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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
Extending the Layout to Implicit Differentiation and Diagnosing Setup Errors

Why Differentiating a y-Term Produces a dy/dx Factor

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Every composite you have differentiated so far had an innermost layer made of plain x, so the last factor of the block was something like 2x. Here is a term with no visible x in it at all: y cubed.
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The problem

Every composite you have differentiated so far had an innermost layer that was a plain expression in \(x\), so the last factor of the chain rule block was something like \(2x\) or \(3\). In an implicit equation the variables are mixed, and a term such as \(y^3\) contains no visible \(x\) at all. The layout still applies, but now you have to decide what the innermost layer is.

y is a function of x

An implicit equation such as \(x^2+y^2=25\) does not hand you a formula for \(y\), but near a point on the curve it still defines \(y\) as a differentiable function of \(x\). That is the whole idea behind implicit differentiation: treat \(y\) as shorthand for \(y(x)\), an unknown function of \(x\).

Once you read it that way, \(y^3\) is a composite:

  • outer function: \((;)^{3}\), the cubing operation
  • inner function: \(y\), the unknown function of \(x\)

The layer inventory has exactly the same shape as the one for \((x^2+1)^3\). Only the inner function has changed.

The block

Apply the established layout: differentiate the outer function with the inner left intact, then multiply by the derivative of the inner function.

\[\frac{d}{dx}\left[y^3\right]=3y^2\cdot\frac{dy}{dx}\]

The first factor, \(3y^2\), is the outer derivative evaluated at the inner function. The second factor, \(\frac{dy}{dx}\), is the innermost layer derivative, because the innermost layer is \(y\) itself and its derivative with respect to \(x\) is \(\frac{dy}{dx}\).

Why the factor is multiplied, not added

The chain rule is a product of one factor per layer. A y-term has the same number of layers as the matching x-term, so it produces the same number of factors. The only difference is that the last factor is written \(\frac{dy}{dx}\) instead of a formula in \(x\).

Common y-terms

Term Outer derivative (inner intact) Innermost factor Block
\(y^3\) \(3y^2\) \(\frac{dy}{dx}\) \(3y^2\frac{dy}{dx}\)
\(\sin y\) \(\cos y\) \(\frac{dy}{dx}\) \(\cos y,\frac{dy}{dx}\)
\(e^{y}\) \(e^{y}\) \(\frac{dy}{dx}\) \(e^{y}\frac{dy}{dx}\)
\(\ln y\) \(\frac{1}{y}\) \(\frac{dy}{dx}\) \(\frac{1}{y}\frac{dy}{dx}\)

In every row the \(\frac{dy}{dx}\) factor appears last, exactly where the innermost layer derivative belongs.

Caution

A term that is already a plain function of \(x\), such as \(x^2\), gets no \(\frac{dy}{dx}\) factor. The factor shows up only when the innermost layer is \(y\).

References

  1. [1]3.6: The Chain Rule - Mathematics LibreTextsmath.libretexts.org
  2. [2]Chain Rule Explained — Step-by-Step with Examples | CalculusConcepts.comcalculusconcepts.com
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