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Lösung 1.2:2e

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 3: Zeile 3:
To begin with, we have a product of <math>x</math> and <math>(2x+1)^4</math> so the product rule gives that
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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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]
&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[5pt]
Zeile 11: Zeile 11:
We can differentiate the expression <math>(2x+1)^4</math> by viewing it as "something raised to the 4",
We can differentiate the expression <math>(2x+1)^4</math> by viewing it as "something raised to the 4",
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{{Displayed math||<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\,\textrm{.}</math>}}
The chain rule then gives
The chain rule then gives
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{d}{dx}\,\bigl[\bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,4}\bigr]
\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]
&= 4\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,3}\cdot \bbox[#FFEEAA;,1.5pt]{\phantom{(2x+1)}\!\!}^{\,\prime}\,,\\[5pt]
Zeile 23: Zeile 23:
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>}}
+
{{Abgesetzte Formel||<math>(2x+1)' = 2\,\textrm{.}</math>}}
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}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt]
&= (x)'\cdot (2x+1)^4 + x\cdot \bigl((2x+1)^4\bigr)'\\[2pt]
Zeile 37: Zeile 37:
Both terms contain a common factor <math>(2x+1)^3</math> which we can take out to get an answer in factorized form,
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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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
\frac{d}{dx}\,\bigl[x(2x+1)^4\bigr]
&= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt]
&= (2x+1)^3\bigl((2x+1)+8x\bigr)\\[5pt]
&= (2x+1)^3(10x+1)\,\textrm{.}
&= (2x+1)^3(10x+1)\,\textrm{.}
\end{align}</math>}}
\end{align}</math>}}

Version vom 12:53, 10. Mär. 2009

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“