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Calculus / English

Chain Rule For Finding Derivatives

The Organic Chemistry Tutor · YouTube · 18:39

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Practice the chain rule by identifying outer and inner functions in power and trigonometric examples.

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Chapters

0:20Differentiate a composition1:30Composite powers3:00Trigonometric compositions

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The chain rule is [f(g(x))]′=f′(g(x))g′(x)[f(g(x))]'=f'(g(x))g'(x). Differentiate the outer function at the unchanged inner expression, then multiply by the inner derivative.

For (5x+3)4(5x+3)^4, the outside derivative contributes 4(5x+3)34(5x+3)^3 and the inside derivative contributes 5. Their product is 20(5x+3)320(5x+3)^3.

The same rule gives [cos⁡(x2)]′=−2xsin⁡(x2)[\cos(x^2)]'=-2x\sin(x^2) and [tan⁡(x3)]′=3x2sec⁡2(x3)[\tan(x^3)]'=3x^2\sec^2(x^3) wherever the original functions are defined.

Knowledge cards

01

Differentiate a composition

The chain rule is [f(g(x))]′=f′(g(x))g′(x)[f(g(x))]'=f'(g(x))g'(x). Differentiate the outer function at the unchanged inner expression, then multiply by the inner derivative.

02

Composite powers

For (5x+3)4(5x+3)^4, the outside derivative contributes 4(5x+3)34(5x+3)^3 and the inside derivative contributes 5. Their product is 20(5x+3)320(5x+3)^3.

03

Trigonometric compositions

The same rule gives [cos⁡(x2)]′=−2xsin⁡(x2)[\cos(x^2)]'=-2x\sin(x^2) and [tan⁡(x3)]′=3x2sec⁡2(x3)[\tan(x^3)]'=3x^2\sec^2(x^3) wherever the original functions are defined.

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  • Chain rule ExplanationAt 0:20
    Why this connection?

    The chain rule is [f(g(x))]′=f′(g(x))g′(x)[f(g(x))]'=f'(g(x))g'(x). Differentiate the outer function at the unchanged inner expression, then multiply by the inner derivative. For (5x+3)4(5x+3)^4, the outside derivative contributes 4(5x+3)34(5x+3)^3 and the inside derivative contributes 5. Their product is 20(5x+3)320(5x+3)^3. The same rule gives [cos⁡(x2)]′=−2xsin⁡(x2)[\cos(x^2)]'=-2x\sin(x^2) and [tan⁡(x3)]′=3x2sec⁡2(x3)[\tan(x^3)]'=3x^2\sec^2(x^3) wherever the original functions are defined.