CalculusIntermediate

Chain Rule Applications

The chain rule is the most important differentiation rule in calculus. Once you understand it mechanically, the next step is applying it to two powerful techniques: implicit differentiation (differentiating equations where y isn't isolated) and related rates (finding how fast one quantity changes when another does).

1 Quick Chain Rule Review

The chain rule: d/dx[f(g(x))] = f'(g(x)) · g'(x). Derivative of the outside times derivative of the inside. This applies whenever you differentiate a composite function , anything where one function is nested inside another.

The chain rule in Leibniz notation

If y = f(u) and u = g(x), then dy/dx = (dy/du)(du/dx). You can think of the du's as canceling, though this is only a mnemonic , it reflects why the chain rule works intuitively.

2 Implicit Differentiation

Usually we differentiate explicit functions: y = f(x). But sometimes the relationship between x and y is implicit , like x² + y² = 25 (a circle). You can't easily solve for y, so you differentiate both sides for x, treating y as a function of x and applying the chain rule whenever y appears.

The key: d/dx[y²] = 2y · (dy/dx), not just 2y. The chain rule adds the dy/dx factor because y is a function of x.

Implicit Differentiation

Find dy/dx for x² + y² = 25 (the circle of radius 5)

1Differentiate both sides with respect to x:

2d/dx[x²] + d/dx[y²] = d/dx[25]

32x + 2y(dy/dx) = 0

4Solve for dy/dx: 2y(dy/dx) = −2x

5dy/dx = −2x/2y = −x/y

Answer: dy/dx = −x/y , the slope at any point (x, y) on the circle

More Complex Implicit Differentiation

Find dy/dx for x³ + y³ = 6xy

1Differentiate both sides: 3x² + 3y²(dy/dx) = 6y + 6x(dy/dx)

2Collect dy/dx terms: 3y²(dy/dx) − 6x(dy/dx) = 6y − 3x²

3Factor: dy/dx(3y² − 6x) = 6y − 3x²

4dy/dx = (6y − 3x²)/(3y² − 6x) = (2y − x²)/(y² − 2x)

Answer: dy/dx = (2y − x²)/(y² − 2x)

4 More Worked Examples

Implicit: Tangent Line

Find the equation of the tangent line to x² + y² = 25 at point (3, 4)

1dy/dx = −x/y (from earlier example)

2At (3, 4): slope = −3/4

3Tangent line: y − 4 = −(3/4)(x − 3)

4y = −(3/4)x + 9/4 + 4 = −(3/4)x + 25/4

Answer: y = −(3/4)x + 25/4

5 Problem-Solving Strategy

For implicit differentiation: (1) Differentiate both sides for x. (2) Every time you differentiate a y term, multiply by dy/dx (chain rule). (3) Collect all dy/dx terms on one side. (4) Factor out dy/dx and divide.

For related rates: (1) Draw a diagram and label all variables. (2) Write an equation relating the variables. (3) Differentiate both sides for time t. (4) Substitute all known values including known rates. (5) Solve for the unknown rate.

Forgetting the chain rule on y terms

When differentiating implicitly, d/dx[y³] = 3y²(dy/dx), NOT just 3y². The dy/dx is essential , it's what the chain rule adds because y depends on x.

Practice Problems

Find dy/dx for x² − 2y² = 4

Differentiate: 2x − 4y(dy/dx) = 0. Solve: dy/dx = x/(2y)

A spherical balloon is inflated so its radius grows at 2 cm/sec. How fast is the volume increasing when r = 5 cm? (V = 4/3 πr³)

dV/dt = 4πr²(dr/dt) = 4π(25)(2) = 200π ≈ 628.3 cm³/sec

Find the slope of x³ + y² = 9 at point (2, 1)

Implicit diff: 3x² + 2y(dy/dx) = 0. dy/dx = −3x²/2y. At (2,1): dy/dx = −12/2 = −6

Sources & Further Reading

The explanations on this page draw on the following established sources. We link to primary and secondary sources so you can verify claims and go deeper on any topic.