If `f(x)=cot^(-1)((x^(x)-x^(-x))/(2))` then f'(1) equals

Options

A. `-1`
B. 1
C. `log_(e)2`
D. `-log_(e)2`

Answer

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

### Step 1: Differentiate the function
We start by differentiating \( f(x) \) using the chain rule. The derivative of \( \cot^{-1}(u) \) is given by:
\[ f'(x) = -\frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \frac{x^x - x^{-x}}{2} \).

### Step 2: Find \( u \) and its derivative
First, we need to differentiate \( u \):
\[ u = \frac{x^x - x^{-x}}{2} \]
To differentiate \( u \), we use the product rule and the chain rule. The derivative of \( x^x \) is:
\[ \frac{d}{dx}(x^x) = e^{x \ln x} \cdot (1 + \ln x) = x^x (1 + \ln x) \]
The derivative of \( x^{-x} \) is:
\[ \frac{d}{dx}(x^{-x}) = -x^{-x} (1 + \ln x) \]
Thus, the derivative of \( u \) is:
\[ \frac{du}{dx} = \frac{1}{2} \left( x^x (1 + \ln x) + x^{-x} (1 + \ln x) \right) \]

### Step 3: Substitute \( u \) and \( \frac{du}{dx} \) into \( f'(x) \)
Now substituting \( u \) and \( \frac{du}{dx} \) into the derivative:
\[ f'(x) = -\frac{1}{1 + \left(\frac{x^x - x^{-x}}{2}\right)^2} \cdot \frac{1}{2} \left( x^x (1 + \ln x) + x^{-x} (1 + \ln x) \right) \]

### Step 4: Evaluate \( f'(1) \)
Now we need to evaluate \( f'(1) \):
1. Calculate \( u \) at \( x = 1 \):
\[ u(1) = \frac{1^1 - 1^{-1}}{2} = \frac{1 - 1}{2} = 0 \]
2. Calculate \( \frac{du}{dx} \) at \( x = 1 \):
\[ \frac{du}{dx}(1) = \frac{1}{2} \left( 1 (1 + 0) + 1 (1 + 0) \right) = \frac{1}{2} \cdot 2 = 1 \]
3. Substitute \( u(1) \) into \( f'(x) \):
\[ f'(1) = -\frac{1}{1 + 0^2} \cdot 1 = -1 \]

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