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Given Expression:

$\text{Let } y = x^{x^{x^{...}}}$ $y = x^y$

We are asked to differentiate this expression with respect to $x$.

Solution:

To differentiate $y = x^y$, we'll use implicit differentiation.

1. Take the natural logarithm of both sides: $\ln(y) = \ln(x^y)$ $\ln(y) = y \cdot \ln(x)$

2. Differentiate both sides with respect to $x$: $\frac{1}{y} \cdot \frac{dy}{dx} = \ln(x) \cdot \frac{dy}{dx} + \frac{y}{x}$

3. Isolate $\frac{dy}{dx}$: $\frac{1}{y} \cdot \frac{dy}{dx} - \ln(x) \cdot \frac{dy}{dx} = \frac{y}{x}$ $\frac{dy}{dx} \cdot \left(\frac{1}{y} - \ln(x)\right) = \frac{y}{x}$ $\frac{dy}{dx} = \frac{y^2}{x(1 - y \ln(x))}$

Thus, the derivative of $y$ with respect to $x$ is: $\frac{dy}{dx} = \frac{y^2}{x(1 - y \ln(x))}$

Would you like more details on any of these steps, or do you have any questions?

Expanding Questions:

1. How does implicit differentiation work in general?
2. Can you explain why we took the natural logarithm of both sides?
3. What is the role of logarithmic differentiation in solving this problem?
4. How does the chain rule apply to the differentiation in this problem?
5. What happens to the expression if $x = e$?

Tip:

When differentiating complex functions, especially those involving exponents with variables in both the base and exponent, logarithmic differentiation is often a powerful technique to simplify the process.