Question
For $$x\,I\,{ R},\,f\left( x \right) = \left| {\log \,2 - \sin \,x} \right|$$ and $$g\left( x \right) = f\left( {f\left( x \right)} \right),$$ then :
A.
$$g'\left( 0 \right) = - \cos \left( {\log \,2} \right)$$
B.
$$g$$ is differentiable at $$x = 0$$ and $$g'\left( 0 \right) = - \sin \left( {\log \,2} \right)$$
C.
$$g$$ is not differentiable at $$x =0$$
D.
$$g'\left( 0 \right) = \cos \left( {\log \,2} \right)$$
Answer :
$$g'\left( 0 \right) = \cos \left( {\log \,2} \right)$$
Solution :
$$g\left( x \right) = f\left( {f\left( x \right)} \right)$$
In the neighborhood of $$x = 0,$$
$$\eqalign{
& f\left( x \right) = \left| {\log \,2 - \sin \,x} \right| = \left( {\log \,2 - \sin \,x} \right) \cr
& \backslash \,\,g\left( x \right) = \left| {\log \,2 - \sin \,\left| {\log \,2 - \sin \,x} \right|} \right| \cr
& = \left( {\log \,2 - \sin \left( {\log \,2 - \sin \,x} \right)} \right) \cr
& \backslash \,\,g\left( x \right) = {\text{is differentiable}} \cr
& {\text{and }}g'\left( x \right) = - \cos \left( {\log \,2 - \sin \,x} \right)\left( { - \cos \,x} \right) \cr
& \Rightarrow g'\left( 0 \right) = \cos \left( {\log \,2} \right) \cr} $$