1. **State the problem:** Differentiate implicitly the equation $$x^2 + y^2 = \log(xy)$$ with respect to $x$. 2. **Recall the rules:** - Derivative of $x^2$ is $2x$. - Derivative of $y^2$ is $2y \frac{dy}{dx}$ by chain rule. - Derivative of $\log(xy)$ uses the chain rule and product rule: $$\frac{d}{dx} \log(xy) = \frac{1}{xy} \frac{d}{dx}(xy)$$ - Derivative of $xy$ is $x \frac{dy}{dx} + y$ by product rule. 3. **Differentiate both sides:** $$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx} \log(xy)$$ $$2x + 2y \frac{dy}{dx} = \frac{1}{xy} (x \frac{dy}{dx} + y)$$ 4. **Multiply both sides to clear denominator and group $\frac{dy}{dx}$ terms:** $$2x + 2y \frac{dy}{dx} = \frac{x}{xy} \frac{dy}{dx} + \frac{y}{xy}$$ $$2x + 2y \frac{dy}{dx} = \frac{1}{y} \frac{dy}{dx} + \frac{1}{x}$$ 5. **Bring all $\frac{dy}{dx}$ terms to one side:** $$2y \frac{dy}{dx} - \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} - 2x$$ 6. **Factor out $\frac{dy}{dx}$:** $$\frac{dy}{dx} \left(2y - \frac{1}{y}\right) = \frac{1}{x} - 2x$$ 7. **Simplify the factor:** $$2y - \frac{1}{y} = \frac{2y^2 - 1}{y}$$ 8. **Solve for $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{\frac{1}{x} - 2x}{\frac{2y^2 - 1}{y}} = \left(\frac{1}{x} - 2x\right) \cdot \frac{y}{2y^2 - 1}$$ 9. **Final answer:** $$\boxed{\frac{dy}{dx} = \frac{y}{2y^2 - 1} \left(\frac{1}{x} - 2x\right)}$$ This is the implicit derivative of the given equation.