Topic 3.2 Implicit Differentiation: find the derivative of a relation defined implicitly by an equation in x and y.

What this topic is asking

The College Board (Topic 3.2) introduces implicit differentiation, a method for finding $\frac{dy}{dx}$ when an equation relates $x$ and $y$ without $y$ being solved for, such as $x^2 + y^2 = 25$ or $x^2 + xy + y^2 = 7$. The key idea is to treat $y$ as an (unknown) function of $x$ and differentiate both sides, using the chain rule every time $y$ appears.

The core idea

A reliable procedure

The method is mechanical once you commit to the chain-rule factors. Differentiate every term, gather all terms containing $\frac{dy}{dx}$ on one side, factor it out, and divide.

When and why you use it

Implicit differentiation is the tool of choice whenever $y$ is entangled with $x$ in a way that makes solving for $y$ messy or impossible - curves like circles, ellipses, and the folium $x^3 + y^3 = 6xy$ above, where $y$ is not a single function of $x$ at all. Even when you could solve for $y$, implicit differentiation is often faster. The result $\frac{dy}{dx}$ typically depends on both $x$ and $y$, which is expected: a point on an implicit curve needs both coordinates to pin down the slope, since the curve may pass through a given $x$-value more than once (a vertical line can cross a circle twice, with opposite slopes). This is also why AP tangent-line questions always give you a specific point $(a, b)$ to substitute.

Reading the answer and the geometry

A derivative like $\frac{dy}{dx} = -\frac{x}{y}$ carries geometric information worth extracting. The tangent is horizontal where the numerator is zero (here $x = 0$) and vertical where the denominator is zero (here $y = 0$), provided the other coordinate satisfies the original equation. AP free-response questions frequently ask exactly this: "find the points where the tangent line is horizontal", which means set the numerator of $\frac{dy}{dx}$ to zero and then solve simultaneously with the original curve. Keeping $\frac{dy}{dx}$ as a single fraction makes these questions clean, because horizontal and vertical tangents read off the numerator and denominator directly. When a question asks for a second derivative $\frac{d^2y}{dx^2}$ on an implicit curve, differentiate $\frac{dy}{dx}$ again implicitly and substitute the first derivative back in - a natural bridge to higher-order derivatives.