For $$x \in R$$, $$f(x) = |\log 2 - \sin x|$$ and $$g(x) = f(f(x))$$, then

We have been given two real‐valued functions

$$f(x)=\left|\log 2-\sin x\right|\;,\qquad g(x)=f\!\bigl(f(x)\bigr).$$

Our task is to investigate the differentiability of $$g$$ at $$x=0$$ and, if possible, compute $$g'(0).$$

First of all we inspect the inner expression of the absolute value that defines $$f(x).$$ Put

$$h(x)=\log 2-\sin x.$$

At $$x=0$$ we get

$$h(0)=\log 2-\sin 0=\log 2-0=\log 2.$$

Because $$\log 2\approx0.6931>0,$$ the quantity $$h(x)$$ is strictly positive for all $$x$$ sufficiently close to $$0$$ (in fact whenever $$|\sin x|\lt\log 2,$$ i.e. for all small $$x$$).

Whenever $$h(x)>0,$$ the absolute value does nothing, so in that neighbourhood

$$f(x)=h(x)=\log 2-\sin x.$$

Therefore, for $$|x|$$ small, $$f$$ is an ordinary differentiable function and we may differentiate term by term:

$$f'(x)=0-\cos x=-\cos x.$$

Evaluating this derivative at $$x=0$$ we obtain

$$f'(0)=-\cos 0=-1.$$

Next, in order to differentiate the composite function $$g(x)=f(f(x)),$$ we will need the derivative of $$f$$ at the point $$y=f(0).$$ We first compute that point:

$$y=f(0)=\left|\log 2-\sin 0\right|=\left|\log 2\right|=\log 2.$$

Just as before, we must check the sign of the expression inside the absolute value when the input is $$\log 2.$$ We calculate

$$h(\log 2)=\log 2-\sin(\log 2).$$

Numerically, $$\sin(\log 2)\approx\sin(0.6931)\approx0.6390,$$ so

$$h(\log 2)\approx0.6931-0.6390\approx0.0541>0.$$

Since the difference is again positive, the absolute value once more leaves the sign unchanged near $$x=\log 2,$$ and thus around that point we still have

$$f(x)=\log 2-\sin x,$$

whence

$$f'(x)=-\cos x\quad\text{for inputs near }x=\log 2.$$

In particular,

$$f'(\log 2)=-\cos(\log 2).$$

We are now ready to apply the chain rule. Recall the chain rule for derivatives of a composite function:

$$\bigl(F\circ H\bigr)'(x)=F'\bigl(H(x)\bigr)\,H'(x).$$

In our case $$F=f$$ and $$H=f,$$ so for every $$x$$ at which both derivatives exist,

$$g'(x)=f'\bigl(f(x)\bigr)\,f'(x).$$

We have already established that both derivatives exist at $$x=0,$$ so substituting $$x=0$$ gives

$$g'(0)=f'\bigl(f(0)\bigr)\,f'(0)=f'(\log 2)\times(-1).$$

Replacing $$f'(\log 2)$$ by the value found above, we get

$$g'(0)=\bigl(-\cos(\log 2)\bigr)\times(-1)=\cos(\log 2).$$

Because a single, well‐defined value for the derivative exists, the function $$g$$ is certainly differentiable at $$x=0.$$ Hence the numerical value of the derivative is

$$g'(0)=\cos(\log 2).$$

Among the listed alternatives this matches Option D.

Hence, the correct answer is Option D.