Lösung 1.2:1c

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Zeile 1: Zeile 1:
The expression is a quotient of two polynomials, <math>x^2+1</math> and <math>x+1</math>, and we therefore use the quotient rule for differentiation,
The expression is a quotient of two polynomials, <math>x^2+1</math> and <math>x+1</math>, and we therefore use the quotient rule for differentiation,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\Bigl(\frac{x^2+1}{x+1}\Bigr)'
\Bigl(\frac{x^2+1}{x+1}\Bigr)'
&= \frac{(x^2+1)'\cdot (x+1) - (x^2+1)\cdot (x+1)'}{(x+1)^2}\\[5pt]
&= \frac{(x^2+1)'\cdot (x+1) - (x^2+1)\cdot (x+1)'}{(x+1)^2}\\[5pt]
Zeile 12: Zeile 12:
Note: It is possible to rewrite the numerator by completing the square,
Note: It is possible to rewrite the numerator by completing the square,
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{{Displayed math||<math>x^2+2x-1 = (x+1)^{2} - 1^2 - 1 = (x+1)^2 - 2</math>}}
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{{Abgesetzte Formel||<math>x^2+2x-1 = (x+1)^{2} - 1^2 - 1 = (x+1)^2 - 2</math>}}
and then the answer can be written as
and then the answer can be written as
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{{Displayed math||<math>\frac{x^2+2x-1}{(x+1)^2} = \frac{(x+1)^2-2}{(x+1)^2} = 1-\frac{2}{(x+1)^2}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{x^2+2x-1}{(x+1)^2} = \frac{(x+1)^2-2}{(x+1)^2} = 1-\frac{2}{(x+1)^2}\,\textrm{.}</math>}}

Version vom 12:52, 10. Mär. 2009

The expression is a quotient of two polynomials, \displaystyle x^2+1 and \displaystyle x+1, and we therefore use the quotient rule for differentiation,

\displaystyle \begin{align}

\Bigl(\frac{x^2+1}{x+1}\Bigr)' &= \frac{(x^2+1)'\cdot (x+1) - (x^2+1)\cdot (x+1)'}{(x+1)^2}\\[5pt] &= \frac{2x\cdot (x+1) - (x^2+1)\cdot 1}{(x+1)^2}\\[5pt] &= \frac{2x^2+2x-x^2-1}{(x+1)^2}\\[5pt] &= \frac{x^2+2x-1}{(x+1)^2}\,\textrm{.} \end{align}


Note: It is possible to rewrite the numerator by completing the square,

\displaystyle x^2+2x-1 = (x+1)^{2} - 1^2 - 1 = (x+1)^2 - 2

and then the answer can be written as

\displaystyle \frac{x^2+2x-1}{(x+1)^2} = \frac{(x+1)^2-2}{(x+1)^2} = 1-\frac{2}{(x+1)^2}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:1c“