If `f(x) = ((2 + x)/(1 + x))^(1 + x)`, then `f'(0)` is equal to

Options

A. 2 log 2
B. log 2
C. 3 log 2 - 1
D. 2 log 2 - 1

Answer

To find \( f'(0) \) for the function \( f(x) = \left( \frac{2 + x}{1 + x} \right)^{1 + x} \), we will follow these steps:

### Step 1: Rewrite the function
Let \( y = f(x) = \left( \frac{2 + x}{1 + x} \right)^{1 + x} \).

### Step 2: Take the natural logarithm
Taking the natural logarithm of both sides gives:
\[ \log y = (1 + x) \log \left( \frac{2 + x}{1 + x} \right) \]

### Step 3: Differentiate both sides
Now, differentiate both sides with respect to \( x \):
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left[ (1 + x) \log \left( \frac{2 + x}{1 + x} \right) \right] \]

Using the product rule on the right-hand side:
\[ \frac{d}{dx} \left[ (1 + x) \log \left( \frac{2 + x}{1 + x} \right) \right] = \log \left( \frac{2 + x}{1 + x} \right) + (1 + x) \frac{d}{dx} \left[ \log \left( \frac{2 + x}{1 + x} \right) \right] \]

### Step 4: Differentiate the logarithm
Using the chain rule:
\[ \frac{d}{dx} \left[ \log \left( \frac{2 + x}{1 + x} \right) \right] = \frac{1}{\frac{2 + x}{1 + x}} \cdot \frac{d}{dx} \left( \frac{2 + x}{1 + x} \right) \]
Using the quotient rule:
\[ \frac{d}{dx} \left( \frac{2 + x}{1 + x} \right) = \frac{(1)(1 + x) - (2 + x)(1)}{(1 + x)^2} = \frac{1 + x - 2 - x}{(1 + x)^2} = \frac{-1}{(1 + x)^2} \]
Thus,
\[ \frac{d}{dx} \left[ \log \left( \frac{2 + x}{1 + x} \right) \right] = \frac{-1}{\frac{2 + x}{1 + x}} \cdot \frac{-1}{(1 + x)^2} = \frac{1 + x}{(2 + x)(1 + x)^2} = \frac{1}{(2 + x)(1 + x)} \]

### Step 5: Substitute back into the derivative
Now substituting back:
\[ \frac{1}{y} \frac{dy}{dx} = \log \left( \frac{2 + x}{1 + x} \right) + (1 + x) \cdot \frac{1}{(2 + x)(1 + x)} \]

### Step 6: Evaluate at \( x = 0 \)
Now we need to evaluate \( f'(0) \):
1. Calculate \( y \) at \( x = 0 \):
\[ y = f(0) = \left( \frac{2 + 0}{1 + 0} \right)^{1 + 0} = 2 \]
2. Calculate \( \log \left( \frac{2 + 0}{1 + 0} \right) \):
\[ \log(2) \]
3. Substitute \( x = 0 \) into the derivative:
\[ \frac{1}{2} \frac{dy}{dx} = \log(2) + 1 \cdot \frac{1}{(2)(1)} = \log(2) + \frac{1}{2} \]
Therefore,
\[ \frac{dy}{dx} = 2 \left( \log(2) + \frac{1}{2} \right) = 2 \log(2) + 1 \]

### Final Answer
Thus, \( f'(0) = 2 \log(2) + 1 \).

---