1. **State the problem:** Find the derivative of the function $f(x) = x^{2x}$. 2. **Recall the formula:** When differentiating functions of the form $y = u(x)^{v(x)}$, use logarithmic differentiation. 3. **Apply logarithmic differentiation:** Take the natural logarithm of both sides: $$\ln y = \ln \left(x^{2x}\right) = 2x \ln x$$ 4. **Differentiate both sides with respect to $x$:** $$\frac{1}{y} \frac{dy}{dx} = 2 \ln x + 2x \cdot \frac{1}{x} = 2 \ln x + 2$$ 5. **Solve for $\frac{dy}{dx}$:** $$\frac{dy}{dx} = y \left(2 \ln x + 2\right) = x^{2x} \left(2 \ln x + 2\right)$$ 6. **Final answer:** $$\boxed{\frac{d}{dx} x^{2x} = x^{2x} \left(2 \ln x + 2\right)}$$