Solution 1.2:3d
From Förberedande kurs i matematik 2
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| - | {{ | + | 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>}} |
| + | |||
| + | and the next differentiation becomes | ||
| + | |||
| + | {{Displayed math||<math>\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}</math>}} | ||
| + | |||
| + | The answer is thus | ||
| + | |||
| + | {{Displayed math||<math>\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}</math>}} | ||
Current revision
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} |
