Method guideCalculus IFoundational9 min read

The Chain Rule as a structure-reading skill

Differentiate the outside function, keep the inside intact, then multiply by the inside rate.

ddxf(g(x))=f(g(x))g(x)\frac{d}{dx}f(g(x))=f'(g(x))g'(x)
Method guide

How to recognize the method, run it, and know when it is the wrong choice.

Reviewed July 11, 2026
PatternStart here
(f(g(x)))=f(g(x))g(x)\boxed{\bigl(f(g(x))\bigr)'=f'(g(x))g'(x)}

Use the Chain Rule whenever one varying expression has been substituted inside another function.

01

See layers, not symbols

Ask what the last operation performed is. In sin(x²), squaring happens first and sine happens last, so sine is the outside layer and x² is the inside layer.

Peel one layer at a time. Deep compositions may require the rule repeatedly.

02

Why the inner derivative appears

The outer output changes according to its own slope, but its input is moving at the rate g′(x). Multiplying the rates accounts for both parts of the dependency.

Units tell the same story: output per inner unit times inner units per x-unit gives output per x-unit.

03

A dependable notation

Temporarily name the inside u when the structure is crowded. Differentiate with respect to u, then multiply by du/dx. The substitution is organizational, not a separate theorem.

Worked exampleDifferentiate three layers

Differentiate (1 + e^(3x))⁵.

15(1+e3x)45(1+e^{3x})^4

Differentiate the fifth-power layer.

2e3x\cdot e^{3x}

Differentiate the 1 + exponential layer.

33\cdot3

Differentiate the exponent 3x.

Result15e3x(1+e3x)4\boxed{15e^{3x}(1+e^{3x})^4}
Watch for

Common mistakes

  1. Differentiating the inside but forgetting the outside derivative.
  2. Replacing the inside with its derivative instead of multiplying.
  3. Stopping after one layer in a nested composition.
Keep

Three takeaways

  1. Identify the last operation first.
  2. Keep the inside intact while differentiating the outside.
  3. Multiply by every inner rate encountered.