If `y=x^(x^(x^(x^(x^(...))))),"then"(dy)/(dx)"is equal to"`

Options

A. `yx^(y-1)`
B. `(y^(2))/(x(1-ylogx))`
C. `(y)/(x(1+ylogx))`
D. none of these

Explanation

To find the derivative \(\frac{dy}{dx}\) for the function defined as \(y = x^{x^{x^{x^{\cdots}}}}\), we can follow these steps:

### Step 1: Rewrite the Function
Since \(y\) is defined as \(x\) raised to the power of itself infinitely, we can express this as:
\[ y = x^y \]

### Step 2: Take the Natural Logarithm
Taking the natural logarithm of both sides gives us:
\[ \log y = \log(x^y) \]
Using the property of logarithms, this simplifies to:
\[ \log y = y \log x \]

### Step 3: Differentiate Both Sides
Now, we differentiate both sides with respect to \(x\). Using implicit differentiation, we have:
\[ \frac{d}{dx}(\log y) = \frac{d}{dx}(y \log x) \]
The left side becomes:
\[ \frac{1}{y} \frac{dy}{dx} \]
The right side requires the product rule:
\[ \frac{d}{dx}(y \log x) = \frac{dy}{dx} \log x + y \frac{1}{x} \]
Thus, we can write:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \log x + \frac{y}{x} \]

### Step 4: Rearrange the Equation
We can rearrange this equation to isolate \(\frac{dy}{dx}\):
\[ \frac{1}{y} \frac{dy}{dx} - \frac{dy}{dx} \log x = \frac{y}{x} \]
Factoring out \(\frac{dy}{dx}\) from the left side gives:
\[ \frac{dy}{dx} \left(\frac{1}{y} - \log x\right) = \frac{y}{x} \]

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

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