(log(2+sinx))^2を微分する問題

$${\hspace{10pt}\{f(\hspace{1pt}g(x))\}' = f'(\hspace{1pt}g(x))\hspace{1pt}g'(x)\hspace{10pt}}$$

$$\begin{aligned} \hspace{10pt}y' & = 2 (\log (2+\sin x)) \cdot (\log (2+\sin x))'\hspace{10pt}\\[1em] & = 2 (\log (2+\sin x)) \cdot \frac{1}{2+\sin x} (2+\sin x)' \hspace{10pt}\\[1em] & = 2 (\log (2+\sin x)) \cdot \frac{1}{2+\sin x} (\cos x) \hspace{10pt}\\[1em] & = \frac{2 \cos x \log (2+\sin x)}{2+\sin x} \hspace{10pt}\\[1em] \end{aligned}$$