Use the chain rule with inner and outer functions

• Identify inner and outer functions from nested notation.
• Differentiate the outer form while preserving the inner expression.
• Multiply by the inner derivative and verify the result when expansion is available.

Build the idea

Worked example

1. Step 1

   Let u equal three x minus two. The outer expression becomes u to the fourth and the inner derivative is three.

   $$u=3x-2,\quad\frac{du}{dx}=3$$

2. Step 2

   The derivative of u to the fourth with respect to u is four u cubed. Keep the inner expression intact at this stage.

   $$\frac{dy}{du}=4u^3$$

3. Step 3

   Multiply the outer derivative by three and replace u with three x minus two. The scalar factors combine to twelve.

   $$\frac{dy}{dx}=12(3x-2)^3$$

• Applying the power rule but forgetting to multiply by the inner derivative.

  The outer rate is measured per inner unit. Multiply by inner units per x unit to obtain the total rate per x.

• Differentiating the inside before preserving it in the outer derivative.

  Keep the full inner expression inside the differentiated outer form, then multiply by its derivative as a separate factor.

Differentiate the square root of the quantity five x plus one, end quantity, and state where the derivative formula is defined.

This reviewed reading order covers the lesson explanation, equations, example, and checks without claiming that a video exists.

1. Part 1

   Lesson overview

   The chain rule multiplies the outer rate with respect to its input by the inner input's rate with respect to x, connecting both stages of nested change. A nested function changes in stages. The inner function changes as x changes, and the outer function responds to that inner output. Identify inner and outer functions from nested notation.

2. Part 2

   Visual model and equations

   Rates across stages multiply because each unit of x creates some units of the inner quantity, and each inner unit creates some units of the outer quantity. An x gear turns an inner-output gear, which turns an outer-output gear. The total rotation per x turn is the product of the two stage ratios. The derivative of f of g of x equals f prime at g of x times g prime of x. d y over d x equals d y over d u times d u over d x.

   • The derivative of f of g of x equals f prime at g of x times g prime of x.
   • d y over d x equals d y over d u times d u over d x.
3. Part 3

   Worked example

   Differentiate the quantity three x minus two, end quantity, to the fourth power, and verify the derivative by expanding enough to compare. Treat three x minus two as the inner function, differentiate the fourth-power outer form, and multiply by the inner derivative. Let u equal three x minus two. The outer expression becomes u to the fourth and the inner derivative is three. The derivative of u to the fourth with respect to u is four u cubed. Keep the inner expression intact at this stage. Multiply the outer derivative by three and replace u with three x minus two. The scalar factors combine to twelve. The derivative is twelve times the quantity three x minus two, cubed. Expanding the original fourth power and differentiating term by term yields the same cubic after refactoring, so the inner factor three has not been lost.

   • u equals three x minus two, and d u over d x equals three.
   • d y over d u equals four u cubed.
   • d y over d x equals twelve times the quantity three x minus two, cubed.
4. Part 4

   Checks, practice, and scope

   Writing a temporary inner variable makes the layers visible. After differentiating the outer rule, substitute the original inner expression and multiply by its derivative. Applying the power rule but forgetting to multiply by the inner derivative. The outer rate is measured per inner unit. Multiply by inner units per x unit to obtain the total rate per x. Differentiating the inside before preserving it in the outer derivative. Keep the full inner expression inside the differentiated outer form, then multiply by its derivative as a separate factor. Differentiate the square root of the quantity five x plus one, end quantity, and state where the derivative formula is defined. Write the square root as a one-half power and multiply by the inner derivative five. The derivative is five divided by the quantity two times the square root of the quantity five x plus one, end square root, end denominator, defined for x greater than negative one fifth. Nested change proceeds through inner and outer stages whose rates multiply. Temporary substitution makes the chain-rule layers easier to identify. The outer derivative keeps the inner expression before the inner derivative is multiplied. The lesson assumes differentiability of both stages at the relevant values; discontinuities can invalidate the rule. Implicit differentiation and multivariable chain rules use related ideas but require additional notation.

Scope and limitations

• The lesson assumes differentiability of both stages at the relevant values; discontinuities can invalidate the rule.
• Implicit differentiation and multivariable chain rules use related ideas but require additional notation.