For x in mathbb{R},f(x) = |log 2 - sin x| and | vedclass

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(D) Given $f(x) = |\log 2 - \sin x|$. Since $\log 2 \approx 0.693  0$. Thus,$f(x) = \log

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

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(D) Given $f(x) = |\log 2 - \sin x|$. Since $\log 2 \approx 0.693  0$. Thus,$f(x) = \log 2 - \sin x$ in the neighborhood of $x=0$.Then $g(x) = f(f(x)) = \log 2 - \sin(f(x)) = \log 2 - \sin(\log 2 - \sin x)$.Since $f(x)$ is differentiable at $x=0$ and the composition of differentiable functions is differentiable,$g(x)$ is differentiable at $x=0$.Now,$g'(x) = -\cos(\log 2 - \sin x) \cdot (-\cos x) = \cos(\log 2 - \sin x) \cdot \cos x$.Evaluating at $x=0$,$g'(0) = \cos(\log 2 - \sin 0) \cdot \cos 0 = \cos(\log 2) \cdot 1 = \cos(\log 2)$.

For $x \in \mathbb{R}$,$f(x) = |\log 2 - \sin x|$ and $g(x) = f(f(x))$,then which of the following is true?

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