if `y^x-x^y=1` then the value of `(dy)/(dx)` at `x=1` is

Options

A. `2(1-log2)`
B. `2(1+log2)`
C. `2-log2`
D. `2+log2`

Explanation

To solve the problem \( y^x - x^y = 1 \) and find the value of \( \frac{dy}{dx} \) at \( x = 1 \), we will follow these steps:

### Step 1: Differentiate the equation implicitly
We start with the equation:
\[ y^x - x^y = 1 \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(y^x) - \frac{d}{dx}(x^y) = 0 \]

### Step 2: Apply the product rule and chain rule
Using the chain rule and product rule for differentiation:
- For \( y^x \):
\[ \frac{d}{dx}(y^x) = y^x \ln(y) + y^x \frac{dy}{dx} \cdot x \]
- For \( x^y \):
\[ \frac{d}{dx}(x^y) = x^y \ln(x) + x^y \frac{dy}{dx} \cdot y \]

So, we have:
\[ y^x \ln(y) + y^x x \frac{dy}{dx} - \left( x^y \ln(x) + x^y y \frac{dy}{dx} \right) = 0 \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ y^x \ln(y) + y^x x \frac{dy}{dx} = x^y \ln(x) + x^y y \frac{dy}{dx} \]

### Step 4: Isolate \( \frac{dy}{dx} \)
Now we can isolate \( \frac{dy}{dx} \):
\[ y^x x \frac{dy}{dx} - x^y y \frac{dy}{dx} = x^y \ln(x) - y^x \ln(y) \]
Factoring out \( \frac{dy}{dx} \):
\[ \left( y^x x - x^y y \right) \frac{dy}{dx} = x^y \ln(x) - y^x \ln(y) \]
Thus,
\[ \frac{dy}{dx} = \frac{x^y \ln(x) - y^x \ln(y)}{y^x x - x^y y} \]

### Step 5: Evaluate at \( x = 1 \)
Now we need to find \( y \) when \( x = 1 \):
\[ y^1 - 1^y = 1 \implies y - 1 = 1 \implies y = 2 \]

Substituting \( x = 1 \) and \( y = 2 \) into the expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1^2 \ln(1) - 2^1 \ln(2)}{2^1 \cdot 1 - 1^2 \cdot 2} \]
Calculating the terms:
- \( \ln(1) = 0 \)
- \( 1^2 = 1 \)
- \( 2^1 = 2 \)

Thus,
\[ \frac{dy}{dx} = \frac{0 - 2 \ln(2)}{2 - 2} = \frac{-2 \ln(2)}{0} \]

### Step 6: Conclusion
Since the denominator is zero, we need to analyze the limit or behavior around this point. However, if we analyze the original equation, we see that as \( x \) approaches 1, the function behaves consistently, leading to a vertical tangent. Thus, we conclude that:
\[ \frac{dy}{dx} \text{ is undefined at } x = 1. \]