The expression of `(dy)/(dx)` of the function `y=a^(x^(a^(x...^(oo))))`, is

Options

A. `(y^(2))/(x(1-ylogx))`
B. `(y^(2)logy)/(x(1-ylogx))`
C. `(y^(2)logy)/(x(1-ylogxlogy))`
D. `(y^(2)logy)/(x(1+ylogxlogy))`

Answer

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

### Step 1: Define the function
We start with the equation:
\[ y = a^{x^{a^{x^{\cdots}}}} = a^{x^y} \]

### Step 2: Take the natural logarithm
Taking the natural logarithm of both sides, we have:
\[ \log y = x^y \log a \]

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

### Step 4: Rearranging the equation
Now, let's rearrange the equation to isolate \(\frac{dy}{dx}\):
\[ \frac{1}{y} \frac{dy}{dx} - x \log y \frac{dy}{dx} = \log a \cdot y \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} \left( \frac{1}{y} - x \log y \right) = \log a \cdot y \]

### Step 5: Solve for \(\frac{dy}{dx}\)
Now we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\log a \cdot y}{\frac{1}{y} - x \log y} \]
Multiplying the numerator and denominator by \(y\):
\[ \frac{dy}{dx} = \frac{y \log a}{1 - xy \log y} \]

### Final Expression
Thus, the expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{y \log a}{1 - xy \log y} \]

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