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 \( y = x^{x^{x^{...}}} \) (where the exponentiation continues infinitely), we can start by recognizing that the expression can be rewritten as:

\[ y = x^y \]

### Step 1: Take the logarithm of both sides
Taking the natural logarithm of both sides gives us:

\[ \log y = \log(x^y) \]

### Step 2: Simplify using logarithmic properties
Using the property of logarithms that states \( \log(a^b) = b \log a \), we can rewrite the equation as:

\[ \log y = y \log x \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \). We will use implicit differentiation on the left side and the product rule on the right side:

\[ \frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx} \]

For the right side, applying the product rule:

\[ \frac{d}{dx}(y \log x) = \frac{dy}{dx} \log x + y \cdot \frac{1}{x} \]

So we have:

\[ \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \log x + \frac{y}{x} \]

### Step 4: Rearranging the equation
Now, we can rearrange the equation to isolate \( \frac{dy}{dx} \):

\[ \frac{1}{y} \frac{dy}{dx} - \frac{dy}{dx} \log x = \frac{y}{x} \]

Factoring out \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} \left( \frac{1}{y} - \log x \right) = \frac{y}{x} \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now, we can solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{y}{x} \cdot \frac{1}{\frac{1}{y} - \log x} \]

### Step 6: Substitute back to find \( x(1 - y \log x) \frac{dy}{dx} \)
Now we need to find \( x(1 - y \log x) \frac{dy}{dx} \):

Substituting \( \frac{dy}{dx} \):

\[ x(1 - y \log x) \frac{dy}{dx} = x(1 - y \log x) \cdot \frac{y}{x} \cdot \frac{1}{\frac{1}{y} - \log x} \]

This simplifies to:

\[ (1 - y \log x) \cdot y \cdot \frac{1}{\frac{1}{y} - \log x} \]

### Final Result
Thus, we have:

\[ x(1 - y \log x) \frac{dy}{dx} = \frac{(1 - y \log x) y}{\frac{1}{y} - \log x} \]