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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

Building a Two-Layer Derivative Product Step by Step

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Let's run the whole thing on three x squared plus one, all raised to the fifth. Two layers, so two slots. Fill the left one first: the power rule brings down the five and drops the exponent to four, and the inside stays exactly as it was.
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Take \(f(x)=(3x^2+1)^5\). The layer inventory is already written: outer \((\square)^5\), inner \(3x^2+1\), two layers.

Step 1 — Outer factor. Differentiate the outer power, keeping the inner block intact:

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

Step 2 — Inner factor. Differentiate the inner function:

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

Step 3 — Write the full product.

\[f'(x)=5(3x^2+1)^4\cdot 6x\]

Each factor carries a label naming the layer it came from:

Factor Layer Built from
\(5(3x^2+1)^4\) outer derivative of \((\square)^5\) at the inner function
\(6x\) inner derivative of \(3x^2+1\)

Step 4 — Simplify last. Only now combine the constants: \(f'(x)=30x(3x^2+1)^4\).

The order matters. Writing the outer factor first guarantees the inner block is still intact when it is copied; if you differentiate the inner function first you are tempted to substitute \(6x\) into the outer factor, which is the classic \(4(6x)^3\) error.

References

  1. [1]Chain Rule: Practice Problems & Solutionsmatheno.com
  2. [2]The Chain Rule Made Easy: Examples and Solutionsmathsathome.com
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