Calculus Differentiation

Chain rule

Q: What is the chain rule?

A rule for composite functions: the derivative of f of g of x equals f prime of g of x times g prime of x, the derivative of the outer function times the derivative of the inner function.

Q: How do you know when to use the chain rule?

Whenever a function is inside another, a power of an expression, a root of an expression, or a trig, exponential, or log of an expression.

Q: What is the most common chain-rule mistake?

Forgetting to multiply by the inner derivative g prime of x. Differentiating 2x plus 1 cubed must include the extra factor of 2.

Q: What is the power form of the chain rule?

The derivative of g of x to the n equals n times g of x to the n minus one times g prime of x: the ordinary power rule times the derivative of the inside.

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The Chain Rule

California Calculus • Standard 5.0 • Differentiation

The Chain Rule is a topic in Differentiation in the California Calculus Standards. It is aligned to Standard 5.0, which requires students to know the chain rule and its proof and apply it to the derivatives of composite functions.

The chain rule differentiates a composite function \(f(g(x))\) as \(f'(g(x))\cdot g'(x)\) — the derivative of the outer function times the derivative of the inner function.

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Theory

The chain rule differentiates a composite function \(f(g(x))\): take the derivative of the outer function (leaving the inside alone) and multiply by the derivative of the inner function.

A composite function \(y=f(g(x))\) applies one function to the output of another. The chain rule says its derivative is the product of the two rates:

\[\dfrac{d}{dx}f(g(x))=f^{\prime}(g(x))\cdot g^{\prime}(x)\]

Think of it as “derivative of the outside (with the inside unchanged) times the derivative of the inside.”

With \(u=g(x)\) the same rule reads \(\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}\) — the intermediate rates multiply.

Key idea: the most common slip is forgetting the inner factor \(g'(x)\). Every time you differentiate an inside function, a new factor must appear.

\(x\to g\to u\to f\to y\): the rates \(\dfrac{dy}{du}\) and \(\dfrac{du}{dx}\) multiply.

The chain rule gives the slope of a composite curve.

The chain rule, in both notations:

\[\dfrac{d}{dx}f(g(x))=f^{\prime}(g(x))\,g^{\prime}(x),\qquad\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}\]

chain rule: derivative of outer times derivative of inner

The common power form (outer function is a power):

\[\dfrac{d}{dx}\big[g(x)\big]^{n}=n\big[g(x)\big]^{n-1}\,g^{\prime}(x)\]

power form: n times g to the n minus one times g prime

Do not forget \(g^{\prime}(x)\). Differentiating \((2x+1)^3\) gives \(3(2x+1)^2\) times \(2\), not just \(3(2x+1)^2\).

How to use the chain rule

Identify the inner function \(g(x)\) (what is inside the power, root, or function).
Differentiate the outer function, keeping the inside unchanged.
Multiply by \(g'(x)\), the derivative of the inside.

Nested composites just repeat the rule: peel one layer at a time, multiplying by each inner derivative.

Example 1 — Power form

Differentiate \(y=(2x+1)^3\).

Solution

Outer \(u^3\), inner \(u=2x+1\): \(\dfrac{dy}{dx}=3u^2\cdot u'\).

\(y'\)	\(=\)	\(3(2x+1)^2\cdot 2\)
	\(=\)	\(6(2x+1)^2\)

derivative is 6 times 2x plus 1 squared

Example 2 — Chain rule with a root

Differentiate \(y=\sqrt{x^2+9}\).

Solution

Write \((x^2+9)^{1/2}\); the inner derivative is \(2x\).

\(y'\)	\(=\)	\(\dfrac12(x^2+9)^{-1/2}\cdot 2x\)
	\(=\)	\(\dfrac{x}{\sqrt{x^2+9}}\)

derivative is x over root x squared plus nine

Example 3 — Coefficient inside

Differentiate \(y=(3x^2-1)^5\).

Solution

Inner \(3x^2-1\), inner derivative \(6x\).

\(y'\)	\(=\)	\(5(3x^2-1)^4\cdot 6x\)
	\(=\)	\(30x(3x^2-1)^4\)

derivative is 30x times 3x squared minus one to the fourth

Example 4 — Evaluate at a point

For \(y=(x^2+3)^4\), find \(\dfrac{dy}{dx}\) at \(x=1\).

Solution

Differentiate, then substitute \(x=1\).

\(y'\)	\(=\)	\(4(x^2+3)^3\cdot 2x=8x(x^2+3)^3\)
\(y'(1)\)	\(=\)	\(8(1)(4)^3=8\cdot 64=512\)

derivative at x equals 1 is 512

Common pitfalls

Forgetting the inner derivative. The factor \(g'(x)\) must appear every time; \((2x+1)^3\) differentiates to \(6(2x+1)^2\), not \(3(2x+1)^2\).

Do not differentiate the inside twice. Leave the inside unchanged in the outer derivative; only the extra factor \(g'(x)\) differentiates it.

Rewrite roots as powers. \(\sqrt{g(x)}=[g(x)]^{1/2}\), then the power form applies cleanly.

Frequently asked questions

What is the chain rule?

A rule for composite functions: \(\dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)\) — the derivative of the outer function times the derivative of the inner function.

How do you know when to use the chain rule?

Whenever a function is inside another — a power of an expression, a root of an expression, or a trig, exponential, or log of an expression.

What is the most common chain-rule mistake?

Forgetting to multiply by the inner derivative \(g'(x)\). Differentiating \((2x+1)^3\) must include the extra factor of \(2\).

What is the power form of the chain rule?

\(\dfrac{d}{dx}[g(x)]^{n}=n[g(x)]^{n-1}g'(x)\): the ordinary power rule times the derivative of the inside.

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