if ` x^y . Y^x =1 ` then ` (dy)/(dx)` =

Options

A. `(y(y+x log y))/( x ( y log x +x))`
B. `(y (x +y log x))/( x (y +x log y))`
C. ` -(y (y +x log y))/(x( x+y log x))`
D. None of these

Answer

To solve the problem \( x^y \cdot y^x = 1 \) and find \( \frac{dy}{dx} \), we will use logarithmic differentiation. Here’s the step-by-step solution:

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

### Step 2: Simplify using logarithmic properties
Using the property of logarithms that states \( \ln(a \cdot b) = \ln a + \ln b \), we can rewrite the left side:
\[ \ln(x^y) + \ln(y^x) = 0 \]
Now, applying the power rule of logarithms \( \ln(a^b) = b \ln a \):
\[ y \ln x + x \ln y = 0 \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y \ln x) + \frac{d}{dx}(x \ln y) = 0 \]
Using the product rule for differentiation:
\[ \frac{dy}{dx} \ln x + y \cdot \frac{1}{x} + \left( \ln y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} \right) = 0 \]

### Step 4: Rearranging the equation
This simplifies to:
\[ \frac{dy}{dx} \ln x + \frac{y}{x} + \ln y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} = 0 \]
Now, we can group the terms involving \( \frac{dy}{dx} \):
\[ \left( \ln x + \frac{x}{y} \right) \frac{dy}{dx} + \frac{y}{x} + \ln y = 0 \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now, isolate \( \frac{dy}{dx} \):
\[ \left( \ln x + \frac{x}{y} \right) \frac{dy}{dx} = -\left( \frac{y}{x} + \ln y \right) \]
Thus,
\[ \frac{dy}{dx} = -\frac{\frac{y}{x} + \ln y}{\ln x + \frac{x}{y}} \]

### Final Result
The final expression for \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -\frac{y + x \ln y}{x(\ln x + \frac{x}{y})} \]