The Chain Rule
Differentiate a composition by multiplying the rate of the outer function by the rate of the inner one. The rule is what makes substitution, related rates and implicit differentiation possible.
What a learner can do afterwards
- Differentiate (3x² + 1)⁵ and name the inner and outer functions
- Apply the rule twice on a triple composition
- Explain the rule as two rates multiplying, using units
1 · Read
For f(g(x)), differentiate the outside first with the inside untouched, then multiply by the derivative of the inside. In symbols: f'(g(x)) times g'(x). With (3x squared + 1)^5, the outside gives 5(3x squared + 1)^4 and the inside gives 6x, so the answer is 30x(3x squared + 1)^4.
Three layers need the rule twice. For e^(sin(x squared)), work inward: keep e^(sin(x squared)), multiply by cos(x squared), then by 2x. The trailing 2x factor makes the gradient 0 at x = 0. Likewise sin(ln(x squared)) gives cos(ln(x squared)) times 2/x.
Rates multiply, and units show it: dy/dx times dx/dt gives dy/dt. In related rates, differentiate with respect to time before touching the numbers. For a balloon with V = (4/3) pi r cubed, dV/dt = 4 pi r squared times dr/dt, so r = 3 with dr/dt = 0.5 gives 18 pi.
Two slips spoil answers. Never drop the inner derivative, and keep the inside intact: cos(x squared) gives -2x sin(x squared), never -sin(2x). Check by differentiating your answer: the chain rule should bring back the start.
Outside first with inside untouched, times the inside derivative, once per layer.
2 · Watch
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Where it sits
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.