If `xlog_(e)y+ylog_(e)x=5`, then `(dy)/(dx)` is

Options

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

Answer

To find \(\frac{dy}{dx}\) from the equation \(x \log_e y + y \log_e x = 5\), we will differentiate both sides of the equation with respect to \(x\).

### Step-by-Step Solution:

1. **Differentiate both sides of the equation**:
\[ \frac{d}{dx}(x \log_e y + y \log_e x) = \frac{d}{dx}(5) \]
The right side differentiates to 0 since 5 is a constant.

2. **Apply the product rule**:
For the term \(x \log_e y\):
\[ \frac{d}{dx}(x \log_e y) = x \cdot \frac{d}{dx}(\log_e y) + \log_e y \cdot \frac{d}{dx}(x) \]
The derivative of \(\log_e y\) with respect to \(x\) is \(\frac{1}{y} \frac{dy}{dx}\) (using the chain rule), and the derivative of \(x\) is 1. Thus,
\[ \frac{d}{dx}(x \log_e y) = x \cdot \frac{1}{y} \frac{dy}{dx} + \log_e y \]

For the term \(y \log_e x\):
\[ \frac{d}{dx}(y \log_e x) = y \cdot \frac{d}{dx}(\log_e x) + \log_e x \cdot \frac{d}{dx}(y) \]
The derivative of \(\log_e x\) is \(\frac{1}{x}\), so we have:
\[ \frac{d}{dx}(y \log_e x) = y \cdot \frac{1}{x} + \log_e x \cdot \frac{dy}{dx} \]

3. **Combine the derivatives**:
Now we can combine the derivatives:
\[ x \cdot \frac{1}{y} \frac{dy}{dx} + \log_e y + y \cdot \frac{1}{x} + \log_e x \cdot \frac{dy}{dx} = 0 \]

4. **Rearrange the equation**:
Group the terms involving \(\frac{dy}{dx}\):
\[ \left( x \cdot \frac{1}{y} + \log_e x \right) \frac{dy}{dx} + \log_e y + \frac{y}{x} = 0 \]

5. **Isolate \(\frac{dy}{dx}\)**:
Move the constant terms to the right side:
\[ \left( x \cdot \frac{1}{y} + \log_e x \right) \frac{dy}{dx} = -\left( \log_e y + \frac{y}{x} \right) \]
Now, divide both sides by \(\left( x \cdot \frac{1}{y} + \log_e x \right)\):
\[ \frac{dy}{dx} = -\frac{\log_e y + \frac{y}{x}}{x \cdot \frac{1}{y} + \log_e x} \]

6. **Simplify the expression**:
We can rewrite the expression:
\[ \frac{dy}{dx} = -\frac{y \log_e y + y^2/x}{x + y \log_e x} \]
This can be factored further:
\[ \frac{dy}{dx} = -\frac{y}{x} \cdot \frac{x \log_e y + y}{x + y \log_e x} \]

### Final Answer:
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = -\frac{y}{x} \left( x \log_e y + y \log_e x \right) \]