Implicit differentiation of tan(y) = x^2 + y

Differentiate both sides of $\tan(y) = x^2 + y$ with respect to $x$. The left side requires the chain rule since $y$ is a function of $x$: $\dfrac{d}{dx}[\tan(y)] = \sec^2(y)\cdot \dfrac{dy}{dx}$. The right side differentiates term by term: $\dfrac{d}{dx}[x^2 + y] = 2x + \dfrac{dy}{dx}$. This gives the equation $\sec^2(y)\cdot \dfrac{dy}{dx} = 2x + \dfrac{dy}{dx}$.

Now collect all terms containing $\dfrac{dy}{dx}$ on one side: $\sec^2(y)\cdot \dfrac{dy}{dx} - \dfrac{dy}{dx} = 2x$. Factor out $\dfrac{dy}{dx}$ to get $\dfrac{dy}{dx}\left(\sec^2(y) - 1\right) = 2x$. Solving for $\dfrac{dy}{dx}$ gives $\dfrac{dy}{dx} = \dfrac{2x}{\sec^2(y) - 1}$, which can also be written as $\dfrac{2x}{\tan^2(y)}$ using the identity $\sec^2(y) - 1 = \tan^2(y)$.