We are asked to find the derivative of the function $y = x^{2x}$ using logarithmic differentiation. Let's break it down step by step:

Step 1: Take the natural logarithm of both sides.

Start by applying the natural logarithm to both sides of the equation:

$\ln(y) = \ln(x^{2x})$

Step 2: Simplify the logarithmic expression.

Using the logarithmic property $\ln(a^b) = b \ln(a)$, we can simplify the right-hand side:

$\ln(y) = 2x \ln(x)$

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

Now, we differentiate both sides using implicit differentiation. On the left-hand side, we differentiate $\ln(y)$ using the chain rule. On the right-hand side, we apply the product rule to differentiate $2x \ln(x)$.

$\frac{d}{dx}[\ln(y)] = \frac{1}{y} \cdot \frac{dy}{dx}$

$\frac{d}{dx}[2x \ln(x)] = 2 \ln(x) + 2 \cdot \frac{1}{x}$

So, we have:

$\frac{1}{y} \cdot \frac{dy}{dx} = 2 \ln(x) + 2 \cdot \frac{1}{x}$

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

Now, we multiply both sides by $y$ to isolate $\frac{dy}{dx}$:

$\frac{dy}{dx} = y \left( 2 \ln(x) + \frac{2}{x} \right)$

Step 5: Substitute $y = x^{2x}$ back into the equation.

Since $y = x^{2x}$, we can substitute this back in:

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

Final Answer:

The derivative is:

$\frac{d}{dx}\left( x^{2x} \right) = x^{2x} \left( 2 \ln(x) + \frac{2}{x} \right)$

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Would you like more details or clarifications on any step? Here are some related questions for further exploration:

1. How does logarithmic differentiation work in general?
2. Can logarithmic differentiation be used for other exponential functions?
3. How does the product rule apply to differentiation here?
4. What are some other common examples where logarithmic differentiation is useful?
5. What is the chain rule, and how is it used in this problem?

Tip: Logarithmic differentiation is particularly helpful when you have variables both in the base and the exponent, like in this case.