Calculus I › Differentiation rules › full formula sheet
Implicit differentiation
When y can’t be isolated, differentiate the whole equation anyway — every y-term earns a dy/dx via the chain rule, then solve.
Notation: dy/dx (or y′) is the derivative of the implicitly defined function y(x). Every time you differentiate a y-term, multiply by dy/dx.
Before this lesson: Chain rule · Power rule
Where it comes from
The circle x² + y² = 25 isn’t a function — it has two y-values for most x. Solving gives y = ±√(25−x²): two functions, each differentiable separately. But splitting every equation is miserable (try solving x² + xy + y³ = 7 for y). Implicit differentiation skips the solving: differentiate the equation as written, treating y as a function of x.
The wrong move: treating y as a constant. d/dx [x²+y²] = 2x + 0 = 2x, set equal to d/dx [25] = 0, gives 2x = 0 — claiming the circle’s slope is zero only at x = 0 and nonsense elsewhere. At (3, 4) the true tangent slope is −3/4 (we’ll compute it); the treat-y-as-constant approach can’t even produce a slope formula. Dead on arrival.
The key idea: y does depend on x (locally, near any point where the curve isn’t vertical), so the chain rule fires on every y-term: d/dx [y²] = 2y·dy/dx — exactly like d/dx [sin²x] = 2 sin x cos x, with y playing the role of sin x.
Derivation
There’s no new theorem — implicit differentiation is the chain rule, applied to an unknown function. The justification, in key steps:
Before reading on: at (3, 4) the circle x² + y² = 25 is descending left-to-right. Should the slope be positive or negative there — and can “treat y as a constant” produce a slope formula at all?
Before reading on: d/dx [yn] treats y as the “inside” of a composition. What does the chain rule say the answer must look like — and why is the result always linear in dy/dx?
Why it’s always linear in dy/dx: each differentiation produces at most one dy/dx per term (chain rule), so the result is a linear equation in dy/dx — always solvable by factoring.
How to use it
The procedure, every time:
- Differentiate every term with respect to x — both sides of the equation.
- y-terms get ×dy/dx (chain rule): d/dx [y³] = 3y²·dy/dx. Write it immediately — never “later.”
- Mixed terms need product/chain too: d/dx [xy] = 1·y + x·dy/dx; d/dx [sin(xy)] = cos(xy)·(y + x·dy/dx).
- Collect all dy/dx terms on one side, factor dy/dx out, divide. The answer usually contains both x and y — that’s normal.
Judgment calls
Use it when solving for y is hard or impossible: x²+xy+y³ = 7, sin(xy) = x, exy = x+y. Don’t use it when y is already isolated: y = x³+2x differentiates directly — implicit would just add dy/dx bookkeeping for nothing.
Worked examples
Four problems, easiest first. dy/dx on every y-term — no exceptions.
Example 1 — the circle: x² + y² = 25
- Differentiate both sides: 2x + 2y·dy/dx = 0. (Why ×dy/dx? Chain rule on y².)
- Solve: 2y·dy/dx = −2x, so dy/dx = −x/y.
- At (3, 4): dy/dx = −3/4. (Why negative? The circle descends left-to-right in the first quadrant.)
- Check via explicit: top half y = √(25−x²), y′ = −x/√(25−x²) = −x/y. Matches ✓
Your turn: The circle: x² + y² = 9
Answer: dy/dx = −x/y
Differentiate both sides: 2x + 2y·dy/dx = 0. Solve: dy/dx = −x/y. At (0, 3): −0/3 = 0 — the top of the circle is flat.
Example 2 — x² + xy + y³ = 7 (product in disguise)
- Term by term: d/dx [x²] = 2x; d/dx [xy] = y + x·dy/dx (product rule!); d/dx [y³] = 3y²·dy/dx; d/dx [7] = 0.
- Equation: 2x + y + x·dy/dx + 3y²·dy/dx = 0.
- Collect dy/dx: dy/dx·(x + 3y²) = −2x − y.
- Solve: dy/dx = −(2x+y)/(x+3y²).
Your turn: x² − xy + y² = 4 (mixed terms need both rules)
Answer: dy/dx = (y − 2x)/(2y − x)
Term by term: 2x − (y + x·dy/dx) + 2y·dy/dx = 0. Collect dy/dx: dy/dx·(2y − x) = y − 2x. Solve: dy/dx = (y − 2x)/(2y − x).
Example 3 — sin(xy) = x (chain inside chain)
- Left side: d/dx [sin(xy)] = cos(xy)·d/dx [xy] = cos(xy)·(y + x·dy/dx).
- Right side: d/dx [x] = 1.
- Equation: cos(xy)·(y + x·dy/dx) = 1.
- Solve: y + x·dy/dx = sec(xy), so dy/dx = [sec(xy) − y]/x = sec(xy)/x − y/x.
Your turn: exy = x (chain inside chain, exponential version)
Answer: dy/dx = (1 − xy)/x²
Left side: exy·(y + x·dy/dx); right side: 1. Since exy = x: x(y + x·dy/dx) = 1, so xy + x²·dy/dx = 1 and dy/dx = (1 − xy)/x².
Example 4 — tangent line to x² + y² = 25 at (3, 4)
- Slope: from Example 1, dy/dx = −x/y = −3/4 at (3, 4).
- Point-slope: y − 4 = −(3/4)(x − 3).
- Tidy: 3x + 4y = 25. (Why nice? Multiply out: 4y = −3x + 25.)
- Check: (3, 4) satisfies 9 + 16 = 25 ✓; the radius to (3,4) has slope 4/3, and −3/4 is its negative reciprocal — perpendicular, as a tangent must be ✓
Your turn: Tangent line to x² − xy + y² = 4 at (2, 2)
Answer: x + y = 4
Differentiate as above: dy/dx = (y − 2x)/(2y − x). At (2, 2): (2−4)/(4−2) = −1. Point-slope: y − 2 = −1(x − 2), so x + y = 4. Check: (2, 2) satisfies 4 − 4 + 4 = 4.
Memorization tips
- dy/dx on every y-term: d/dx [yn] = n·yn−1·dy/dx. Write it the instant you differentiate — “later” means “never.”
- y is the inside: treat y like sin x in a chain — the power rule fires outside, dy/dx inside. Same move, new letter.
- Mixed terms need both rules: xy → product (y + x·dy/dx); sin(xy) → chain then product. Peel layer by layer.
- It’s always linear: dy/dx appears to the first power only — collect, factor, divide. If you see (dy/dx)², something went wrong.
- Answers contain y: dy/dx = −x/y is finished — don’t try to eliminate y. The slope depends on where you stand.
- Differentiate first, plug in last: find the dy/dx formula with x and y, then substitute the point. Numbers too early kill the formula.
Final challenge
Five mixed questions — the dy/dx discipline, products in disguise, and tangent lines. Score 5/5 and it’s yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
What is implicit differentiation?
A technique for equations where y can’t (or shouldn’t) be isolated: differentiate both sides with respect to x, applying the chain rule to every y-term (each earns a ×dy/dx factor), then solve the resulting linear equation for dy/dx.
Why does every y-term get ×dy/dx?
Because y is a function of x (locally), so yⁿ is a composition — exactly like sinⁿx. The chain rule gives d/dx[yⁿ] = n·yⁿ⁻¹·dy/dx. Forgetting it treats y as a constant, which breaks the equation.
How do I differentiate xy?
With the product rule: d/dx[xy] = 1·y + x·dy/dx = y + x·dy/dx. It needs both the product rule (two factors) and the chain rule (y is a function of x).
When should I use implicit differentiation?
When solving for y is hard or impossible: x²+y² = 25, x²+xy+y³ = 7, sin(xy) = x. If y is already isolated (y = x³+2x), differentiate directly — implicit only adds bookkeeping.
Why is the answer allowed to contain y?
Because the slope genuinely depends on the point: on the circle x²+y² = 25, dy/dx = −x/y varies with y. Eliminating y would require solving — the thing we’re avoiding.
More from the codex
Support the codex
This page is free, with no account and no ads. If it helped you learn, consider supporting the indie dev behind it.
Questions or a bug to report? Email [email protected].