Difficulty: Intermediate to Advanced | Prerequisites: All derivative rules including chain rule, critical points and extrema (Doc 1), second derivative test (Doc 2).
So far you have differentiated functions written as y = f(x), where y is isolated on one side. Implicit differentiation handles equations where y is tangled up with x (such as y² − x³ = 1) and you cannot or do not want to solve for y first. Related rates take this one step further: both x and y change over time, and you need to find how fast one quantity is changing given information about the other. These problems are among the most applied on the exam, often involving geometric shapes expanding, sliding, or inflating.
Implicit differentiation means differentiating both sides of an equation with respect to x, treating y as a function of x (so every y-term picks up a dy/dx via the chain rule), then solving for dy/dx. Related rates apply the same idea to quantities that change over time: differentiate a geometric or algebraic relationship with respect to t, substitute known values, and solve for the unknown rate.
Implicit differentiation
Differentiating an equation that defines y implicitly (y is not isolated). You differentiate every term with respect to x, applying the chain rule to y-terms, then solve algebraically for dy/dx. Think of it as: y is secretly a function of x, so whenever you differentiate a y-term, you tack on a dy/dx.
Related rates
Problems where two or more quantities are linked by an equation and each changes over time. You differentiate the linking equation with respect to t to relate the rates of change. Think of it as: the chain rule applied to time, connecting how fast different measurements are changing simultaneously.
Tangent line to an implicitly defined curve
Once you have dy/dx via implicit differentiation, evaluate it at the given point (x₀, y₀) to get the slope, then use point-slope form: y − y₀ = m(x − x₀).
Differentiate both sides of the equation with respect to x.
Every time you differentiate a term containing y, multiply by dy/dx (chain rule).
Collect all dy/dx terms on one side.
Factor out dy/dx and solve.
Worked example (problem 24): y² − 3xy = 2
Differentiate: 2y(dy/dx) − 3[y + x(dy/dx)] = 0
Expand: 2y(dy/dx) − 3y − 3x(dy/dx) = 0
Collect: (2y − 3x)(dy/dx) = 3y
Solve: dy/dx = 3y / (2y − 3x)
Worked example (problem 26): 1/x + 1/y = 2
Rewrite: x⁻¹ + y⁻¹ = 2
Differentiate: −x⁻² − y⁻²(dy/dx) = 0
Solve: dy/dx = −y²/x²
Worked example (problem 27): e^(pq) = C (find dp/dq)
Differentiate with respect to q: e^(pq) · [p + q(dp/dq)] = 0
Since e^(pq) is never zero: p + q(dp/dq) = 0
dp/dq = −p/q
Use dy/dx evaluated at the point, then apply point-slope form.
Worked example (problem 28): y² − x³ = 1 at (2, 3)
Differentiate: 2y(dy/dx) − 3x² = 0, so dy/dx = 3x²/(2y)
At (2, 3): dy/dx = 3(4)/(2·3) = 12/6 = 2
Tangent: y − 3 = 2(x − 2), or y = 2x − 1
Worked example (problem 29): y² + y + 1 = x at (1, −1)
Differentiate: 2y(dy/dx) + dy/dx = 1, so (2y + 1)(dy/dx) = 1, dy/dx = 1/(2y + 1)
At (1, −1): dy/dx = 1/(−2 + 1) = 1/(−1) = −1
Tangent: y − (−1) = −1(x − 1), or y = −x
Worked example (problem 30): x² + y² = xy + 7 at (3, 2)
Differentiate: 2x + 2y(dy/dx) = y + x(dy/dx)
Rearrange: (2y − x)(dy/dx) = y − 2x
dy/dx = (y − 2x)/(2y − x)
At (3, 2): dy/dx = (2 − 6)/(4 − 3) = −4/1 = −4
Tangent: y − 2 = −4(x − 3), or y = −4x + 14
Draw a picture and label variables.
Write down the equation that relates the variables.
Differentiate both sides with respect to time t.
Substitute all known values (including the rates you know).
Solve for the unknown rate.
Worked example (problem 14): Spherical balloon, dV/dt = 100 cm³/sec. Find dr/dt when r = 50.
V = (4/3)πr³
dV/dt = 4πr²(dr/dt)
100 = 4π(50)²(dr/dt) = 10,000π(dr/dt)
dr/dt = 100/(10,000π) = 1/(100π) ≈ 0.00318 cm/sec
Worked example (problem 19): A 13-ft ladder, base sliding away at 5 ft/sec when base is 12 ft from the house.
x² + y² = 13² = 169 (Pythagorean theorem)
When x = 12: y² = 169 − 144 = 25, so y = 5
Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0
2(12)(5) + 2(5)(dy/dt) = 0
120 + 10(dy/dt) = 0
dy/dt = −12 ft/sec. The top of the ladder is sliding down at 12 ft/sec.
Worked example (problem 17): R(x) = 50x − 0.4x², C(x) = 5x + 15, x = 40, dx/dt = 10.
(a) dR/dt = R'(x)·(dx/dt) = (50 − 0.8x)·10 = (50 − 32)·10 = 180 dollars/day
(b) dC/dt = C'(x)·(dx/dt) = 5·10 = 50 dollars/day
(c) dP/dt = dR/dt − dC/dt = 180 − 50 = 130 dollars/day
Implicit differentiation rule of thumb: when you see y, differentiate it normally and attach dy/dx.
Common geometric relationships for related rates:
Circle: A = πr²
Sphere: V = (4/3)πr³, surface area = 4πr²
Pythagorean theorem: x² + y² = L² (ladders, distances)
Rectangle: A = lw, perimeter = 2l + 2w
Point-slope form for tangent lines: y − y₀ = m(x − x₀), where m = dy/dx evaluated at the point.
Related rates with revenue/cost: dR/dt = R'(x) · dx/dt, and similarly for cost and profit.
Related rates problems model real situations where multiple quantities change simultaneously. Engineers use these to track how fast a tank is filling, how quickly a shadow lengthens, or how rapidly pressure changes in a system. The balloon problem is a classic example from physics, and the ladder problem models structural safety considerations. The revenue/cost rate problem mirrors how a business might track daily changes in profitability as production scales.
Students often forget to multiply by dy/dx when differentiating y-terms. Every y-term needs the chain rule applied, because y depends on x.
In related rates, students sometimes plug in numerical values before differentiating. You must differentiate first, then substitute. Plugging in early turns variables into constants and kills the derivative.
Students confuse the sign of dy/dt in ladder problems. If the top is sliding down, dy/dt is negative. Let the algebra tell you the sign rather than guessing.
When finding tangent lines, students sometimes use the wrong point. Make sure the point (x₀, y₀) actually satisfies the original equation.
⚠️ Problems 24 through 26 are pure implicit differentiation. Practise until the chain-rule-on-y step is automatic.
⚠️ Problems 28 through 30 combine implicit differentiation with tangent lines. You need both skills in one problem.
⚠️ Problem 14 (balloon) and problem 19 (ladder) are classic related rates. Memorise the setup patterns.
⚠️ Problem 17 is a business-context related rate. Differentiate R, C, and P with respect to t using the chain rule: dR/dt = R'(x) · dx/dt.
⚠️ Problem 18 (running track) and problem 20 (box) are optimisation problems with geometry constraints. Build the objective function carefully.
Fill in the blank: When differentiating y³ with respect to x, the result is ______ · dy/dx.
3y².
True or False: In a related rates problem, you should substitute known values before differentiating.
False. Always differentiate first, then substitute.
Fill in the blank: The slope of the tangent line to an implicit curve at a point is found by evaluating ______ at that point.
dy/dx.
True or False: If a ladder slides so that dx/dt > 0, then dy/dt must be negative.
True (assuming the ladder length is fixed and the top is against the wall). As the base moves out, the top moves down.
Q: Find dy/dx for y⁴ + x⁴ − 2x²y² = 9.
A: Differentiate: 4y³(dy/dx) + 4x³ − 4xy² − 4x²y(dy/dx) = 0. Collect: (4y³ − 4x²y)(dy/dx) = 4xy² − 4x³. Simplify: dy/dx = (4xy² − 4x³)/(4y³ − 4x²y) = x(y² − x²) / [y(y² − x²)] = x/y (when y² ≠ x²).
Q: A spherical balloon's volume increases at 100 cm³/sec. How fast is the radius changing when r = 50 cm?
A: V = (4/3)πr³, dV/dt = 4πr²(dr/dt). Substituting: 100 = 4π(2500)(dr/dt), dr/dt = 1/(100π) ≈ 0.00318 cm/sec.
Q: A 13-ft ladder has its base 12 ft from the house, sliding out at 5 ft/sec. How fast is the top sliding down?
A: x² + y² = 169. When x = 12, y = 5. Differentiate: 2(12)(5) + 2(5)(dy/dt) = 0. dy/dt = −120/10 = −12 ft/sec. The top slides down at 12 ft/sec.
Q: Find dp/dq for e^(pq) = C.
A: Differentiate with respect to q: e^(pq)[p + q(dp/dq)] = 0. Since e^(pq) ≠ 0: p + q(dp/dq) = 0, so dp/dq = −p/q.
Implicit differentiation is the foundation for related rates: in both cases you differentiate an equation where the variables are not isolated. The optimisation problems from the second doc (running track, box construction) combine constraint equations with objective functions, which is essentially the same "reduce to one variable" technique used in implicit differentiation. Elasticity of demand (extra credit topic) also uses implicit differentiation to relate price and quantity.
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