Lösung 1.2:3d
Aus Online Mathematik Brückenkurs 2
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} |
