In the neighbourhood of x = 0

f(x) = log 2 – sin x

g(x) = f(f(x)) = log 2 – sin (log 2 – sin x)

$\therefore$g'(x) = – cos(log 2 – sin x) ( – cos x)

⇒            g'(0) = cos (log 2)

$\mathrm{Aliter}:\mathrm{g}\text{'}\left(0^{+}\right)=\underset{\mathrm{h}\to 0}{\mathrm{Lim}}\frac{\mathrm{f}\left(\mathrm{f}(0+\mathrm{h})\right)-\mathrm{f}\left(\mathrm{f}\left(0\right)\right)}{\mathrm{h}}=\underset{\mathrm{h}\to 0}{\mathrm{Lim}}\frac{\mathrm{f}(\log 2-\sin \mathrm{h})-|\log 2-\sin (\log 2\left)\right|}{\mathrm{h}}$

$=\underset{\mathrm{h}\to 0}{\mathrm{Lim}}\frac{|\log 2-\sin (\log 2-\sin \mathrm{h}\left)\right|-|\log 2-\sin (\log 2\left)\right|}{\mathrm{h}}=$

$=\underset{\mathrm{h}\to 0}{\mathrm{Lim}}\frac{|\log 2-\sin (\log 2-\sin \mathrm{h}\left)\right|-|\log 2-\sin (\log 2\left)\right|}{\mathrm{h}}=\underset{\mathrm{h}\to 0}{\mathrm{Lim}}\frac{\sin \left(\log 2\right)-\sin (\log 2-\sin \mathrm{h})}{\mathrm{h}}=$

$=\underset{\mathrm{h}\to 0}{\mathrm{Lim}}\frac{2\cos \left(\frac{2\log 2-\sin \mathrm{h}}{2}\right)\sin \left(\frac{\sin \mathrm{h}}{2}\right)}{\mathrm{h}}=\cos \left(\log 2\right)$

$\mathrm{g}\text{'}\left(0^{-}\right)=\underset{\mathrm{h}\to 0}{\mathrm{Lim}}\frac{\mathrm{f}\left(\mathrm{f}(0-\mathrm{h})\right)-\mathrm{f}\left(\mathrm{f}\left(0\right)\right)}{-\mathrm{h}}=\underset{\mathrm{h}\to 0}{\mathrm{Lim}}\frac{\mathrm{f}(\log 2+\sin \mathrm{h})-\mathrm{f}\left(\log 2\right)}{-\mathrm{h}}$

$=\underset{\mathrm{h}\to 0}{\mathrm{Lim}}\frac{|\log 2-\sin (\log 2+\sin \mathrm{h}\left)\right|-|\log 2-\sin (\log 2\left)\right|}{-\mathrm{h}}$

$=\underset{\mathrm{h}\to 0}{\mathrm{Lim}}\frac{\sin (\log 2+\sin \mathrm{h})-\sin \left(\log 2\right)}{\mathrm{h}}$

$=\underset{\mathrm{h}\to 0}{\mathrm{Lim}}\frac{2\cos \left(\frac{2\log 2+\sin \mathrm{h}}{2}\right)\sin \left(\frac{\sin \mathrm{h}}{2}\right)}{\mathrm{h}}=\cos \left(\log 2\right).$