Implicit differentiation of a logarithmic equation

Differentiate both sides with respect to $x$. The left side requires the chain rule since $\ln(x+y)$ is a composite function: $\frac{1}{x+y}\left(1 + \frac{dy}{dx}\right) = 2x - \frac{dy}{dx}.$ Here the inner function is $x+y$ and $\frac{d}{dx}[y] = \frac{dy}{dx}$ by the implicit differentiation rule.

Expand the left side to get $\dfrac{1}{x+y} + \dfrac{1}{x+y}\dfrac{dy}{dx} = 2x - \dfrac{dy}{dx}$. Move all $\dfrac{dy}{dx}$ terms to one side: $\dfrac{1}{x+y}\dfrac{dy}{dx} + \dfrac{dy}{dx} = 2x - \dfrac{1}{x+y}$. Factor out $\dfrac{dy}{dx}$: $\dfrac{dy}{dx}\left(\dfrac{1}{x+y}+1\right) = 2x - \dfrac{1}{x+y}$. Simplify the left side by combining fractions: $\dfrac{dy}{dx}\left(\dfrac{1+x+y}{x+y}\right) = 2x - \dfrac{1}{x+y}$. Multiply both sides by $(x+y)$: $\frac{dy}{dx}(1+x+y) = 2x(x+y) - 1.$ Therefore: $\frac{dy}{dx} = \frac{2x(x+y) - 1}{x+y+1}.$

The answer is the chain rule. Since $y$ is treated as a function of $x$, we apply the chain rule: $\frac{d}{dx}[y] = \frac{dy}{dx}$. This is the core mechanism of implicit differentiation—applying the chain rule to expressions involving $y$ because $y$ is implicitly a function of $x$.