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How is the chain rule derived using the three-tiered number line model with specific values?

By tracking a small input change dxdx through nested mappings x→h(x)→g(h(x))x \to h(x) \to g(h(x)). For x=1.5x=1.5, h=x2h=x^2, and g=sin⁡g=\sin, the inner rate is h′(1.5)=3h'(1.5)=3, scaling dxdx to dh≈3dxdh \approx 3dx. The outer rate is g′(h(1.5))=cos⁡(2.25)g'(h(1.5))=\cos(2.25), scaling dhdh to dg≈cos⁡(2.25)dhdg \approx \cos(2.25)dh. Combining them gives dg≈3cos⁡(2.25)dxdg \approx 3\cos(2.25)dx, leading to the derivative 2xcos⁡(x2)2x\cos(x^2).

Conditions

  • Inner function h(x)h(x) is differentiable at xx
  • Outer function g(u)g(u) is differentiable at u=h(x)u=h(x)
  • Example uses h(x)=x2h(x)=x^2 and g(u)=sin⁡ug(u)=\sin u near x=1.5x=1.5

Reasoning, step by step

  1. Start with input x=1.5x=1.5 and apply a tiny nudge dxdx.
  2. Map to the middle tier h=x2h=x^2. Calculate local scaling factor h′(x)=2x=3h'(x) = 2x = 3.
  3. Compute intermediate change dh≈3dxdh \approx 3 dx.
  4. Map to the top tier g=sin⁡hg=\sin h. Evaluate outer derivative at the *current* value h(1.5)=2.25h(1.5)=2.25.
  5. Calculate outer scaling factor g′(2.25)=cos⁡(2.25)g'(2.25) = \cos(2.25).
  6. Compute final change dg≈cos⁡(2.25)dhdg \approx \cos(2.25) dh.
  7. Substitute dhdh: dg≈cos⁡(2.25)(3dx)dg \approx \cos(2.25) (3 dx).
  8. Generalize to formula: ddx[g(h(x))]=g′(h(x))h′(x)\frac{d}{dx}[g(h(x))] = g'(h(x))h'(x).

Example

Script at 600s describes: 'Near x=1.5x=1.5 the inner rate is 2x=32x=3, while the outer rate is cos⁡(2.25)\cos (2.25)... The overall derivative is 2xcos⁡(x2)2x\cos(x^2): evaluate the outer derivative at x², not at x.'

Common misconceptions

  • Evaluating the outer derivative g′g' at xx instead of at the inner output h(x)h(x).
  • Confusing the additive nature of the sum rule with the multiplicative nature of the chain rule.

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