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For $x\epsilon R$ , $f(x)=\mid \log2-\sin x\mid$ and $g(x)=f((f(x)), then$  

Answer:

Explanation:

We have $f(x)=\mid\log2-\sin x$  and $g(x)=f(f(x)),x\epsilon R$

   Note that , for $x\rightarrow0, \log2>\sin x$

 $f(x)=\log 2-\sin x$

 $\Rightarrow g(x)=\log 2-\sin(f(x))$

 Clearly g(x) is differentiable at x=0 as sin x is diffenerntiable

 Now    $g'(x)=-\cos (\log 2-\sin x)(-\cos x)$

               =$\cos x.\cos(\log2-\sin x)$

  $\Rightarrow g'(0)=1.\cos(\log 2)$