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Business Mathematics and Basic Statistics · Ch 10 — Limits and Derivatives

The Chain Rule

8

The Chain Rule

None of §7's five formulae, on their own, can differentiate a composite function — a function of a function, such as e3x2+1e^{3x^2+1} or ln⁡(5x−2)\ln(5x-2), where the "outer" function (e(⋅)e^{(\cdot)} or ln⁡(⋅)\ln(\cdot)) is applied not to xx directly but to another function of xx. The chain rule is the tool for exactly this situation.

Note

The Chain Rule, Stated

If y=f(u)y = f(u) where u=g(x)u = g(x) (so yy is a function of xx through the intermediate quantity uu), then

dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}

Method:

  1. Identify the "outer" function and the "inner" function — write y=f(u)y=f(u) where uu is whatever expression the outer function is applied to.
  2. Differentiate the outer function with respect to uu, using the standard formulae of §7, to get dydu\dfrac{dy}{du}.
  3. Differentiate the inner function uu with respect to xx to get dudx\dfrac{du}{dx}.
  4. Multiply the two results together, then substitute back so the final answer is expressed in xx alone (not uu).

For example, to differentiate y=e3x2+1y = e^{3x^2+1}: the outer function is e(⋅)e^{(\cdot)} applied to u=3x2+1u = 3x^2+1; dydu=eu\dfrac{dy}{du}=e^u and dudx=6x\dfrac{du}{dx}=6x, so dydx=eu⋅6x=6x e3x2+1\dfrac{dy}{dx} = e^u \cdot 6x = 6x\,e^{3x^2+1}.

Note

Stay Within the Chain Rule …

Definition 11Chain rule

For y=f(u)y=f(u) with u=g(x)u=g(x): dydx=dydu⋅dudx\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}. Used to differentiate a composite function (a function nested inside another), such a …