Let's tackle the problem step by step.

Part 1: Using the formula $a^b = e^{b \ln a}$, find $D(x^x)$.

We are tasked with finding the derivative of $y = x^x$. Using logarithmic differentiation, we start by expressing $x^x$ as an exponential:

$x^x = e^{x \ln x}$

Now, apply the chain rule to differentiate $y = e^{x \ln x}$.

1. First, differentiate $e^{x \ln x}$ with respect to $x$: $\frac{d}{dx} e^{x \ln x} = e^{x \ln x} \cdot \frac{d}{dx}(x \ln x)$

2. Now, differentiate $x \ln x$: $\frac{d}{dx}(x \ln x) = \ln x + 1$

3. Combine the results: $\frac{d}{dx}(x^x) = e^{x \ln x} (\ln x + 1)$

4. Recall that $e^{x \ln x} = x^x$, so: $\frac{d}{dx}(x^x) = x^x (\ln x + 1)$

Thus, the derivative of $x^x$ is: $\boxed{x^x (\ln x + 1)}$

Part 2: Logarithmic differentiation of the following functions.

1. $y = \frac{x}{\sqrt{x}}$

Simplify the function first: $y = \frac{x}{\sqrt{x}} = x^{1 - \frac{1}{2}} = x^{\frac{1}{2}}$

Now, differentiate: $\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}}$ Thus: $\boxed{\frac{dy}{dx} = \frac{1}{2\sqrt{x}}}$

2. $y = (x - 1)^2 (x - 3)^5$

Take the natural logarithm of both sides for logarithmic differentiation: $\ln y = \ln\left( (x - 1)^2 (x - 3)^5 \right)$ Using the logarithm property $\ln(ab) = \ln a + \ln b$, we get: $\ln y = 2 \ln(x - 1) + 5 \ln(x - 3)$

Now differentiate both sides: $\frac{1}{y} \frac{dy}{dx} = \frac{2}{x - 1} + \frac{5}{x - 3}$

Multiply both sides by $y = (x - 1)^2 (x - 3)^5$: $\frac{dy}{dx} = (x - 1)^2 (x - 3)^5 \left( \frac{2}{x - 1} + \frac{5}{x - 3} \right)$

Simplify the expression: $\frac{dy}{dx} = (x - 1)(x - 3)^5 \cdot 2 + (x - 1)^2 (x - 3)^4 \cdot 5$

Thus, the derivative is: $\boxed{\frac{dy}{dx} = 2(x - 1)(x - 3)^5 + 5(x - 1)^2 (x - 3)^4}$

Do you need further clarification or details on any of these steps?

Here are 5 related questions:

1. Can you apply logarithmic differentiation to other power functions like $x^{f(x)}$?
2. How would you approach differentiating a product of more than two functions?
3. What is the chain rule, and how does it apply in logarithmic differentiation?
4. Can logarithmic differentiation be used for implicit differentiation problems?
5. How would the derivative change if the function were x^x^x?

Tip: Logarithmic differentiation is particularly useful when dealing with complicated products or powers, as it turns multiplication into addition and powers into simpler forms for differentiation.