Difficulty: Intermediate to Advanced | Prerequisites: Sections 1 through 4 of Chapter 2 (limit definition, all differentiation rules, trig derivatives). Related rates also requires comfort with geometry formulas.
These three topics build on each other. The chain rule lets you differentiate composite functions (functions inside functions). Implicit differentiation uses the chain rule on equations where y is not isolated. Related rates use implicit differentiation on equations relating physical quantities that change with time. Together they unlock the hardest derivative problems in the course.
The chain rule says: to differentiate f(g(x)), take the derivative of the outer function evaluated at the inner, then multiply by the derivative of the inner. Implicit differentiation applies the chain rule to both sides of an equation, treating y as a function of x, then solves for dy/dx. Related rates translate a word problem into an equation connecting changing quantities, differentiate with respect to time, and solve for the unknown rate.
Composition of functions
f(g(x)), also written (f ∘ g)(x). The output of g becomes the input of f. Think of it as one function eating another.
Chain rule
If F(x) = f(g(x)), then F'(x) = f'(g(x)) · g'(x). In Leibniz notation: dF/dx = (dF/du) · (du/dx) where u = g(x). In simple terms, differentiate the outside, leave the inside alone, then multiply by the derivative of the inside.
Implicit differentiation
A method of differentiating both sides of an equation with respect to x (even when the equation is not solved for y), then solving for y' = dy/dx. Think of it as treating y as a hidden function of x and applying the chain rule every time you differentiate a term containing y.
Related rates
Problems where two or more quantities are linked by an equation and all change with time. You differentiate the equation with respect to t, plug in known values and rates, and solve for the unknown rate. In simple terms, it is implicit differentiation applied to real-world motion and growth problems.
Before applying the chain rule, identify the outer and inner functions. For h(x) = sin(x² + x), the outer function is f(u) = sin u and the inner function is g(x) = x² + x.
For h(x) = (7x³ + 2x² − x + 3)⁵, the outer is f(u) = u⁵ and the inner is g(x) = 7x³ + 2x² − x + 3.
F'(x) = f'(g(x)) · g'(x)
In words: differentiate the outer function, evaluate it at the inner function, then multiply by the derivative of the inner function.
In Leibniz notation with u = g(x): dF/dx = (dF/du) · (du/dx). It looks as though the du terms cancel, which is a handy way to remember the rule (though it is not a true fraction).
h(x) = sin(x² + x): outer = sin u, inner = x² + x. h'(x) = cos(x² + x) · (2x + 1).
f(x) = (7x³ + 2x² − x + 3)⁵: h'(x) = 5(7x³ + 2x² − x + 3)⁴ · (21x² + 4x − 1).
g(x) = tan(3x² + 5x + 8): g'(x) = sec²(3x² + 5x + 8) · (6x + 5).
y = √(x + √x): Rewrite as (x + x^(1/2))^(1/2). Outer = u^(1/2), inner = x + x^(1/2). y' = (1/2)(x + √x)^(−1/2) · (1 + 1/(2√x)).
Given a table of values for f, g, f', g' at specific x-values, you can compute derivatives of compositions by plugging into the chain rule formula. For h(x) = f(g(x)), h'(1) = f'(g(1)) · g'(1). Read g(1) from the table, find f' at that value, multiply by g'(1).
For deeply composed functions, apply the chain rule from the outside in. f(x) = 3·tan(2·cos(5x)): the outermost is 3·tan(u), the middle is u = 2·cos(v), the innermost is v = 5x. f'(x) = 3·sec²(2·cos(5x)) · (−2·sin(5x)) · 5 = −30·sin(5x)·sec²(2·cos(5x)).
Some curves are defined by equations like x² + y² = 2 or x³ + y³ = 6xy that cannot easily (or at all) be solved for y as a function of x. Implicit differentiation lets you find dy/dx without isolating y first.
Differentiate both sides of the equation with respect to x.
Every time you differentiate a term involving y, apply the chain rule: the derivative of y^n is n·y^(n−1)·(dy/dx), because y is implicitly a function of x.
Collect all terms containing dy/dx on one side.
Factor out dy/dx and solve.
Differentiate both sides: 3x² + 3y²·(dy/dx) = 6y + 6x·(dy/dx).
Collect dy/dx terms: 3y²·(dy/dx) − 6x·(dy/dx) = 6y − 3x².
Factor: (dy/dx)(3y² − 6x) = 6y − 3x².
Solve: dy/dx = (6y − 3x²) / (3y² − 6x) = (2y − x²) / (y² − 2x).
At the point (3,3): dy/dx = (6 − 9)/(9 − 6) = −1. Tangent line: y − 3 = −1(x − 3).
Horizontal tangent in the first quadrant: set dy/dx = 0, so 2y − x² = 0, giving y = x²/2. Substitute into the original equation and solve.
For y·sin(2x) = x·cos(2y), differentiate both sides using the product rule and chain rule. Every derivative of a "y term" picks up a dy/dx factor. Then isolate dy/dx.
To find y''(x) implicitly, differentiate the expression for dy/dx with respect to x, substituting the known dy/dx back in where it appears.
Write an equation relating the quantities involved (geometry formulas, Pythagorean theorem, trig relationships).
Differentiate both sides with respect to time t, using implicit differentiation and the chain rule.
Substitute known values and rates, then solve for the unknown rate.
Circle: A = πr², C = 2πr
Sphere: V = (4/3)πr³, SA = 4πr²
Cone: V = (1/3)πr²h
Rectangle: A = lw, P = 2l + 2w
Right triangle: a² + b² = c²
Trig: tan θ = opposite/adjacent
Expanding sphere: Volume increases at 10 in³/s. Find how fast the radius changes when r = 2. Differentiate V = (4/3)πr³ with respect to t: dV/dt = 4πr²·(dr/dt). Plug in dV/dt = 10 and r = 2, solve for dr/dt.
Sliding ladder: A 13-ft ladder slides down a wall. The top moves down at 3 ft/s. How fast does the bottom move when it is 5 ft from the wall? Use x² + y² = 13². Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0. When x = 5, y = 12. Plug in dy/dt = −3, solve for dx/dt.
Filling cone: Water fills a cone-shaped tank at 9 ft³/min. The tank is 10 ft tall with base radius 5 ft. By similar triangles, r = h/2. Substitute into V = (1/3)πr²h to get V = πh³/12. Differentiate and solve for dh/dt when h = 6.
Rising balloon: A person 500 ft away watches a balloon rise. The angle of inclination θ increases at 1/10 rad/min when θ = π/4. Use tan θ = (height)/500. Differentiate: sec²θ · (dθ/dt) = (1/500)·(dh/dt). Solve for dh/dt.
Two cars diverging: Cars leave the same point, one south at 20 mi/h, one west at 30 mi/h. After 2 hours, how fast is the distance between them increasing? Use x² + y² = z². After 2 h: x = 60, y = 40, z = √(3600+1600) = √5200. Differentiate and solve for dz/dt.
Students often forget to multiply by dy/dx (or g'(x)) when applying the chain rule. Every inner function that is not just "x" needs its derivative multiplied in.
In implicit differentiation, students sometimes differentiate y² as 2y instead of 2y·(dy/dx). Always attach dy/dx when differentiating a y-term.
In related rates, a common mistake is substituting numerical values before differentiating. Differentiate first, then plug in numbers. Substituting early kills the terms you need.
Students confuse which rates are positive and which are negative. A quantity that is decreasing has a negative rate: dy/dt < 0 when y is shrinking.
⚠️ Chain rule problems appear on every exam. Practise nested compositions (chain rule applied twice).
⚠️ Implicit differentiation is tested both as a standalone skill (find dy/dx, find the tangent line) and as part of related rates.
⚠️ Related rates problems are among the highest-weighted exam questions. Memorise the geometry formulas. Draw a diagram. Label every variable and rate before touching the algebra.
Q: If h(x) = f(g(x)) and f(1) = 3, f(2) = 1, f(3) = 7, g(1) = 2, g(2) = 8, g(3) = 2, f'(1) = 4, f'(2) = 5, f'(3) = 7, g'(1) = 6, g'(2) = 7, g'(3) = 9, find h'(1).
A: h'(1) = f'(g(1)) · g'(1) = f'(2) · 6 = 5 · 6 = 30.
Q: Find dy/dx for the curve x² + 4xy + y² = 13.
A: Differentiate: 2x + 4y + 4x(dy/dx) + 2y(dy/dx) = 0. Solve: dy/dx = −(2x + 4y) / (4x + 2y) = −(x + 2y) / (2x + y).
Q: A 13-ft ladder leans against a wall. The top slides down at 3 ft/s. How fast does the bottom move when it is 5 ft from the wall?
A: x² + y² = 169. When x = 5, y = 12. Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0. With dy/dt = −3: 10(dx/dt) + 24(−3) = 0, so dx/dt = 72/10 = 36/5 ft/s.
Q: Find the tangent line to y·sin(2x) = x·cos(2y) at (π/2, π/4).
A: Differentiate implicitly using the product rule on both sides. At the given point, evaluate each term (sin(π) = 0, cos(π/2) = 0, cos(π) = −1, sin(π/2) = 1). Solve for dy/dx and write point-slope form.
chain rule, composition of functions, f(g(x)), inner function, outer function, Leibniz notation chain rule, implicit differentiation, dy/dx, implicit equation, related rates, related rates strategy, sliding ladder problem, expanding sphere, filling cone, similar triangles, Pythagorean theorem, tangent line implicit, MTH 132, Calculus I, Michigan State, Chapter 2