1. **Problem:** Prove that $$\frac{dy}{dx} = x^{\sin x - 1} \left( \sin x + x \cos x \ln x \right)$$ is the derivative of $$y = x^{\sin x}$$. 2. **Formula and rules:** For functions of the form $$y = u^v$$ where both $$u$$ and $$v$$ depend on $$x$$, use logarithmic differentiation: $$\ln y = v \ln u$$ Differentiate both sides using the product and chain rules: $$\frac{1}{y} \frac{dy}{dx} = \frac{dv}{dx} \ln u + v \frac{1}{u} \frac{du}{dx}$$ 3. **Apply to given function:** Let $$y = x^{\sin x}$$, so $$u = x$$ and $$v = \sin x$$. 4. **Differentiate:** $$\ln y = \sin x \cdot \ln x$$ Differentiate both sides: $$\frac{1}{y} \frac{dy}{dx} = \cos x \ln x + \sin x \cdot \frac{1}{x}$$ 5. **Solve for $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} = y \left( \cos x \ln x + \frac{\sin x}{x} \right) = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right)$$ 6. **Rewrite expression:** Multiply inside the parentheses by $$\frac{x}{x}$$ to get a common denominator: $$x^{\sin x} \cdot \frac{x \cos x \ln x + \sin x}{x} = x^{\sin x - 1} \left( \sin x + x \cos x \ln x \right)$$ 7. **Conclusion:** Thus, the derivative is $$\frac{dy}{dx} = x^{\sin x - 1} \left( \sin x + x \cos x \ln x \right)$$ which matches the given expression. Final answer: $$\boxed{\frac{dy}{dx} = x^{\sin x - 1} \left( \sin x + x \cos x \ln x \right)}$$