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

Options

A. `-1`
B. `1`
C. `log 2`
D. `- log 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 \( f(x) \)

Using the chain rule, we know that 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 \( \frac{du}{dx} \)

First, we need to differentiate \( u \):

\[ u = \frac{x^x - x^{-x}}{2} \]

To differentiate \( u \), we apply the quotient rule. The derivative of \( x^x \) is:

\[ \frac{d}{dx}(x^x) = x^x (1 + \log x) \]

And the derivative of \( x^{-x} \) is:

\[ \frac{d}{dx}(x^{-x}) = -x^{-x} (1 + \log x) \]

Thus, we have:

\[ \frac{du}{dx} = \frac{1}{2} \left( x^x (1 + \log x) + x^{-x} (1 + \log x) \right) \]

### Step 3: Substitute \( u \) and \( \frac{du}{dx} \) into \( f'(x) \)

Now substituting \( u \) and \( \frac{du}{dx} \) into the derivative formula, we get:

\[ f'(x) = -\frac{1}{1 + \left( \frac{x^x - x^{-x}}{2} \right)^2} \cdot \frac{1}{2} \left( x^x (1 + \log x) + x^{-x} (1 + \log x) \right) \]

### 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 \( x = 1 \):

\[ \frac{du}{dx} \bigg|_{x=1} = \frac{1}{2} \left( 1^1 (1 + \log 1) + 1^{-1} (1 + \log 1) \right) = \frac{1}{2} \left( 1(1 + 0) + 1(1 + 0) \right) = \frac{1}{2} (1 + 1) = 1 \]

3. Substitute \( u(1) \) and \( \frac{du}{dx} \) into \( f'(1) \):

\[ f'(1) = -\frac{1}{1 + 0^2} \cdot \frac{1}{2} \cdot 1 = -\frac{1}{1} \cdot \frac{1}{2} = -\frac{1}{2} \]

### Final Answer

Thus, \( f'(1) = -\frac{1}{2} \).

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