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A Repeatable Setup System for Chain Rule Problems

1Naming the Layers Before Differentiating2Writing the Derivative as a Labeled Product3Running the Layout on the Standard Function Families4Keeping the Layout Intact Inside Products and Quotients5Extending the Layout to Implicit Differentiation and Diagnosing Setup Errors
Writing the Derivative as a Labeled Product

Why the Chain Rule Gives One Factor per Layer

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You already have the layer inventory written down. Now watch what that inventory turns into. Each arrow in the diagram — x into the inner box, inner box into the outer box — is a rate of change, and the chain rule multiplies those rates.
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The layer diagram from the previous chapter showed a chain of dependencies: an input x enters the inner function g and produces an intermediate value u = g(x), which then enters the outer function f and produces y = f(u).

Each arrow in that diagram is a rate of change:

  • \(\dfrac{du}{dx}\) is how fast the intermediate value changes as x changes.
  • \(\dfrac{dy}{du}\) is how fast the output changes as the intermediate value changes.

The chain rule says these rates multiply:

\[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]

In function notation, with \(y=f(g(x))\), the same statement is

\[\frac{d}{dx}\big[f(g(x))\big]=f'(g(x))\cdot g'(x)\]

Why the factor count equals the layer count. Every arrow in the dependency diagram contributes exactly one rate, and every layer of the composite is exactly one arrow. So a two-layer composite produces a two-factor product, and a three-layer composite produces three factors:

\[\frac{d}{dx}\big[f(g(h(x)))\big]=f'\big(g(h(x))\big)\cdot g'\big(h(x)\big)\cdot h'(x)\]

This is the bookkeeping rule that makes the setup reliable: count the layers first, then expect that many factors. If the written product has fewer factors than the layer count, a layer was skipped.

References

  1. [1]3.6 The Chain Rule – Calculus Volume 1 (OpenStax)openstax.org
  2. [2]Chain rule – Wikipediaen.wikipedia.org
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