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Grade 12 Calculus · Lesson 2.490 min

The Chain Rule · See the Layers

Composite functions have layers. Read from outside to inside, differentiate one layer at a time, and never lose the derivative of the inside.

Use Mr. Azzam’s inside/outside method for composite functions, powers, radicals, trigonometric compositions, and mixed differentiation rules.

Recognize compositionu-substitutionGeneral Power RuleTrig + ChainMixed rules
Key takeaway

Mr. Azzam’s Chain Rule routine

1 · Read the layers

Identify the outside operation and complete inside expression.

2 · Keep the inside

Differentiate the outside while leaving the inside unchanged.

3 · Differentiate inside

Find the derivative of the inner function.

4 · Multiply + check

Multiply and check that no inside derivative is missing.

See the layers before you differentiate

The Chain Rule is for a function inside another function. Before differentiating, identify what is outside and what is inside.

ExplanationOutsidey=(x)5y=(\boxed{\phantom{x}})^5
ExplanationInsideu=3x2−1u=3x^2-1
Read from outside → inside
y=(3x2−1)5y=(3x^2-1)^5u=3x2−1u=3x^2-1y=u5y=u^5
ExplanationLayer test

Ask: What happens last? The operation that happens last is the outside layer.

Your turn

For y=(3x2−1)5y=(3x^2-1)^5, what is the inside function?

The Chain Rule

dydx=dydu⋅dudx\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
ddx[f(g(x))]=f′(g(x))g′(x)\frac{d}{dx}[f(g(x))]=f'(g(x))g'(x)
Key takeawayMr. Azzam’s idea

Differentiate the outside, keep the inside unchanged, then multiply by the derivative of the inside.

Your turn

If y=f(g(x))y=f(g(x)), which structure matches the Chain Rule?

Mr. Azzam’s method · use u to expose the structure

STEP 1
Choose the inside
u=g(x)u=g(x)
STEP 2
Rewrite the outside
y=f(u)y=f(u)
STEP 3
Differentiate the outside
dydu\frac{dy}{du}
STEP 4
Differentiate the inside
dudx\frac{du}{dx}
STEP 5
Multiply
dydx=dydududx\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}
STEP 6
Substitute u back
Write the final answer in xx.
Interactive lab

Chain Rule Builder

Predict, identify the layers, then build and check the derivative. Advance only when you choose to.
y=(3x2−1)5y=(3x^2-1)^5

Step 1 / 6

Which rule controls this nested function?

General Power Rule

ddx[u(x)]n=n[u(x)]n−1u′(x)\frac{d}{dx}[u(x)]^n=n[u(x)]^{n-1}u'(x)
Power outside · polynomial inside
y=(2x3+1)4y=(2x^3+1)^4y′=4(2x3+1)3(6x2)y'=4(2x^3+1)^3(6x^2)y′=24x2(2x3+1)3y'=24x^2(2x^3+1)^3
Radical · rewrite first
y=x2+3x+1=(x2+3x+1)1/2y=\sqrt{x^2+3x+1}=(x^2+3x+1)^{1/2}y′=12(x2+3x+1)−1/2(2x+3)y'=\frac12(x^2+3x+1)^{-1/2}(2x+3)y′=2x+32x2+3x+1y'=\frac{2x+3}{2\sqrt{x^2+3x+1}}
Common mistake. Do not lower the power and stop. The derivative of the inside function must also appear.
Key takeawayFast self-check

Identify the inside before differentiating. Your final derivative should include the derivative of that inside as a factor.

Your turn

Question 1 of 2
For y=(2x3+1)4y=(2x^3+1)^4, which factor is lost in the incorrect derivative 4(2x3+1)34(2x^3+1)^3?

Trigonometric functions + Chain Rule

Sine with an inside function
y=sin⁡(5x2)y=\sin(5x^2)u=5x2u=5x^2y=sin⁡uy=\sin udydu=cos⁡u,dudx=10x\frac{dy}{du}=\cos u,\qquad\frac{du}{dx}=10xy′=10xcos⁡(5x2)y'=10x\cos(5x^2)

Your turn

Differentiate y=sin⁡(4x)y=\sin(4x).

Chain Rule can live inside another rule

First identify the structure of the whole expression. Product or Quotient Rule may organize the outside problem while the Chain Rule is used inside one of its factors.

Product Rule + Chain Rule
y=5x3sec⁡(3x2+x)y=5x^3\sec(3x^2+x)y′=15x2sec⁡(3x2+x)+5x3[sec⁡(3x2+x)tan⁡(3x2+x)](6x+1)y'=15x^2\sec(3x^2+x)+5x^3[\sec(3x^2+x)\tan(3x^2+x)](6x+1)
Common mistake. Do not apply one rule blindly to the whole expression. Read the structure first: product, quotient, then the composite layers inside.
Key takeawayExam strategy

Choose the outermost structure first. If the whole expression is a product or quotient, organize that rule first, then use Chain Rule inside the composite factors.

Your turn

For y=x2sin⁡(3x)y=x^2\sin(3x), which derivative uses both required rules?

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