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

Options

A. g'(0)= cos `(log_(e)2)`
B. `g'(0) =- cos (log_(e)2)`
C. g is differentiable at x=0 and g'(0) =- sin `( log_(e)2)`
D. g is not differentiable at x=0

Answer

To solve the problem step by step, we need to find the value of \( g'(0) \) where \( g(x) = f(f(x)) \) and \( f(x) = |\log 2 - \sin x| \).

### Step 1: Define the function \( f(x) \)
We have:
\[ f(x) = |\log 2 - \sin x| \]

### Step 2: Find \( f(f(x)) \)
Next, we need to find \( g(x) = f(f(x)) \). This means we need to substitute \( f(x) \) into itself:
\[ g(x) = f(f(x)) = f(|\log 2 - \sin x|) \]
This requires evaluating \( f(y) \) where \( y = |\log 2 - \sin x| \):
\[ g(x) = |\log 2 - \sin(|\log 2 - \sin x|)| \]

### Step 3: Differentiate \( g(x) \)
To find \( g'(x) \), we will use the chain rule. We need to differentiate \( g(x) \):
\[ g'(x) = f'(f(x)) \cdot f'(x) \]

### Step 4: Find \( f'(x) \)
To differentiate \( f(x) \), we consider two cases based on the value of \( \log 2 - \sin x \):
1. If \( \log 2 - \sin x \geq 0 \), then \( f(x) = \log 2 - \sin x \) and \( f'(x) = -\cos x \).
2. If \( \log 2 - \sin x < 0 \), then \( f(x) = -(\log 2 - \sin x) = \sin x - \log 2 \) and \( f'(x) = \cos x \).

### Step 5: Evaluate \( g'(0) \)
Now we need to evaluate \( g'(0) \):
1. Calculate \( f(0) \):
\[ f(0) = |\log 2 - \sin(0)| = |\log 2 - 0| = \log 2 \]
2. Calculate \( f'(0) \):
\[ f'(0) = -\cos(0) = -1 \]
3. Now substitute into \( g'(x) \):
\[ g'(0) = f'(f(0)) \cdot f'(0) = f'(\log 2) \cdot (-1) \]

### Step 6: Find \( f'(\log 2) \)
We need to determine if \( \log 2 - \sin(\log 2) \) is positive or negative:
- If \( \log 2 - \sin(\log 2) \geq 0 \), then \( f'(\log 2) = -\cos(\log 2) \).
- If \( \log 2 - \sin(\log 2) < 0 \), then \( f'(\log 2) = \cos(\log 2) \).

Assuming \( \log 2 - \sin(\log 2) \geq 0 \) (which we can verify numerically), we have:
\[ g'(0) = -(-\cos(\log 2)) \cdot (-1) = \cos(\log 2) \]

### Final Result
Thus, the final answer is:
\[ g'(0) = \cos(\log 2) \]