Lösung 1.2:3d

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K (Lösning 1.2:3d moved to Solution 1.2:3d: Robot: moved page)
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We differentiate the function successively, one part at a time,
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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
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<math>\begin{align}
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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 \sin x\centerdot \cos x \\
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\end{align}</math>
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The answer is thus
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<math>\begin{align}
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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\centerdot \sin \sin x\centerdot \cos x \\
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\end{align}</math>

Version vom 13:18, 12. Okt. 2008

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


\displaystyle \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 },

and the next differentiation becomes


\displaystyle \begin{align} & \frac{d}{dx}\cos \left\{ \left. \sin x \right\} \right.=-\sin \left\{ \left. \sin x \right\} \right.\centerdot \left( \sin x \right)^{\prime } \\ & =-\sin \sin x\centerdot \cos x \\ \end{align}


The answer is thus


\displaystyle \begin{align} & \frac{d}{dx}\sin \cos \sin x=\cos \cos \sin x\centerdot \left( -\sin \sin x\centerdot \cos x \right) \\ & =-\cos \cos \sin x\centerdot \sin \sin x\centerdot \cos x \\ \end{align}

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