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Lesson 2: The Chain Rule

Estimated time: 35-45 minutes

Learning Objectives

By the end of this lesson, you will be able to:

Composite Functions Review

A composite function has the form f(g(x)): you first apply g (the inner function), then f (the outer function). For example, (x² + 1)5 is f(g(x)) where g(x) = x² + 1 (inner) and f(u) = u5 (outer).

To differentiate a composite function, you need the Chain Rule.

The Chain Rule

Chain Rule

If y = f(g(x)) and both f and g are differentiable, then:

dy/dx = f'(g(x)) · g'(x)

In words: "the derivative of the outer function evaluated at the inner function, times the derivative of the inner function."

Leibniz Notation

If y = f(u) and u = g(x), then: dy/dx = (dy/du) · (du/dx)

This form makes the Chain Rule look like fraction cancellation (though dy/du and du/dx are not true fractions).

Example 1: Power of a Function

Find d/dx [(x² + 1)5].

Step 1: Outer function: u5. Inner function: u = x² + 1.

Step 2: Derivative of outer: 5u4. Derivative of inner: 2x.

Step 3: Chain Rule: 5(x² + 1)4 · 2x = 10x(x² + 1)4

Example 2: Trig of a Linear Function

Find d/dx [sin(3x)].

Step 1: Outer: sin u, inner: u = 3x.

Step 2: Derivative of outer: cos u. Derivative of inner: 3.

Step 3: Chain Rule: cos(3x) · 3 = 3 cos(3x)

Example 3: Exponential Composition

Find d/dx [e].

Step 1: Outer: eu, inner: u = x².

Step 2: Derivative of outer: eu. Derivative of inner: 2x.

Step 3: Chain Rule: e · 2x = 2x e

The General Power Rule

The Chain Rule combined with the power rule gives us a very useful pattern:

General Power Rule

d/dx [g(x)]n = n [g(x)]n−1 · g'(x)

Example 4: Square Root Composition

Find d/dx [√(3x² + 7)].

Rewrite: (3x² + 7)1/2

Apply General Power Rule: (1/2)(3x² + 7)−1/2 · 6x = 3x / √(3x² + 7)

Chain Rule with Other Rules

Example 5: Chain Rule + Product Rule

Find d/dx [x² sin(3x)].

Product Rule first: d/dx [x²] · sin(3x) + x² · d/dx [sin(3x)]

Chain Rule for sin(3x): d/dx [sin(3x)] = 3 cos(3x)

Combine: 2x sin(3x) + x² · 3 cos(3x) = 2x sin(3x) + 3x² cos(3x)

Example 6: Double Chain Rule

Find d/dx [sin²(4x)] = d/dx [(sin(4x))²].

Outer: u², middle: sin v, inner: v = 4x.

Apply: 2 sin(4x) · cos(4x) · 4 = 8 sin(4x) cos(4x) = 4 sin(8x)

Key Takeaways

Check Your Understanding

1. Find d/dx [(2x − 1)7].

Answer: 7(2x − 1)6 · 2 = 14(2x − 1)6.

2. Find d/dx [cos(x³)].

Answer: −sin(x³) · 3x² = −3x² sin(x³).

3. Find d/dx [esin x].

Answer: esin x · cos x = cos x · esin x.

4. Find d/dx [ln(x² + 5)].

Answer: [1/(x² + 5)] · 2x = 2x / (x² + 5).

Ready for More?

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Lesson 3 covers the derivatives of all six trigonometric functions.

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Module Progress

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