Lösung 1.1:2e

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With the help of the square rule, we can expand the quadratic as
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We expand the quadratic expression as
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{{Displayed math||<math>\begin{align}
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f(x) &= \bigl(x^2-1\bigr)^2\\[5pt]
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&= \bigl(x^2\bigr)^2 - 2\cdot x^2\cdot 1 + 1^2\\[5pt]
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&= x^4 - 2x^2 + 1\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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When the function is written in this form, it is easy to differentiate term by term,
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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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{{Displayed math||<math>\begin{align}
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When the function is written in this form, it is easy to differentiate term by term:
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f^{\,\prime}(x) &= \frac{d}{dx}\,\bigl(x^4-2x^2+1\bigr)\\[5pt]
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&= \frac{d}{dx}\,x^4 - 2\frac{d}{dx}\,x^2 + \frac{d}{dx}\,1\\[5pt]
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&= 4\cdot x^{4-1} - 2\cdot 2x^{2-1} + 0\\[5pt]
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<math>\begin{align}
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&= 4x^{3} - 4x\\[5pt]
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& {f}'\left( x \right)=\frac{d}{dx}\left( x^{4}-2x^{2}+1 \right) \\
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&= 4x(x^2-1)\,\textrm{.}
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& =\frac{d}{dx}x^{4}-2\frac{d}{dx}x^{2}+\frac{d}{dx}1 \\
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\end{align}</math>}}
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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>
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Version vom 12:24, 14. Okt. 2008

We expand the quadratic expression as

\displaystyle \begin{align}

f(x) &= \bigl(x^2-1\bigr)^2\\[5pt] &= \bigl(x^2\bigr)^2 - 2\cdot x^2\cdot 1 + 1^2\\[5pt] &= x^4 - 2x^2 + 1\,\textrm{.} \end{align}

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

\displaystyle \begin{align}

f^{\,\prime}(x) &= \frac{d}{dx}\,\bigl(x^4-2x^2+1\bigr)\\[5pt] &= \frac{d}{dx}\,x^4 - 2\frac{d}{dx}\,x^2 + \frac{d}{dx}\,1\\[5pt] &= 4\cdot x^{4-1} - 2\cdot 2x^{2-1} + 0\\[5pt] &= 4x^{3} - 4x\\[5pt] &= 4x(x^2-1)\,\textrm{.} \end{align}

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