If `f(x)=(x+1)tan^(-1)(e^(-2x))`, then f'(0) is

Options

A. `(pi)/(2)+1`
B. `(pi)/(4)-1`
C. `(pi)/(6)+5`
D. `(pi)/(4)+1`

Answer

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

### Step 1: Differentiate \( f(x) \)
We will use the product rule for differentiation, which states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then:
\[ (uv)' = u'v + uv' \]
In our case, let:
- \( u(x) = x + 1 \)
- \( v(x) = \tan^{-1}(e^{-2x}) \)

Now we need to find \( u' \) and \( v' \).

### Step 2: Find \( u' \)
\[ u' = \frac{d}{dx}(x + 1) = 1 \]

### Step 3: Find \( v' \)
To differentiate \( v(x) = \tan^{-1}(e^{-2x}) \), we apply the chain rule:
\[ v' = \frac{1}{1 + (e^{-2x})^2} \cdot \frac{d}{dx}(e^{-2x}) \]
Now, differentiate \( e^{-2x} \):
\[ \frac{d}{dx}(e^{-2x}) = e^{-2x} \cdot (-2) = -2e^{-2x} \]
Thus,
\[ v' = \frac{-2e^{-2x}}{1 + e^{-4x}} \]

### Step 4: Apply the product rule
Now we can substitute \( u' \), \( v \), \( u \), and \( v' \) into the product rule:
\[ f'(x) = u'v + uv' = 1 \cdot \tan^{-1}(e^{-2x}) + (x + 1) \cdot \left(\frac{-2e^{-2x}}{1 + e^{-4x}}\right) \]
This simplifies to:
\[ f'(x) = \tan^{-1}(e^{-2x}) - \frac{2(x + 1)e^{-2x}}{1 + e^{-4x}} \]

### Step 5: Evaluate \( f'(0) \)
Now we need to evaluate \( f'(0) \):
\[ f'(0) = \tan^{-1}(e^{-2 \cdot 0}) - \frac{2(0 + 1)e^{-2 \cdot 0}}{1 + e^{-4 \cdot 0}} \]
This simplifies to:
\[ f'(0) = \tan^{-1}(1) - \frac{2 \cdot 1 \cdot 1}{1 + 1} \]
Since \( \tan^{-1}(1) = \frac{\pi}{4} \) and \( e^0 = 1 \):
\[ f'(0) = \frac{\pi}{4} - \frac{2}{2} = \frac{\pi}{4} - 1 \]

### Final Answer
Thus, the value of \( f'(0) \) is:
\[ \boxed{\frac{\pi}{4} - 1} \]