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Lösung 1.2:3d

Aus Online Mathematik Brückenkurs 2

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We differentiate the function successively, one part at a time,
We differentiate the function successively, one part at a time,
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{{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>}}
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<math>\frac{d}{dx}\sin \left\{ \left. \cos \sin x \right\} \right.=\cos \left\{ \left. \cos \sin x \right\} \right.\centerdot \left( \left\{ \left. \cos \sin x \right\} \right. \right)^{\prime }</math>,
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and the next differentiation becomes
and the next differentiation becomes
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\frac{d}{dx}\,\cos \bbox[#FFEEAA;,1.5pt]{\sin x}
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& \frac{d}{dx}\cos \left\{ \left. \sin x \right\} \right.=-\sin \left\{ \left. \sin x \right\} \right.\centerdot \left( \sin x \right)^{\prime } \\
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&= -\sin \bbox[#FFEEAA;,1.5pt]{\sin x}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sin x}\bigr)'\\[5pt]
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& =-\sin \sin x\centerdot \cos x \\
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&= -\sin \sin x\cdot \cos x\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}
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The answer is thus
The answer is thus
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\frac{d}{dx}\,\sin \cos \sin x
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& \frac{d}{dx}\sin \cos \sin x=\cos \cos \sin x\centerdot \left( -\sin \sin x\centerdot \cos x \right) \\
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&= \cos \cos \sin x\cdot ( -\sin \sin x\cdot \cos x)\\[5pt]
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& =-\cos \cos \sin x\centerdot \sin \sin x\centerdot \cos x \\
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&= -\cos \cos \sin x\cdot \sin \sin x\cdot \cos x\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}

Version vom 13:04, 15. Okt. 2008

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

ddxsincossinx=coscossinxcossinx 

and the next differentiation becomes

ddxcossinx=−sinsinxsinx=−sinsinxcosx.

The answer is thus

ddxsincossinx=coscossinx(−sinsinxcosx)=−coscossinxsinsinxcosx.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:3d“