Source: MTH 132, Michigan State University, Sections 2.6/2.8
Difficulty: Intermediate | Prerequisites: Sections 2.1 through 2.5 (basic differentiation rules, product rule, chain rule, trigonometric derivatives).
Implicit differentiation is one of the core techniques in Calculus I, and it appears any time you cannot (or do not want to) solve an equation for y before differentiating. It extends the chain rule to equations where y is tangled up with x in ways that make isolating y impractical or impossible. You should already be comfortable with the product rule, the chain rule, and the derivatives of sine, cosine, and tangent. If those feel shaky, revisit Sections 2.3 through 2.5 before diving in here.
When an equation mixes x and y and you cannot isolate y easily, differentiate both sides with respect to x, treat every y as a function of x (so each y term picks up a dy/dx via the chain rule), then solve algebraically for dy/dx. The same idea lets you find tangent lines to curves that are not the graph of any single function.
Implicit equation
An equation relating x and y that is not solved for one variable in terms of the other, e.g. x² + y² = 25.
In simple terms, y is "trapped" inside the equation rather than sitting alone on one side.
Implicit differentiation
The process of differentiating both sides of an implicit equation with respect to x, applying the chain rule to every term containing y, and then solving for dy/dx.
Think of it as: "Differentiate everything in sight, remember y depends on x, then isolate dy/dx."
Explicit function
A function written in the form y = f(x), where y is isolated on one side.
In simple terms, you can read off the output directly: plug in x, get y.
Chain rule (in implicit context)
When differentiating a function of y with respect to x, multiply by dy/dx. For example, d/dx[y³] = 3y² · (dy/dx).
Think of it as: the chain rule is the entire engine behind implicit differentiation. Every time you differentiate a y-term, the chain rule forces a dy/dx to appear.
Tangent line (to an implicitly defined curve)
A line that just touches the curve at a given point, with slope equal to dy/dx evaluated at that point. Its equation is y − y₀ = m(x − x₀).
Start with an equation involving x and y (do not try to solve for y).
Differentiate every term on both sides with respect to x.
Pure x-terms: differentiate normally.
Pure y-terms: differentiate with respect to y, then multiply by dy/dx (chain rule).
Mixed xy-terms: use the product rule, and any factor of y still picks up a dy/dx.
Collect all terms containing dy/dx on one side of the equation.
Factor out dy/dx.
Divide both sides by the remaining factor to isolate dy/dx.
Differentiate each term with respect to x:
d/dx[x² cos(y)] requires the product rule: 2x cos(y) + x²(−sin(y)) · dy/dx
d/dx[sin(2y)] requires the chain rule: cos(2y) · 2 · dy/dx
d/dx[xy] requires the product rule: y + x · dy/dx
Putting it together:
2x cos(y) − x² sin(y) · dy/dx + 2 cos(2y) · dy/dx = y + x · dy/dx
Collect dy/dx terms on one side:
−x² sin(y) · dy/dx + 2 cos(2y) · dy/dx − x · dy/dx = y − 2x cos(y)
Factor out dy/dx:
dy/dx · [−x² sin(y) + 2 cos(2y) − x] = y − 2x cos(y)
Solve:
dy/dx = (y − 2x cos(y)) / (−x² sin(y) + 2 cos(2y) − x)
Differentiate each term:
d/dx[3x²y⁵]: product rule gives 6xy⁵ + 3x² · 5y⁴ · dy/dx = 6xy⁵ + 15x²y⁴ · dy/dx
d/dx[tan(y)] = sec²(y) · dy/dx
d/dx[5 sin(x)] = 5 cos(x)
Full equation:
6xy⁵ + 15x²y⁴ · dy/dx + sec²(y) · dy/dx = 5 cos(x)
Collect and factor:
dy/dx · [15x²y⁴ + sec²(y)] = 5 cos(x) − 6xy⁵
Solve:
dy/dx = (5 cos(x) − 6xy⁵) / (15x²y⁴ + sec²(y))
Differentiate:
2x + 4y + 4x · dy/dx + 2y · dy/dx = 0
Collect dy/dx terms:
dy/dx · (4x + 2y) = −2x − 4y
Solve:
dy/dx = (−2x − 4y) / (4x + 2y)
Plug in (2, 1):
dy/dx = (−4 − 4) / (8 + 2) = −8/10 = −4/5
Tangent line using point-slope form:
y − 1 = (−4/5)(x − 2)
Simplified: y = (−4/5)x + 13/5
The implicit differentiation formula pattern
For any term f(y), the derivative with respect to x is:
d/dx[f(y)] = f′(y) · dy/dx
Product rule reminder (for mixed terms)
d/dx[u · v] = u′v + uv′
When one factor involves y, its derivative picks up a dy/dx.
Tangent line at a point (x₀, y₀)
y − y₀ = (dy/dx)|₍ₓ₀,y₀₎ · (x − x₀)
Evaluate dy/dx at the specific point after finding the general expression.
Implicit differentiation is how engineers and physicists work with equations of curves and surfaces that have no tidy "y = ..." form. Pressure-volume-temperature relationships in thermodynamics, orbital mechanics equations, and the shapes of lenses in optics all involve implicit relations where you still need slopes or rates of change.
Students often forget to attach dy/dx when differentiating a y-term. Every single y-term needs it, no exceptions.
Students sometimes apply the product rule but forget that the y-factor's derivative also requires dy/dx. For instance, in d/dx[x²y³], the y³ piece becomes 3y² · dy/dx, not just 3y².
Students occasionally try to solve for y first and then differentiate explicitly. This works for simple equations but fails or gets ugly for most implicit problems, and defeats the purpose of the technique.
Students mix up the sign when collecting dy/dx terms. Careful algebra at the "collect and factor" step prevents most errors.
⚠️ Implicit differentiation problems appear on nearly every Calc I midterm. Expect at least one "find dy/dx" and one "find the tangent line" problem.
⚠️ The tangent-line variant often pairs with verifying that the given point lies on the curve. Plug the point into the original equation first as a quick sanity check.
⚠️ Trig-heavy implicit equations (as in Problems 1a and 1b) are a favourite testing ground because they combine chain rule, product rule, and trig derivatives in a single problem.
True or False: When differentiating y² with respect to x, the result is 2y. Answer: False. It is 2y · dy/dx.
Fill in the blank: d/dx[sin(y)] = ______ · dy/dx. Answer: cos(y)
True or False: You must solve for y before you can find dy/dx. Answer: False. That is exactly what implicit differentiation avoids.
Fill in the blank: To find the tangent line at a point, you need the ______ and the ______ . Answer: slope (dy/dx evaluated at the point) and the point itself.
True or False: d/dx[xy] = x · dy/dx. Answer: False. By the product rule it is y + x · dy/dx.
Q: Find dy/dx for the equation x³ + y³ = 6xy.
A: Differentiate both sides: 3x² + 3y² · dy/dx = 6y + 6x · dy/dx. Collect: (3y² − 6x) · dy/dx = 6y − 3x². Solve: dy/dx = (6y − 3x²) / (3y² − 6x), which simplifies to dy/dx = (2y − x²) / (y² − 2x).
Q: The curve xy + y² = 5 passes through (1, −1+√6). Without computing √6, explain how you would set up the tangent line at that point.
A: Differentiate: y + x · dy/dx + 2y · dy/dx = 0. Solve: dy/dx = −y / (x + 2y). Evaluate at the given point, then write y − y₀ = m(x − 1).
Q: Why does the chain rule matter in implicit differentiation?
A: Because y is a function of x (even though it is not written as one), every time you differentiate a y-expression with respect to x, the chain rule produces a factor of dy/dx. Without it, you would be treating y as a constant and losing the derivative entirely.
Q: Given x² + 4xy + y² = 13, verify that (2, 1) lies on the curve and find the tangent line there.
A: Check: 4 + 8 + 1 = 13. ✓ Differentiating gives dy/dx = (−2x − 4y)/(4x + 2y). At (2,1): dy/dx = −8/10 = −4/5. Tangent line: y − 1 = (−4/5)(x − 2).
This connects directly to Related Rates (Section 2.8), which applies the same chain-rule-with-respect-to-time logic: instead of dy/dx you solve for dy/dt or dx/dt. If implicit differentiation feels comfortable, related rates is the same machinery pointed at a different variable.
It also connects back to the chain rule (Section 2.5), which is the single rule doing all the heavy lifting here. Later in the course, implicit differentiation will reappear in logarithmic differentiation and in finding derivatives of inverse trig functions.
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