Question

If \( y = x^{x^{x^{.... \text{ad inf}}}} \), then prove that \( x\frac{dy}{dx} = \frac{y^2}{1 - y \log x} \).

Hint:

Logarithmic differentiation is the standard technique for functions of the form \( y = u(x)^{v(x)} \). The trick for infinite nested functions (like power towers or continued fractions) is to express the whole function in terms of itself to get a simpler, non-infinite equation.

Answer 1

Step 1: Understanding the Concept:
The given function is an infinite power tower. The key to solving such problems is to recognize the repeating pattern. Since the tower is infinite, the exponent of the first x is the same as the entire expression y. This allows us to write a simpler implicit equation, which can then be differentiated.
Step 2: Key Formula or Approach:
1. Rewrite the function using its self-similar property: \( y = x^y \).
2. Use logarithmic differentiation because the function is of the form \( [f(x)]^{g(x)} \).
3. Take the natural logarithm of both sides.
4. Differentiate implicitly with respect to x.
5. Rearrange the resulting equation to find \( \frac{dy}{dx} \).
Step 3: Detailed Explanation or Calculation:
The given function is \( y = x^{x^{x^{....}}} \).
We can write this as:
\[ y = x^y \] Take the natural logarithm (log) of both sides:
\[ \log y = \log(x^y) \] Using the logarithm property \( \log(a^b) = b \log a \):
\[ \log y = y \log x \] Now, differentiate both sides with respect to x, using implicit differentiation and the product rule on the right side:
\[ \frac{d}{dx}(\log y) = \frac{d}{dx}(y \log x) \] \[ \frac{1}{y} \frac{dy}{dx} = \left(\frac{dy}{dx} \cdot \log x\right) + \left(y \cdot \frac{1}{x}\right) \] Rearrange the equation to isolate terms with \( \frac{dy}{dx} \):
\[ \frac{1}{y} \frac{dy}{dx} - \frac{dy}{dx} \log x = \frac{y}{x} \] Factor out \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} \left( \frac{1}{y} - \log x \right) = \frac{y}{x} \] \[ \frac{dy}{dx} \left( \frac{1 - y \log x}{y} \right) = \frac{y}{x} \] Solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{y}{x} \cdot \frac{y}{1 - y \log x} = \frac{y^2}{x(1 - y \log x)} \] Multiply both sides by x to get the required form:
\[ x \frac{dy}{dx} = \frac{y^2}{1 - y \log x} \] Step 4: Final Answer:
The relation \( x\frac{dy}{dx} = \frac{y^2}{1 - y \log x} \) is proved.