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Why is Leibniz notation for the chain rule not justified by simple fraction cancellation?

Leibniz notation dydx=dydududx\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} resembles algebraic cancellation, but derivatives are limits, not ordinary fractions. The rigorous justification relies on expressing increments exactly as Δy=[g′(u)+ϵ]Δu\Delta y = [g'(u)+\epsilon]\Delta u with ϵ→0\epsilon \to 0. This ensures that error terms vanish faster than Δx\Delta x in the limit, preserving equality without dividing by an potentially zero differential dudu.

Conditions

  • Functions are differentiable.
  • The inner derivative du/dxdu/dx can be zero.
  • The argument requires rigorous limit definitions, not informal algebra.

Reasoning, step by step

  1. Acknowledge that dydx\frac{dy}{dx} looks like a fraction.
  2. Recall that derivatives are defined as limits of difference quotients.
  3. Note that if du/dx=0du/dx = 0, dividing by dudu is invalid in standard algebra.
  4. Use the exact increment formula: Δy=g′(u)Δu+ϵΔu\Delta y = g'(u)\Delta u + \epsilon \Delta u.
  5. Show that as Δx→0\Delta x \to 0, Δu→0\Delta u \to 0 and ϵ→0\epsilon \to 0.
  6. Conclude that the limit yields the product of derivatives without illegal division.

Example

If u=x2u=x^2 and x=0x=0, then du/dx=0du/dx=0. Simple cancellation would involve dividing by zero, but the limit definition handles this correctly, yielding dy/dx=0dy/dx = 0.

Common misconceptions

  • Treating dxdx and dydy as independent infinitesimal numbers that can be freely cancelled.
  • Believing the chain rule fails when the inner derivative is zero.
  • Thinking that Leibniz notation is merely a mnemonic with no rigorous basis.

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