Lösung 1.1:2e

Aus Online Mathematik Brückenkurs 2

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With the help of the square rule, we can expand the quadratic as
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<center> [[Image:1_1_2e.gif]] </center>
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<math>\begin{align}
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& f\left( x \right)=\left( x^{2}-1 \right)^{2}=\left( x^{2} \right)^{2}-2\centerdot x^{2}\centerdot 1+1^{2} \\
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& =x^{4}-2x^{2}+1 \\
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\end{align}</math>
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When the function is written in this form, it is easy to differentiate term by term:
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<math>\begin{align}
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& {f}'\left( x \right)=\frac{d}{dx}\left( x^{4}-2x^{2}+1 \right) \\
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& =\frac{d}{dx}x^{4}-2\frac{d}{dx}x^{2}+\frac{d}{dx}1 \\
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& =4\centerdot x^{-1}-2\centerdot 2x^{-1}+0 \\
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& =4x^{3}-4x=4x\left( x^{2}-1 \right) \\
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\end{align}</math>

Version vom 11:37, 10. Okt. 2008

With the help of the square rule, we can expand the quadratic as


\displaystyle \begin{align} & f\left( x \right)=\left( x^{2}-1 \right)^{2}=\left( x^{2} \right)^{2}-2\centerdot x^{2}\centerdot 1+1^{2} \\ & =x^{4}-2x^{2}+1 \\ \end{align}


When the function is written in this form, it is easy to differentiate term by term:


\displaystyle \begin{align} & {f}'\left( x \right)=\frac{d}{dx}\left( x^{4}-2x^{2}+1 \right) \\ & =\frac{d}{dx}x^{4}-2\frac{d}{dx}x^{2}+\frac{d}{dx}1 \\ & =4\centerdot x^{-1}-2\centerdot 2x^{-1}+0 \\ & =4x^{3}-4x=4x\left( x^{2}-1 \right) \\ \end{align}

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