1. Given functions

We have $f(x)=|\log 2-\sin x|$ and $g(x)=f(f(x)).$

We need to study differentiability of $g$ at $x=0$ and compute $g'(0)$.

2. First compute $f(0)$

Since $\sin 0=0$, $f(0)=|\log 2-0|=\log 2.$ Because $\log 2>0$, this is simply $\log 2$.

3. Simplify $f(x)$ near $x=0$

Since $\sin x$ is small near $0$, and $\log 2\approx 0.693>0$, we have $\log 2-\sin x>0$ for all $x$ sufficiently close to $0$. So near $x=0$, $f(x)=\log 2-\sin x.$ Hence, $f'(x)=-\cos x \quad \text{(near }0\text{)},$ so $f'(0)=-1.$

4. Now study the outer function at the point $f(0)=\log 2$

We need $f'(\log 2)$, because if $f$ is differentiable there, then by chain rule $g'(0)=f'(f(0))\cdot f'(0)=f'(\log 2)\cdot (-1).$

First check the sign of the quantity inside modulus at $x=\log 2$: $\log 2-\sin(\log 2).$ Now since $\sin t<t$ for all $t>0$, $\sin(\log 2)<\log 2.$ Therefore, $\log 2-\sin(\log 2)>0.$ So the modulus is again inactive at $x=\log 2$, and $f$ is differentiable there with $f'(x)=-\cos x$ near $x=\log 2$. Thus, $f'(\log 2)=-\cos(\log 2).$

5. Apply chain rule

Therefore, $g'(0)=f'(f(0))\,f'(0)$ $=\big(-\cos(\log 2)\big)(-1)$ $=\cos(\log 2).$

So $g$ is differentiable at $x=0$, and $g'(0)=\cos(\log 2).$

6. Check options

• A: $g$ is not differentiable at $x=0$ ❌
• B: $g'(0)=\cos(\log 2)$ ✅
• C: $g'(0)=-\cos(\log 2)$ ❌
• D: $g$ is differentiable at $x=0$ and $g'(0)=-\sin(\log 2)$ ❌

Thus the correct option is B.

7. Comparison with stored answer

Stored correct answer: B

Our derived answer matches it.