To solve the problem, we need to differentiate $y = \left( \sqrt{x} \right)^x$ with respect to $x$. Let's proceed step-by-step:

Step 1: Rewrite $y$

We start by rewriting the function: $y = \left( \sqrt{x} \right)^x = \left( x^{1/2} \right)^x = x^{x/2}.$

Step 2: Take the natural logarithm on both sides

Taking the natural logarithm simplifies the power: $\ln y = \ln \left( x^{x/2} \right).$ Using the property of logarithms, $\ln(a^b) = b \ln a$, we rewrite: $\ln y = \frac{x}{2} \ln x.$

Step 3: Differentiate implicitly

Now, differentiate both sides with respect to $x$. Using the chain rule on the left side and the product rule on the right side: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left( \frac{x}{2} \ln x \right).$ The derivative of $\frac{x}{2} \ln x$ is: $\frac{d}{dx} \left( \frac{x}{2} \ln x \right) = \frac{1}{2} \ln x + \frac{x}{2} \cdot \frac{1}{x} = \frac{1}{2} \ln x + \frac{1}{2}.$ Thus: $\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \ln x + \frac{1}{2}.$

Step 4: Solve for $\frac{dy}{dx}$

Multiply through by $y$ to isolate $\frac{dy}{dx}$: $\frac{dy}{dx} = y \left( \frac{1}{2} \ln x + \frac{1}{2} \right).$

Substitute $y = x^{x/2}$: $\frac{dy}{dx} = x^{x/2} \left( \frac{1}{2} \ln x + \frac{1}{2} \right).$

Final Answer:

$\boxed{\frac{dy}{dx} = x^{x/2} \left( \frac{1}{2} \ln x + \frac{1}{2} \right)}$

Would you like a detailed explanation of any step or an example application?

Here are 5 related questions to extend this topic:

1. How do you handle derivatives of functions involving exponents where both base and power depend on $x$?
2. How does the chain rule apply when differentiating implicit functions?
3. Can this method be used to solve $y = \left( \sqrt[n]{x} \right)^x$ for any $n$?
4. What happens to the derivative if $y = \left( x^2 \right)^x$?
5. How does logarithmic differentiation help simplify problems with products or powers?

Tip: For complicated powers, logarithmic differentiation is a powerful tool to simplify the process of differentiation.