3.1 The Chain Rule

Syllabus
2020
Topic
3.1
Level

Differentiate from the Outside In

A composite function has the form y=f(g(x))y=f(g(x)): the inner function gg produces the input for the outer function ff. The total derivative multiplies the rate through both stages.

\frac{d}{dx}f(g(x))=f'(g(x))g'(x)\qquad\text{or}\qquad\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

  1. Identify the inner expression u=g(x)u=g(x).
  2. Differentiate the outer function with respect to uu, leaving uu in place.
  3. Multiply by du/dxdu/dx.
  4. Substitute the original inner expression and simplify.

For y=(3x21)5y=(3x^2-1)^5, let u=3x21u=3x^2-1. Then dy/du=5u4dy/du=5u^4 and du/dx=6xdu/dx=6x, so dydx=5(3x21)4(6x)=30x(3x21)4.\frac{dy}{dx}=5(3x^2-1)^4(6x)=30x(3x^2-1)^4.

Do not stop after differentiating only the outer function: 5(3x21)45(3x^2-1)^4 is missing the inner derivative 6x6x. The extra factor is unnecessary only when the inner function is exactly xx, whose derivative is 11.