Lösung 1.1:2b

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The expression is a sum of two terms which we differentiate one by one:
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The expression is a sum of two terms which we differentiate one by one,
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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f^{\,\prime}(x)
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& {f}'\left( x \right)=\frac{d}{dx}\left( \cos x-\sin x \right) \\
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&= \frac{d}{dx}\,\bigl(\cos x-\sin x\bigr)\\[5pt]
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& =\frac{d}{dx}\cos x-\frac{d}{dx}\sin x=-\sin x-\cos x \\
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&= \frac{d}{dx}\,\cos x - \frac{d}{dx}\,\sin x\\[5pt]
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\end{align}</math>
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&= -\sin x-\cos x\,\textrm{.}
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\end{align}</math>}}

Version vom 11:47, 14. Okt. 2008

The expression is a sum of two terms which we differentiate one by one,

\displaystyle \begin{align}

f^{\,\prime}(x) &= \frac{d}{dx}\,\bigl(\cos x-\sin x\bigr)\\[5pt] &= \frac{d}{dx}\,\cos x - \frac{d}{dx}\,\sin x\\[5pt] &= -\sin x-\cos x\,\textrm{.} \end{align}

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