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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
Running the Layout on the Standard Function Families

Chain Rule on a Power of a Function Composite

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Start with the power composite: three x squared plus one, all raised to the fifth. The layout doesn't care that the inside is a polynomial. The outer layer is the fifth power, the inner layer is three x squared plus one. That's two layers, so two factor slots.
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The layout on a power composite

A power of a function, such as \((3x^2+1)^5\), is a composite with two layers:

  • Outer function: the fifth power, \((\square)^5\).
  • Inner function: the polynomial \(3x^2+1\).

Step 1 — Write the outer factor

Differentiate the outer function while copying the inner expression in unchanged:

\[5(3x^2+1)^4\]

The exponent drops by one, and the base stays exactly \(3x^2+1\). It does not become \(3x^2\) or \(x^2+1\).

Step 2 — Write the inner factor

Differentiate the inner function as its own separate factor:

\[\frac{d}{dx}(3x^2+1) = 6x\]

Step 3 — Multiply, then stop

The complete unsimplified derivative is

\[\frac{d}{dx}(3x^2+1)^5 = 5(3x^2+1)^4 \cdot 6x\]

Two layers, two factors. Simplifying to \(30x(3x^2+1)^4\) is optional and can wait until the factor count has been checked.

Why the layout matters here

The most common error on a power composite is differentiating the outer power and forgetting the inner factor entirely, giving \(5(3x^2+1)^4\) and stopping. Writing the inner factor as a separate slot makes that omission visible: the second slot is empty.

References

  1. [1]Chain Rule Explained — Step-by-Step with Examples | CalculusConcepts.comcalculusconcepts.com
  2. [2]Chain Rule: Practice Problems & Solutionsmatheno.com
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