If `x^(y). y^(x)=16`, then the value of `(dy)/(dx)` at (2, 2) is

Options

A. `-1`
B. 0
C. 1
D. None of these

Answer

To solve the equation \( x^y \cdot y^x = 16 \) and find the value of \( \frac{dy}{dx} \) at the point (2, 2), we can follow these steps:

### Step 1: Take the logarithm of both sides
We start with the equation:
\[ x^y \cdot y^x = 16 \]
Taking the natural logarithm of both sides gives us:
\[ \log(x^y \cdot y^x) = \log(16) \]

### Step 2: Use logarithmic properties
Using the property of logarithms that states \( \log(a \cdot b) = \log(a) + \log(b) \), we can rewrite the left side:
\[ \log(x^y) + \log(y^x) = \log(16) \]
Using the property \( \log(a^b) = b \cdot \log(a) \), we can further simplify:
\[ y \cdot \log(x) + x \cdot \log(y) = \log(16) \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y \cdot \log(x)) + \frac{d}{dx}(x \cdot \log(y)) = \frac{d}{dx}(\log(16)) \]
Since \( \log(16) \) is a constant, its derivative is 0:
\[ \frac{d}{dx}(y \cdot \log(x)) + \frac{d}{dx}(x \cdot \log(y)) = 0 \]

### Step 4: Apply the product rule
Using the product rule, we differentiate each term:
1. For \( y \cdot \log(x) \):
\[ \frac{dy}{dx} \cdot \log(x) + y \cdot \frac{1}{x} \]
2. For \( x \cdot \log(y) \):
\[ \log(y) + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} \]
Putting it all together, we have:
\[ \frac{dy}{dx} \cdot \log(x) + y \cdot \frac{1}{x} + \log(y) + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} = 0 \]

### Step 5: Substitute \( x = 2 \) and \( y = 2 \)
Now we substitute \( x = 2 \) and \( y = 2 \):
\[ \frac{dy}{dx} \cdot \log(2) + 2 \cdot \frac{1}{2} + \log(2) + 2 \cdot \frac{1}{2} \cdot \frac{dy}{dx} = 0 \]
This simplifies to:
\[ \frac{dy}{dx} \cdot \log(2) + 1 + \log(2) + \frac{dy}{dx} = 0 \]

### Step 6: Combine like terms
Combining the terms involving \( \frac{dy}{dx} \):
\[ \left(\log(2) + 1\right) \frac{dy}{dx} + 1 + \log(2) = 0 \]

### Step 7: Solve for \( \frac{dy}{dx} \)
Rearranging gives:
\[ \frac{dy}{dx} \left(\log(2) + 1\right) = - (1 + \log(2)) \]
Thus,
\[ \frac{dy}{dx} = -\frac{1 + \log(2)}{\log(2) + 1} = -1 \]

### Final Answer
The value of \( \frac{dy}{dx} \) at the point (2, 2) is:
\[ \boxed{-1} \]