Question

If \(y = x^{x^{x^{\dots \text{to infinity}}}}\), prove that \(x \frac{dy}{dx} = \frac{y^2}{1 - y \log x}\).

Hint:

Logarithmic differentiation is the go-to method for functions that have variables in both the base and the exponent (like \(f(x)^{g(x)}\)) or for complex products/quotients. The trick for infinite towers or nested radicals is always to express the function in terms of itself, e.g., \(y = f(x, y)\).

Answer 1

Step 1: Understanding the Concept:
This problem involves an infinite power tower. The key to solving such problems is to recognize the self-similar nature of the expression. The expression in the exponent is the same as the original expression for y. This allows us to write a simpler, implicit equation for y, which can then be differentiated using logarithmic differentiation.
Step 2: Key Formula or Approach:
1. Rewrite the infinite tower in a finite form: \(y = x^y\).
2. Take the natural logarithm of both sides to bring the exponent down: \(\log y = y \log x\).
3. Differentiate both sides of the equation implicitly with respect to \(x\).
4. Solve the resulting equation for \(\frac{dy}{dx}\).
5. Rearrange the expression to match the required form.
Step 3: Detailed Explanation:
Given the function: \(y = x^{x^{x^{\dots}}}\).
Due to the infinite nature of the tower, we can write: \[ y = x^y \] Take the natural logarithm on both sides: \[ \log y = \log(x^y) \] Using the logarithm property \(\log(a^b) = b \log a\), we get: \[ \log y = y \log x \] Now, differentiate both sides with respect to \(x\), using the product rule on the right side: \[ \frac{d}{dx}(\log y) = \frac{d}{dx}(y \log x) \] \[ \frac{1}{y} \frac{dy}{dx} = \left(\frac{dy}{dx} \cdot \log x\right) + \left(y \cdot \frac{1}{x}\right) \] Now, we need to solve for \(\frac{dy}{dx}\). Group the terms containing \(\frac{dy}{dx}\) on one side: \[ \frac{1}{y} \frac{dy}{dx} - (\log x) \frac{dy}{dx} = \frac{y}{x} \] Factor out \(\frac{dy}{dx}\): \[ \frac{dy}{dx} \left(\frac{1}{y} - \log x\right) = \frac{y}{x} \] \[ \frac{dy}{dx} \left(\frac{1 - y \log x}{y}\right) = \frac{y}{x} \] Isolate \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{y}{x} \cdot \frac{y}{1 - y \log x} = \frac{y^2}{x(1 - y \log x)} \] Finally, multiply both sides by \(x\) to get the desired form: \[ x \frac{dy}{dx} = \frac{y^2}{1 - y \log x} \] Step 4: Final Answer:
The relation \(x \frac{dy}{dx} = \frac{y^2}{1 - y \log x}\) has been proven.