Difficulty: Intermediate | Prerequisites: basic differentiation rules (power, product, quotient, chain rule), equation of a line (point-slope form).
Most curves you meet in early calculus are written as y = f(x), where y is isolated on one side. Implicit differentiation handles the rest: curves defined by an equation mixing x and y together, where solving for y is awkward or impossible. Once you can differentiate implicitly, you can find tangent and normal lines to these curves at any given point, which is one of the most frequently tested skills in a Calculus I exam. This unit also introduces local linearisation, inverse-trig derivatives, and the derivative of an inverse function, all of which lean on the same chain-rule mechanics you use in implicit differentiation.
Implicit differentiation lets you find dy/dx for any equation that mixes x and y, by differentiating both sides with respect to x and solving for y'. You then plug a specific point into that expression to get the slope of the tangent (or, by taking the negative reciprocal, the normal) line at that point.
Implicit differentiation
The process of differentiating an equation that defines y implicitly (not solved for y) with respect to x, treating y as a function of x and applying the chain rule whenever you differentiate a y-term. Think of it as: every time you differentiate something with y in it, a dy/dx (or y') gets tacked on by the chain rule.
Tangent line
A straight line that just touches a curve at a given point and has the same slope as the curve there. In simple terms, it is the line that best approximates the curve at that exact spot.
Normal line
The line perpendicular to the tangent line at the point of tangency. Its slope is the negative reciprocal of the tangent slope. Think of it as: if the tangent goes "along" the curve, the normal goes "through" it at right angles.
Point-slope form
The equation y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is a known point on the line. This is the go-to format for writing tangent and normal line equations in calculus.
Local linearisation (linear approximation)
Using the tangent line at a known, easy-to-evaluate point to estimate the function's value at a nearby point. In simple terms, you replace the curve with its tangent line for a quick, close-enough answer.
The method follows a fixed sequence every time:
Differentiate every term with respect to x. Whenever a term contains y, apply the chain rule: the derivative of y with respect to x is y' (or dy/dx).
Collect all y' terms on one side of the equation, everything else on the other.
Factor out y' from the collected terms.
Solve for y' by dividing both sides by whatever multiplies it.
Worked example (Problem 4A): given x + xy + 2y² = 6, find dy/dx.
Differentiate term by term: 1 + (xy' + y) + 4yy' = 0
Gather y' terms: xy' + 4yy' = -1 - y
Factor: y'(x + 4y) = -1 - y
Solve: y' = (-1 - y) / (x + 4y)
This expression gives the slope of the curve at any point (x, y) that lies on it.
Once you have y' as an expression in x and y, finding the tangent line at a specific point is straightforward:
Plug the point's coordinates into your y' expression to get a numerical slope m.
Write the tangent line in point-slope form: y - y₁ = m(x - x₁).
Example (Problem 1): find the tangent to x⁴ - 12x² + 8y² = 0 at (2, 2).
Differentiate: 4x³ - 24x + 16yy' = 0, so y' = (-x³ + 6x) / (4y).
At (2, 2): y' = (-8 + 12) / 8 = 1/2.
Tangent line: y - 2 = 1/2 (x - 2).
Example (Problem 5): find the tangent to x² - xy + y² = 7 at (-1, 2).
Differentiate: 2x - xy' - y + 2yy' = 0, so y' = (-2x + y) / (-x + 2y).
At (-1, 2): y' = (2 + 2) / (1 + 4) = 4/5.
Tangent line: y - 2 = 4/5 (x + 1).
Finding other points with the same slope (Problem 4C): set y' equal to the known slope value (here, -1/3), solve the resulting equation simultaneously with the original curve equation. In Problem 4, this yields the additional point (6, -3).
The normal line at a point is perpendicular to the tangent there. If the tangent slope is m, the normal slope is -1/m.
Example (Problem 2): the curve x² + 2xy - 3y² = 0 at (1, 1).
Implicit differentiation gives y' = (-x - y) / (x - 3y).
At (1, 1): y' = (-1 - 1) / (1 - 3) = -2 / -2 = 1.
The tangent slope is 1, so the normal slope is -1.
Normal line: y - 1 = -1(x - 1), which simplifies to y = -x + 2.
To find where this normal intersects the curve again, substitute y = -x + 2 into the original equation and solve. This gives x² + 2x(-x + 2) - 3(-x + 2)² = 0, which factors to -4(x - 1)(x - 3) = 0. The other intersection is at x = 3, y = -1, so the point is (3, -1).
Sometimes a problem asks for the tangent line(s) to a curve that are parallel to a given line. Parallel lines share the same slope, so the strategy is:
Read off the slope m of the given line.
Set y' = m and solve (together with the original equation) for the point(s) of tangency.
Write the tangent line equation at each such point.
Example (Problem 3): find tangent lines to y - xy - 2x = 0 that are parallel to x - 2y = 0.
The line x - 2y = 0 rearranges to y = x/2, so m = 1/2.
Implicit differentiation of y - xy - 2x = 0 gives y' = (y + 2) / (1 - x).
Set (y + 2) / (1 - x) = 1/2 and solve: 2y + 4 = 1 - x, so x = -2y - 3.
Substitute into the original equation and solve: 2y² + 8y + 6 = 0, giving y = -1 (point (-1, -1)) and y = -3 (point (3, -3)).
Tangent lines: y + 1 = 1/2 (x + 1) and y + 3 = 1/2 (x - 3).
Tangent line at (x₁, y₁): y - y₁ = y'(x₁, y₁) · (x - x₁)
Normal line at (x₁, y₁): y - y₁ = -1/y'(x₁, y₁) · (x - x₁)
Implicit differentiation shortcut: differentiate F(x, y) = 0 with respect to x, then y' = -Fx / Fy (where Fx and Fy are the partial derivatives of F with respect to x and y).
Linear approximation: f(a + h) ≈ f(a) + f'(a) · h, where a is a point you can evaluate exactly and h is a small step.
Implicit differentiation is how engineers and physicists handle curves that arise naturally in constraint equations, such as the pressure-volume relationship in thermodynamics or the orbit of a planet described by a conic section. Tangent and normal lines appear in optics (the angle of reflection off a curved mirror is measured from the normal) and in computer graphics (surface normals determine how light bounces off 3D objects).
Students often forget to apply the chain rule when differentiating y-terms. Every time you differentiate a function of y with respect to x, a factor of y' must appear. Missing it is the single most common error.
Students sometimes substitute the point into the original equation before differentiating. You must differentiate first, then substitute.
A normal line is not the same as a tangent line with a negated slope. The normal slope is the negative reciprocal (-1/m), not simply -m.
When a problem says "parallel to a given line," students occasionally set y' equal to the negative reciprocal of that line's slope (confusing parallel with perpendicular).
⚠️ Implicit differentiation appears on nearly every Calculus I exam. Expect at least one free-response problem asking for a tangent or normal line to an implicitly defined curve.
⚠️ "Find all points where the tangent is horizontal / vertical" is a classic variant: set the numerator of y' to zero (horizontal) or the denominator to zero (vertical).
⚠️ Multi-part problems (like Problem 4) that ask for the slope expression, a specific tangent line, and then other points with the same slope are a common exam format. Each part builds on the previous one.
True or false: when differentiating x²y implicitly, the result is 2xy + x²y'. ___
True or false: the normal line has slope equal to the negative of the tangent slope. ___
Fill in the blank: to find tangent lines parallel to a given line, set y' equal to ___.
True or false: you should substitute the point's coordinates before differentiating. ___
Fill in the blank: point-slope form is y - y₁ = ___(x - x₁).
Answers: True; False (it is the negative reciprocal, -1/m); the slope of the given line; False (differentiate first); m (the slope).
Q: Given x + xy + 2y² = 6, write the equation of the tangent line at (2, 1).
A: First, y' = (-1 - y)/(x + 4y). At (2, 1): y' = -2/6 = -1/3. Tangent line: y - 1 = -1/3 (x - 2).
Q: For the curve x² + 2xy - 3y² = 0, the normal at (1, 1) meets the curve again at what point?
A: The tangent slope at (1, 1) is 1, so the normal slope is -1. The normal line is y = -x + 2. Substituting into the curve equation and solving gives the other intersection at (3, -1).
Q: Find the tangent to x⁴ - 12x² + 8y² = 0 at (2, 2).
A: y' = (-x³ + 6x)/(4y). At (2, 2): y' = 4/8 = 1/2. Tangent: y - 2 = 1/2 (x - 2).
Q: For y - xy - 2x = 0, find the tangent lines parallel to x - 2y = 0.
A: The reference line has slope 1/2. Setting y' = (y + 2)/(1 - x) = 1/2 and solving with the original equation gives two tangent lines through (-1, -1) and (3, -3), both with slope 1/2.
Implicit differentiation is the chain rule applied in a specific setting, so strong chain-rule fluency is the prerequisite. This same technique reappears in related rates problems (Unit 5 in most Calculus I courses), where you differentiate an equation relating changing quantities with respect to time. It also forms the foundation for multivariable calculus (partial derivatives and gradients), where every variable is treated similarly to how y is treated here.
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