Most interesting curves cannot be written as y = f(x), and most real machines change several quantities at once. Implicit differentiation is the single manoeuvre — treat y as a function of x, and every y-term grows a y′ factor — behind both of these workhorses of calculus.
1 · The technique: differentiate y-terms as chain rules
When y is defined implicitly by an equation, differentiate both sides with respect to x. Every time you meet y, the chain rule attaches a dy/dx: d/dx(y²) = 2y·y′, d/dx(y³) = 3y²·y′, and d/dx(sin y) = cos y·y′. Then collect y′-terms on one side and solve. Nothing else is new — just patience.
(a) Circle x² + y² = 25 at (3, 4): 2x + 2y·y′ = 0, so y′ = −x/y = −3/4.
(b) Rectangular hyperbola xy = 1: product rule → y + x·y′ = 0, so y′ = −y/x; at (2, ½) the slope is −¼.
(c) Trig implicit sin y = x + y²: cos y·y′ = 1 + 2y·y′ → y′(cos y − 2y) = 1 → y′ = 1/(cos y − 2y).
2 · Finding second derivatives implicitly
Differentiate twice, and the key is substitution, not algebra heroics: after the first pass you own an expression for y′ — feed it back in wherever y′ appears in the second. For y² = 4x: 2y·y′ = 4 gives y′ = 2/y; differentiating again y″ = −2y′/y² = −4/y³, so at (1, 2), y″ = −½ (concave down, as a sideways-opening parabola should be on its upper branch). On the circle, the same loop yields y″ = −r²/y³.
3 · Related rates: the five-step setup
- Draw the situation and label every changing quantity with a variable (never substitute changing numbers yet).
- Write one equation linking the variables — a rate is never in it, a geometry or volume formula always is.
- Check which quantities are constants (ladder length, pole height) and keep them as constants.
- Differentiate every term with respect to time t: x becomes dx/dt, y² becomes 2y·dy/dt.
- Only now substitute the instant's known numbers — position, then rates — and solve.
4 · The four archetypes
(a) Expanding circle: A = πr² → dA/dt = 2πr·dr/dt. If r grows at 3 cm/s, at r = 10 the area grows at 60π ≈ 188.5 cm²/s.
(b) Ladder sliding: x² + y² = 100 (10 ft ladder). If dx/dt = 4 ft/s, then dy/dt = −(x/y)·dx/dt; at x = 6, y = 8, so dy/dt = −3 ft/s — the top descends at 3 ft/s.
(c) Walking person, lamp 6 m, person 2 m, walking at 1.5 m/s: similar triangles give 6s = 2(x + s) where s is shadow length, so s = x/2 and ds/dt = 0.75 m/s. The tip of the shadow moves at dx/dt + ds/dt = 2.25 m/s.
(d) Cone being filled: r = h/2 always (fixed cone), V = πh³/12 → dV/dt = (πh²/4)·dh/dt; at h = 6 with dh/dt = 2 cm/s, dV/dt = 18π ≈ 56.5 cm³/s.
5 · Handling time derivatives
Every variable in a related-rates problem is secretly a function of t: writing dy/dx where the problem means dy/dt is the single most common mechanical slip. Differentiate each piece as a t-derivative — constants survive, variables become a rate — then replace measured values. If the cone in 8.2(d) poured at a known volume rate instead, you would substitute dV/dt and solve for dh/dt.
6 · When the equation will not solve for y
Tough related-rates setups hide behind non-solvable equations — think two boats with the distance between them growing by the law of cosines, or air filling a sphere. The escape is always the same: differentiating implicitly needs no explicit formula, so set up the geometric relation honestly and let the chain rule do the rest.
Substituting known numbers before differentiating silently deletes the rate information — 6 m and 2 m are constants, but x and s are not. Differentiate the symbolic equation first, and only then feed in the snapshot values: position at that instant, not last Tuesday's.
d/dx(yⁿ) = n yⁿ⁻¹·y′, d/dx(sin y) = cos y·y′
x² + y² = r² ⇒ y′ = −x/y, y″ = −r²/y³
related rates: dy/dt = (dy/dx)·(dx/dt)
A = πr² ⇒ dA/dt = 2πr·dr/dt ; V = (1/3)πr²h
second derivative: differentiate twice, substitute y′
Common mistakes
- Forgetting the y′ factor on y-terms — d/dx(y²) is 2y·y′, never 2y.
- Substituting x and y numbers before differentiating, which kills every rate.
- Treating dy/dx as if it were dy/dt (or vice versa) — name the variable before differentiating.
- Using the ladder's constant length as if x or y were constant — only 100 = x² + y² is permanent.
- Confusing the shadow's length rate with the shadow tip's speed, which differ by the walking rate.
- Declaring y′ = 0 at a maximum without checking where the curve genuinely flattens.