Let f : (-1,1) `to` R be a differentiable function with f(0) =-1 and f'(0)=1
Let g(x)=`[f(f(2x)+1)]^2` . Then g'(0)=

Options

A. `4`
B. `-4`
C. log 2
D. `-log2`

Answer

To solve the problem step by step, we need to find the derivative \( g'(0) \) for the function defined as \( g(x) = [f(f(2x) + 1)]^2 \).

### Step 1: Define the function
We start with the function:
\[ g(x) = [f(f(2x) + 1)]^2 \]

### Step 2: Differentiate \( g(x) \)
Using the chain rule for differentiation, we have:
\[ g'(x) = 2[f(f(2x) + 1)] \cdot \frac{d}{dx}[f(f(2x) + 1)] \]

### Step 3: Differentiate the inner function
Next, we need to differentiate \( f(f(2x) + 1) \). Again, we will use the chain rule:
\[ \frac{d}{dx}[f(f(2x) + 1)] = f'(f(2x) + 1) \cdot \frac{d}{dx}[f(2x) + 1] \]
Now, differentiating \( f(2x) + 1 \):
\[ \frac{d}{dx}[f(2x) + 1] = f'(2x) \cdot 2 \]

### Step 4: Combine the derivatives
Now we can combine these results:
\[ g'(x) = 2[f(f(2x) + 1)] \cdot f'(f(2x) + 1) \cdot (2f'(2x)) \]
Thus,
\[ g'(x) = 4[f(f(2x) + 1)] \cdot f'(f(2x) + 1) \cdot f'(2x) \]

### Step 5: Evaluate \( g'(0) \)
Now we need to evaluate \( g'(0) \):
1. Calculate \( f(2 \cdot 0) + 1 = f(0) + 1 = -1 + 1 = 0 \).
2. Therefore, \( g'(0) = 4[f(0)] \cdot f'(0) \cdot f'(0) \).
3. Substitute \( f(0) = -1 \) and \( f'(0) = 1 \):
\[ g'(0) = 4 \cdot (-1) \cdot 1 \cdot 1 = -4 \]

### Final Answer
Thus, the value of \( g'(0) \) is:
\[ \boxed{-4} \]