We are tasked with finding $\frac{dy}{dx}$ for the given function $y = (\sqrt{x})^x$. Let's solve it step-by-step:

Step 1: Rewrite the function

Recall that $\sqrt{x} = x^{1/2}$. So, we rewrite: $y = (\sqrt{x})^x = (x^{1/2})^x = x^{x/2}.$

Step 2: Take the natural logarithm (logarithmic differentiation)

Taking the natural logarithm of both sides: $\ln y = \ln(x^{x/2}).$ Using the logarithm rule $\ln(a^b) = b \ln a$: $\ln y = \frac{x}{2} \ln x.$

Step 3: Differentiate both sides with respect to $x$

Differentiate implicitly: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}\left(\frac{x}{2} \ln x\right).$

Use the product rule for $\frac{x}{2} \ln x$: $\frac{d}{dx}\left(\frac{x}{2} \ln x\right) = \frac{1}{2} \ln x + \frac{x}{2} \cdot \frac{1}{x}.$ Simplify: $\frac{d}{dx}\left(\frac{x}{2} \ln x\right) = \frac{1}{2} \ln x + \frac{1}{2}.$

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

Final Answer:

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

Let me know if you need further explanation! Here are some related questions:

1. What happens to the derivative if $x = 1$?
2. How can you simplify the expression further for large values of $x$?
3. Can we apply this process to other forms of $y = (x^a)^x$?
4. How does logarithmic differentiation help with more complex functions?
5. Can this be solved using an alternative method, such as differentiating without logarithms?

Tip: For functions with variables in both the base and the exponent, logarithmic differentiation is often the simplest and most effective approach.