AP Syllabus focus:
‘Differentiate implicitly defined curves to find dy/dx and use this derivative to determine slopes and equations of tangent lines at specified points.’
Implicitly defined curves often do not solve neatly for y, but differentiation still proceeds through the chain rule, revealing slopes and tangent-line behavior directly from the equation.
Understanding Slopes from Implicit Derivatives
When a relation between x and y is given implicitly—meaning y is not isolated as a function of x—we still seek the derivative dy/dx to determine the slope of the tangent line at a point. Because the variables appear intermixed, implicit differentiation becomes the main tool for uncovering how y changes with respect to x along the curve. This subsubtopic emphasizes using that implicit derivative to compute slopes and form tangent-line equations at specified coordinate points.
Implicit Relations and Differentiation Context
Implicit equations such as circles, ellipses, or more irregular curves define sets of points rather than explicit functional relationships. Still, each differentiable point on such a curve possesses a well-defined tangent line. To uncover its slope, differentiate both sides of the equation with respect to x, apply the chain rule to every y-term, and then isolate dy/dx. This derivative represents the local slope of the implicitly defined curve.
An equation such as x2+y2=25x^2 + y^2 = 25x2+y2=25 can define y implicitly as a function of x, even though the relation itself is not written in the form y=f(x)y = f(x)y=f(x).
A set of graphs illustrating how the relation x2+y2=25x^2 + y^2 = 25x2+y2=25 can represent multiple explicit functions and the full implicit curve. The full circle shows the complete implicit relation, while the semicircles and arcs depict specific explicit branches of yyy. Extra panels extend slightly beyond the syllabus by displaying additional explicit forms derived from the same implicit equation. Source.
The Role of dy/dx in Tangent Lines
A tangent line captures the instantaneous rate of change of the curve at a specified point. Once dy/dx is computed implicitly, evaluating it at an ordered pair that satisfies the original equation yields the slope of that tangent line. Students must recognize that even without an explicit form for y, the slope is still accessible through careful manipulation of the differentiated equation.
Tangent Line Structure
A tangent line has the point–slope form, which depends on the slope value obtained from the implicit derivative and the given point on the curve. Because AP Calculus AB often asks students to justify tangent behavior, understanding exactly how implicit differentiation reveals slope is essential.
= slope of the tangent line at the point
= coordinates where the tangent line is taken
Differentiating Implicit Equations to Obtain Slopes
Executing implicit differentiation requires recognizing that every y-term must be differentiated using the chain rule, because y depends on x. This ensures the derivative captures the variable interplay correctly. The objective is to reorganize the resulting expression so dy/dx is isolated on one side.
Required Use of the Chain Rule
Whenever differentiating an expression containing y, recall that y behaves as an inner function of x. Therefore, its derivative must be multiplied by dy/dx, highlighting the dependence between variables. Failure to apply this step results in incorrect slopes and invalid tangent-line equations.
Implicit Differentiation: A method of differentiating equations involving both x and y by treating y as a function of x and applying the chain rule to each y-term.
A single implicit derivative may involve rearranging multiple algebraic components before dy/dx is isolated. The slope extracted from the final expression is valid only at points belonging to the original equation’s graph.
Determining Tangent Lines at Specified Points
Once dy/dx is obtained, the next task is evaluating it at a specific coordinate point. This point must satisfy the implicit equation; otherwise, no tangent line exists there. Substituting x and y values into the derivative yields a numerical slope, which is then used in the point–slope formula.
After finding dy/dxdy/dxdy/dx implicitly, we substitute the coordinates of the given point into the derivative to obtain the slope of the tangent line at that point.
A diagram of the circle x2+y2=25x^2 + y^2 = 25x2+y2=25 with its tangent line at (3,−4)(3,-4)(3,−4). The slope is computed via the implicit derivative dy/dx=−x/ydy/dx = -x/ydy/dx=−x/y, and the tangent line uses the resulting slope in point–slope form. The explicit equation appears in the image, extending slightly beyond syllabus requirements by showing the final line equation. Source.
Process Overview
• Differentiate the implicit equation with respect to x.
• Apply the chain rule to all y-terms, appending dy/dx appropriately.
• Collect and isolate the dy/dx terms on one side of the equation.
• Solve for dy/dx algebraically.
• Substitute the given point’s coordinates to obtain the specific slope.
• Use the slope and point to form the tangent-line equation.
A critical observation is that some implicit curves may fail to have a tangent line at certain points if the resulting slope is undefined or the derivative expression becomes nonreal. Students should always confirm that the point lies on the curve and that the derivative yields a valid number.
Algebraic Structure of Implicit Derivatives
Implicit derivatives often involve combinations of product, sum, or exponential structures embedded within the relation. While only the chain rule is required conceptually, manipulating the resulting algebra correctly is equally important for obtaining a clean formula for dy/dx. Recognizing opportunities to factor, rearrange, or simplify is part of constructing a usable derivative expression.
= variables related implicitly
Between applying differentiation rules and performing algebra, students strengthen their understanding of how slope behaves on curves that resist explicit functional representation.
Relationship Between Implicit Slopes and Curve Geometry
The slope determined from an implicit derivative tells how steeply the curve rises or falls at a specific point. This geometric interpretation helps connect the algebraic derivative with the curve’s shape in the coordinate plane. A positive slope corresponds to an increasing tangent direction, while a negative slope represents decreasing behavior. Vertical tangents occur when the denominator of the derivative expression approaches zero, signaling extremely steep or undefined slope.
Geometrically, the derivative dy/dxdy/dxdy/dx at a point gives the slope of the unique line that just touches the curve there without cutting across it nearby.
FAQ
A point must satisfy the implicit equation exactly; otherwise, it does not lie on the curve and no tangent line is defined there.
Additionally, the expression for dy/dx must produce a real value. If the derivative yields division by zero or a non-real value, the curve may have a vertical tangent, a cusp, or may be non-differentiable at that point.
A vertical tangent occurs when the denominator of dy/dx equals zero while the numerator does not.
To check this efficiently:
• Compute dy/dx.
• Identify where the denominator becomes zero.
• Verify whether the numerator is non-zero at that point, confirming the tangent is vertical rather than undefined through both numerator and denominator cancelling.
Implicit curves can fail the vertical line test, meaning they may represent more than one y-value for a single x-value.
At a given x-value, different points (different y-values) may lie on the curve, each with its own local slope. As a result, the curve can have distinct tangent lines at multiple points sharing the same x-coordinate.
Non-differentiable points occur when dy/dx cannot be defined or the geometry is too sharp for a tangent line to exist.
Signs of non-differentiability include:
• dy/dx undefined because both numerator and denominator vanish.
• An abrupt change in direction visible when plotting the curve.
• Regions where solving for dy/dx leads to conflicting slopes.
Implicit curves often possess geometric symmetries, such as reflection across axes or rotational symmetry.
Symmetry allows you to infer:
• Whether slopes at symmetric points will be equal or opposite.
• Where horizontal or vertical tangents are likely to appear.
• Whether certain tangent behaviours repeat without full recalculation.
Recognising symmetry can significantly reduce the algebra required when analysing tangent lines.
Practice Questions
Question 1 (1–3 marks)
The curve is defined implicitly by the equation
x² + xy + y² = 7.
(a) Find dy/dx in terms of x and y.
(b) Hence determine the slope of the tangent line at the point (2,1).
Question 1
(a) 2 marks
• Differentiates correctly: 2x + x(dy/dx) + y + 2y(dy/dx) = 0 (1 mark)
• Solves for dy/dx to obtain dy/dx = −(2x + y) / (x + 2y) (1 mark)
(b) 1 mark
• Substitutes (2,1) correctly to obtain slope = −5/4 (1 mark)
Question 2 (4–6 marks)
A curve is defined implicitly by
x³ + y³ = 9xy.
(a) Show that dy/dx = (3y² − 9x) / (9y − 3x²).
(b) The point P(1,2) lies on the curve. Use your result from part (a) to find the equation of the tangent line at P.
(c) State whether the tangent line at P is increasing or decreasing, giving a reason.
Question 2
(a) 2 marks
• Differentiates implicitly: 3x² + 3y²(dy/dx) = 9y + 9x(dy/dx) (1 mark)
• Rearranges to show dy/dx = (3y² − 9x) / (9y − 3x²) (1 mark)
(b) 2 marks
• Substitutes x = 1, y = 2 into derivative to obtain dy/dx = (12 − 9) / (18 − 3) = 3/15 = 1/5 (1 mark)
• Uses point–slope form to give tangent line: y − 2 = (1/5)(x − 1) (1 mark)
(c) 1–2 marks
• States the tangent line is increasing because dy/dx at P is positive (1 mark)
• Clear justification referencing the sign of the derivative at that point (1 mark)
