Implicit differentiation

Some equations define y as a function of x without expressing it explicitly. The circle x² + y² = 25 is a familiar example: solving for y produces two separate square-root branches, which is awkward to differentiate. Implicit differentiation avoids that step entirely by treating y as an unknown function of x and differentiating the equation as it stands.

The method rests on the chain rule. Whenever you differentiate a term containing y with respect to x, you must multiply by dy/dx. So d/dx of y² is 2y · dy/dx, and d/dx of y³ is 3y² · dy/dx. Applying this to x² + y² = 25 gives 2x + 2y · dy/dx = 0, and solving yields dy/dx = −x / y. Terms containing both variables, such as xy, require the product rule as well: d/dx of xy is y + x · dy/dx.

Notice that the resulting derivative usually depends on both x and y, so you evaluate it at a specific point on the curve. At the point (3, 4) on the circle above, dy/dx = −3/4, which is the slope of the tangent line there. The technique also underpins related rates problems, where an equation links two quantities and both are differentiated with respect to time instead of x, and it is how the derivatives of inverse functions such as arcsin and arctan are derived.

Implicit differentiation is a required topic on the AP Calculus AB exam, appearing in the advanced differentiation unit and again in related rates. Expect free-response questions that ask for dy/dx, a tangent line at a given point, or the points where a curve has a horizontal or vertical tangent — horizontal where the numerator of dy/dx is zero, vertical where the denominator is zero.