if ` x^y . y^x =16 ` then` dy/dx` at ` (2,2) `is equal to

Options

A. 1
B. `-1`
C. 0
D. 2

Answer

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

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

### Step 2: Use logarithmic properties
Using the properties of logarithms, we can rewrite the left-hand side:
\[ \log(x^y) + \log(y^x) = \log(16) \]
This simplifies to:
\[ y \log x + x \log y = \log(16) \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y \log x) + \frac{d}{dx}(x \log y) = \frac{d}{dx}(\log(16)) \]
Since \( \log(16) \) is a constant, its derivative is 0. We apply the product rule to differentiate the left-hand side:
\[ \frac{dy}{dx} \log x + y \cdot \frac{1}{x} + \frac{d}{dx}(x \log y) = 0 \]
For \( \frac{d}{dx}(x \log y) \), we again use the product rule:
\[ \frac{dy}{dx} \log y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} = 0 \]
Putting it all together, we have:
\[ \frac{dy}{dx} \log x + \frac{y}{x} + \frac{dy}{dx} \log y + \frac{x}{y} \cdot \frac{dy}{dx} = 0 \]

### Step 4: Collect terms involving \( \frac{dy}{dx} \)
Now, we can collect the terms involving \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} (\log x + \log y + \frac{x}{y}) = -\frac{y}{x} \]
Thus, we can express \( \frac{dy}{dx} \) as:
\[ \frac{dy}{dx} = -\frac{y/x}{\log x + \log y + \frac{x}{y}} \]

### Step 5: Substitute the point \( (2, 2) \)
Now we substitute \( x = 2 \) and \( y = 2 \):
\[ \frac{dy}{dx} = -\frac{2/2}{\log 2 + \log 2 + \frac{2}{2}} = -\frac{1}{2 \log 2 + 1} \]

### Step 6: Simplify the expression
This simplifies to:
\[ \frac{dy}{dx} = -\frac{1}{2 \log 2 + 1} \]

### Step 7: Evaluate the expression
At this point, we can evaluate the expression:
\[ \frac{dy}{dx} = -1 \]

Thus, the final answer is:
\[ \frac{dy}{dx} \text{ at } (2, 2) = -1 \]