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How does the chain rule handle a function inside another function?

Quick answer

For a composite function, differentiate the outside function while keeping the inside expression in place, then multiply by the inside derivative. For example, if h(x) = (3x² + 1)⁴, the inside is 3x² + 1 and the outside is a fourth power. Thus h′(x) = 4(3x² + 1)³ · 6x = 24x(3x² + 1)³. [S1]

LET’S WORK THROUGH IT

The question, the details, the next step.

Question

I see an expression like (something) to a power. How do I know when to use the chain rule instead of just the power rule?

Geez

Look for a function nested inside another function. In (3x² + 1)⁴, the entire expression 3x² + 1 is inside the fourth power. The power rule handles differentiating a power, but because its base is itself a function of x, you also need to account for how that base changes. The chain rule says to differentiate the outside function, evaluate that derivative at the inside expression, and multiply by the inside derivative. [S1]

A useful reminder is “outside derivative, then inside derivative.” Bringing down the exponent handles only the outside; don’t stop there.

Follow-up question

Can you work through that nested-power example and identify each part?

Geez

Let h(x) = (3x² + 1)⁴.

  1. Inner function: g(x) = 3x² + 1.
  2. Outer function: f(u) = u⁴, where u represents the input to the outer function.
  3. Differentiate the outer function: f′(u) = 4u³. Evaluate it at the inner expression: 4(3x² + 1)³.
  4. Differentiate the inner function: g′(x) = 6x.
  5. Multiply: h′(x) = 4(3x² + 1)³ · 6x = 24x(3x² + 1)³. [S1]

The inside expression stays in the base when you differentiate the outside power. The extra factor, 6x, comes from differentiating the inside.

Follow-up question

How can I check that I haven’t forgotten a part or multiplied incorrectly?

Geez

Keep the two derivative steps visible until the end. For this example, write the outside derivative as 4(3x² + 1)³ and the inside derivative as 6x. Then multiply: 4 · 6x = 24x, giving 24x(3x² + 1)³. A common slip is to write only 4(3x² + 1)³; that differentiates the outside but omits the inside derivative.

As a structural check, make sure your result includes both the outside derivative evaluated at the inside expression and the derivative of the inside. The chain rule applies where the relevant derivatives are defined. [S1]

Follow-up question

Can I use the same layout for other nested functions, not just powers?

Geez

Yes. Write the composite function as h(x) = f(g(x)), where g(x) is the inside function and f is the outside function. Then use h′(x) = f′(g(x)) · g′(x): differentiate f while keeping its input represented by g(x), and multiply by the derivative of g. [S1]

A reusable layout is:

Function: h(x) = f(g(x)) Inner: g(x) = [expression] Outer: f(u) = [rule applied to input u] Outside derivative, evaluated at the inside: f′(g(x)) = [expression] Inside derivative: g′(x) = [expression] Product: h′(x) = f′(g(x)) · g′(x)

For the worked example, g(x) = 3x² + 1 and f(u) = u⁴. The outside derivative evaluated at the inside is 4(3x² + 1)³, and the inside derivative is 6x. Multiply those factors. The formula assumes the component derivatives are defined at the inputs involved. [S1]

A practical checklist

  • Identify the full inside expression and label it g(x).

  • Name the outside function and differentiate it while keeping g(x) in place.

  • Differentiate g(x) separately.

  • Multiply the outside derivative by the inside derivative.

  • Check that the result contains both factors.

A few more useful answers

Why do I keep the inside expression when differentiating the outside?

The outside function receives the inside expression as its input. Evaluate its derivative at that same expression, then multiply by the inside derivative.

What is the chain rule formula for a composite function?

If h(x) = f(g(x)), then h′(x) = f′(g(x)) · g′(x), where the relevant derivatives are defined. [S1]

What if the inside function is more complicated than a quadratic?

Use the same steps: differentiate the outside function with the full inside expression in place, then multiply by the derivative of that inside expression.

Sources

References used for this guide.

  1. [S1] OpenStax Calculus Volume 1, Section 3.6: The Chain Rule
    openstax.org · Checked

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