Lösung 1.2:1c

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The expression is a quotient of two polynomials,
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<center> [[Image:1_2_1c.gif]] </center>
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<math>x^{2}+1</math>
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and
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<math>x+1</math>, and we therefore use the quotient rule for differentiation:
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<math>\begin{align}
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& \left( \frac{x^{2}+1}{x+1} \right)^{\prime }=\frac{\left( x^{2}+1 \right)^{\prime }\centerdot \left( x+1 \right)-\left( x^{2}+1 \right)\centerdot \left( x+1 \right)^{\prime }}{\left( x+1 \right)^{2}} \\
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& \\
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& =\frac{2x\centerdot \left( x+1 \right)-\left( x^{2}+1 \right)\centerdot 1}{\left( x+1 \right)^{2}} \\
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& \\
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& =\frac{2x^{2}+2x-x^{2}-1}{\left( x+1 \right)^{2}}=\frac{x^{2}+2x-1}{\left( x+1 \right)^{2}} \\
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\end{align}</math>
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NOTE: it is possible to rewrite the numerator by completing the square,
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<math>x^{2}+2x-1=\left( x+1 \right)^{2}-1^{2}-1=\left( x+1 \right)^{2}-2</math>
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and then the answer can be written as
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<math>\frac{x^{2}+2x-1}{\left( x+1 \right)^{2}}=\frac{\left( x+1 \right)^{2}-2}{\left( x+1 \right)^{2}}=1-\frac{2}{\left( x+1 \right)^{2}}</math>

Version vom 15:06, 12. Sep. 2008

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} & \left( \frac{x^{2}+1}{x+1} \right)^{\prime }=\frac{\left( x^{2}+1 \right)^{\prime }\centerdot \left( x+1 \right)-\left( x^{2}+1 \right)\centerdot \left( x+1 \right)^{\prime }}{\left( x+1 \right)^{2}} \\ & \\ & =\frac{2x\centerdot \left( x+1 \right)-\left( x^{2}+1 \right)\centerdot 1}{\left( x+1 \right)^{2}} \\ & \\ & =\frac{2x^{2}+2x-x^{2}-1}{\left( x+1 \right)^{2}}=\frac{x^{2}+2x-1}{\left( x+1 \right)^{2}} \\ \end{align}


NOTE: it is possible to rewrite the numerator by completing the square,


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


and then the answer can be written as


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

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:1c“