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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 Outer Factor Keeps the Inner Function Intact

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Now the outer slot gets filled, and this is where most chain rule answers break. The outer derivative is not evaluated at x. It is evaluated at the inner function, so the inner expression gets copied in whole.
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The most common chain rule error is writing the outer derivative at x instead of at the inner function. The formula prevents this:

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

The first factor is \(f'\big(g(x)\big)\), not \(f'(x)\). The outer function's derivative is evaluated at whatever the inner function produced.

Worked contrast. For \(y=\ln(x^2)\), the outer function is \(\ln(\square)\) and the inner function is \(x^2\).

  • Wrong: \(\dfrac{1}{x}\cdot 2x=2\). Here \(\dfrac{1}{x}\) evaluated the log derivative at x, not at the inner function.
  • Right: \(\dfrac{1}{x^2}\cdot 2x=\dfrac{2}{x}\). The inner expression \(x^2\) is copied into the denominator unchanged.

The same idea in placeholder form: if the outer function is \(\ln(\square)\), its derivative is \(\dfrac{1}{\square}\), and the placeholder is filled with the inner function exactly as written — nothing differentiated, nothing simplified.

Practical check. After writing the outer factor, look at the expression sitting inside it. It must be character-for-character the inner function you identified before differentiating. If it has been altered, the factor is wrong.

References

  1. [1]Chain Rule | Varsity Tutors AP Calculus ABvarsitytutors.com
  2. [2]3.6 The Chain Rule – Calculus Volume 1 (OpenStax)openstax.org
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