Lösung 1.2:3d

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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We differentiate the function successively, one part at a time,
We differentiate the function successively, one part at a time,
-
{{Displayed math||<math>\frac{d}{dx}\,\sin \bbox[#FFEEAA;,1.5pt]{\cos\sin x} = \cos \bbox[#FFEEAA;,1.5pt]{\cos\sin x}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\cos\sin x}\bigr)'\,,</math>}}
+
{{Abgesetzte Formel||<math>\frac{d}{dx}\,\sin \bbox[#FFEEAA;,1.5pt]{\cos\sin x} = \cos \bbox[#FFEEAA;,1.5pt]{\cos\sin x}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\cos\sin x}\bigr)'\,,</math>}}
and the next differentiation becomes
and the next differentiation becomes
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{d}{dx}\,\cos \bbox[#FFEEAA;,1.5pt]{\sin x}
\frac{d}{dx}\,\cos \bbox[#FFEEAA;,1.5pt]{\sin x}
&= -\sin \bbox[#FFEEAA;,1.5pt]{\sin x}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sin x}\bigr)'\\[5pt]
&= -\sin \bbox[#FFEEAA;,1.5pt]{\sin x}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sin x}\bigr)'\\[5pt]
Zeile 13: Zeile 13:
The answer is thus
The answer is thus
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{d}{dx}\,\sin \cos \sin x
\frac{d}{dx}\,\sin \cos \sin x
&= \cos \cos \sin x\cdot ( -\sin \sin x\cdot \cos x)\\[5pt]
&= \cos \cos \sin x\cdot ( -\sin \sin x\cdot \cos x)\\[5pt]
&= -\cos \cos \sin x\cdot \sin \sin x\cdot \cos x\,\textrm{.}
&= -\cos \cos \sin x\cdot \sin \sin x\cdot \cos x\,\textrm{.}
\end{align}</math>}}
\end{align}</math>}}

Version vom 12:54, 10. Mär. 2009

We differentiate the function successively, one part at a time,

\displaystyle \frac{d}{dx}\,\sin \bbox[#FFEEAA;,1.5pt]{\cos\sin x} = \cos \bbox[#FFEEAA;,1.5pt]{\cos\sin x}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\cos\sin x}\bigr)'\,,

and the next differentiation becomes

\displaystyle \begin{align}

\frac{d}{dx}\,\cos \bbox[#FFEEAA;,1.5pt]{\sin x} &= -\sin \bbox[#FFEEAA;,1.5pt]{\sin x}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sin x}\bigr)'\\[5pt] &= -\sin \sin x\cdot \cos x\,\textrm{.} \end{align}

The answer is thus

\displaystyle \begin{align}

\frac{d}{dx}\,\sin \cos \sin x &= \cos \cos \sin x\cdot ( -\sin \sin x\cdot \cos x)\\[5pt] &= -\cos \cos \sin x\cdot \sin \sin x\cdot \cos x\,\textrm{.} \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:3d“