If `y=x^(x^(x^(x...^(oo))))` , then `x(1-ylogx)(dy)/(dx)`

Options

A. `x^(2)`
B. `y^(2)`
C. `xy^(2)`
D. `xy`

Answer

To solve the problem \( y = x^{x^{x^{\cdots}}} \) (where the exponentiation continues infinitely), we can follow these steps:

### Step 1: Define the equation
Since \( y = x^{y} \), we can express the relationship as:
\[ y = x^y \]

### Step 2: Take the logarithm of both sides
Taking the natural logarithm of both sides gives:
\[ \log y = \log(x^y) \]
Using the logarithmic property \( \log(a^b) = b \log a \), we can rewrite this as:
\[ \log y = y \log x \]

### Step 3: Differentiate both sides
Now, we differentiate both sides with respect to \( x \). Using implicit differentiation:
\[ \frac{d}{dx}(\log y) = \frac{d}{dx}(y \log x) \]
Applying the chain rule on the left side and the product rule on the right side, we get:
\[ \frac{1}{y} \frac{dy}{dx} = \log x \frac{dy}{dx} + y \frac{1}{x} \]

### Step 4: Rearranging the equation
Now, we can rearrange this equation to isolate \( \frac{dy}{dx} \):
\[ \frac{1}{y} \frac{dy}{dx} - \log x \frac{dy}{dx} = \frac{y}{x} \]
Factoring out \( \frac{dy}{dx} \) gives us:
\[ \left(\frac{1}{y} - \log x\right) \frac{dy}{dx} = \frac{y}{x} \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{y}{x} \cdot \frac{1}{\frac{1}{y} - \log x} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{y^2}{x(1 - y \log x)} \]

### Step 6: Multiply by \( x(1 - y \log x) \)
Now, we multiply both sides by \( x(1 - y \log x) \):
\[ x(1 - y \log x) \frac{dy}{dx} = y^2 \]

### Final Result
Thus, we have:
\[ x(1 - y \log x) \frac{dy}{dx} = y^2 \]