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

### Step 1: Differentiate Both Sides
We start with the equation:
\[ x^y \cdot y^x = 16 \]
We will differentiate both sides with respect to \( x \). Using the product rule, we have:
\[ \frac{d}{dx}(x^y \cdot y^x) = \frac{d}{dx}(16) \]
Since the right side is a constant, its derivative is 0.

### Step 2: Apply the Product Rule
Using the product rule on the left side:
\[ \frac{d}{dx}(x^y) \cdot y^x + x^y \cdot \frac{d}{dx}(y^x) = 0 \]

### Step 3: Differentiate \( x^y \) and \( y^x \)
Now we need to differentiate \( x^y \) and \( y^x \):
- For \( x^y \), we use implicit differentiation:
\[ \frac{d}{dx}(x^y) = y \cdot x^{y-1} + x^y \cdot \frac{dy}{dx} \cdot \ln(x) \]
- For \( y^x \), we again use implicit differentiation:
\[ \frac{d}{dx}(y^x) = y^x \cdot \ln(y) + x^{y-1} \cdot \frac{dy}{dx} \]

### Step 4: Substitute Back into the Equation
Substituting these derivatives back into our differentiated equation gives:
\[ \left( y \cdot x^{y-1} + x^y \cdot \frac{dy}{dx} \cdot \ln(x) \right) y^x + x^y \left( y^x \cdot \ln(y) + y^x \cdot \frac{dy}{dx} \cdot \ln(y) \right) = 0 \]

### Step 5: Simplify the Equation
This simplifies to:
\[ y \cdot x^{y-1} \cdot y^x + x^y \cdot y^x \cdot \ln(y) + x^y \cdot y^x \cdot \frac{dy}{dx} \cdot \ln(y) + x^y \cdot y^x \cdot \frac{dy}{dx} \cdot \ln(x) = 0 \]

### Step 6: Solve for \( \frac{dy}{dx} \)
We can factor out \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} \left( x^y \cdot y^x \cdot \ln(y) + x^y \cdot y^x \cdot \ln(x) \right) = -\left( y \cdot x^{y-1} \cdot y^x + x^y \cdot \ln(y) \right) \]
Thus,
\[ \frac{dy}{dx} = -\frac{y \cdot x^{y-1} \cdot y^x + x^y \cdot \ln(y)}{x^y \cdot y^x \cdot (\ln(y) + \ln(x))} \]

### Step 7: Substitute (2, 2) into the Derivative
Now we substitute \( x = 2 \) and \( y = 2 \):
\[ \frac{dy}{dx} = -\frac{2 \cdot 2^{2-1} \cdot 2^2 + 2^2 \cdot \ln(2)}{2^2 \cdot 2^2 \cdot (\ln(2) + \ln(2))} \]
Calculating this gives:
\[ = -\frac{2 \cdot 2 \cdot 4 + 4 \cdot \ln(2)}{16 \cdot 2 \ln(2)} \]
\[ = -\frac{16 + 4 \ln(2)}{32 \ln(2)} \]
\[ = -\frac{1 + \frac{1}{4} \ln(2)}{2 \ln(2)} \]

### Final Result
At the point (2, 2), the value of \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -1 \]