+1)^3\cdot 2\\[5pt]
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&= (2x+1)^4 + 8x(2x+1)^3\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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Both terms contain a common factor <math>(2x+1)^3</math> which we can take out to get an answer in factorized form,
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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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{{Displayed math||<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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\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
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& =\left( 2x+1 \right)^{3}\left( 10x+1 \right) \\
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&= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt]
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\end{align}</math>
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&= (2x+1)^3(10x+1)\,\textrm{.}
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\end{align}</math>}}

Version vom 08:56, 15. Okt. 2008

One way to differentiate the expression could be to expand (2x+1)4 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+1)4 so the product rule gives that

ddxx(2x+1)4=(x)(2x+1)4+x(2x+1)4=1(2x+1)4+x(2x+1)4.

We can differentiate the expression (2x+1)4 by viewing it as "something raised to the 4",

4.

The chain rule then gives

ddx4ddx(2x+1)4=43=4(2x+1)3(2x+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

ddxx(2x+1)4=(x)(2x+1)4+x(2x+1)4=1(2x+1)4+x4(2x+1)3(2x+1)=(2x+1)4+x4(2x+1)32=(2x+1)4+8x(2x+1)3.

Both terms contain a common factor (2x+1)3 which we can take out to get an answer in factorized form,

ddxx(2x+1)4=(2x+1)3(2x+1)+8x=(2x+1)3(10x+1).
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:2e“