Calculus AB — implicit second derivatives and tangents of implicit curves

Implicit differentiation is usually met through one question: the slope of a circle at a point. The AP Calculus AB exam goes further. A free-response question built on a curve such as x² − xy + y² = 3 typically asks for dy/dx, then for the points where the tangent line is horizontal or vertical, then for d²y/dx² at a point and what it says about the shape of the curve. This material trains those later steps, from topics 3.2 and 3.6 of the course framework, together with the higher-order derivatives of ordinary functions.

The twelve quiz questions begin with the ellipse x² + 4y² = 8. You find d²y/dx² by differentiating dy/dx with the quotient rule, substituting dy/dx back in and simplifying with the curve's own equation until it becomes −1/(2y³); then you use that result at a point on the lower half of the ellipse, where the sign of y³ reverses the concavity. The curve x³ + y³ = 9 gives a tangent line at (1, 2) and its second derivative there, where a term containing dy/dx is the one most often dropped.

The curve x² − xy + y² = 3 gives three questions: the points with a horizontal tangent (numerator of dy/dx equal to zero, point on the curve), the points with a vertical tangent (denominator equal to zero), and what a negative second derivative at a horizontal tangent says about the shape of the curve there. The curve y³ − 3y = x shows a curve with vertical tangents and no horizontal tangent at all. Two more implicit curves, sin y = x and y² = 4x, practice the chain rule inside the second derivative. Two explicit functions close the set: the second derivative of x e²ˣ at 0 and the third derivative of sin 3x at 0.

The wrong options are the slips these problems produce: the first derivative offered as the second, a dropped chain-rule factor, a sign lost on a negative y, horizontal and vertical conditions swapped, points that satisfy the tangent condition but not the curve's equation. Each explanation writes the differentiation out step by step and shows where the substitution happens.

The flashcards give the steps of implicit differentiation and of the implicit second derivative, the conditions for horizontal and vertical tangents, the chain and product rules for y terms, the notation for higher derivatives, concavity from the second derivative, and the point-slope form of a tangent line.

This is independent practice written by Zestly, based on the published AP Calculus AB course framework; it is not produced or endorsed by the College Board.

• Compute d²y/dx² for an implicitly defined curve by substituting dy/dx and simplifying with the curve's equation
• Decide the concavity of an implicit curve at a point
• Write the tangent line to an implicit curve
• Find points with horizontal and vertical tangents on an implicit curve
• Use the second derivative at a horizontal tangent of an implicit curve
• Compute second and third derivatives of explicit functions

Practice material written by Zestly, based on the College Board AP Calculus AB course framework (Unit 3: Differentiation: Composite, Implicit, and Inverse Functions, topics 3.2 and 3.6; applied with Unit 5 concavity), 2026–27 course and exam description.