Aus Online Mathematik Brückenkurs 2

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}