The chain rule: gears
Tonight's idea
Three gears in a row. The 36-tooth gear \(x\) drives the 18-tooth gear \(u\), which drives the 6-tooth gear \(y\). Their speed magnitudes multiply: \(2 \times 3 = 6\). Each mesh also reverses direction. If clockwise turns of \(x\) are positive, then \(du/dx = -2\) and \(dy/du = -3\), so \(dy/dx = (-3)(-2) = 6\). Two reversals cancel. Signed rates through a chain multiply.
Functions chain the same way. Suppose \(y = f(u)\) and \(u = g(x)\), so \(y = f(g(x))\): \(x\) goes into \(g\), and what comes out goes into \(f\). Nudge \(x\) by \(\Delta x\). Then \(u\) changes by \(\Delta u\) and \(y\) by \(\Delta y\), and as long as \(\Delta u \ne 0\),
\[\frac{\Delta y}{\Delta x} = \frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x}.\]
This cancellation is useful intuition when \(\Delta u \ne 0\), not a complete proof: \(\Delta u\) may be zero even when \(\Delta x \ne 0\). The derivative definition handles those flat cases and gives:
\(\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}\), or \(\big(f(g(x))\big)' = f'(g(x))\,g'(x)\): differentiate the outer function at the inner value, then multiply by the derivative of the inner.
Gears have constant rates. Functions do not, so the rule is read at a point: \(dy/du\) is evaluated at the current \(u = g(x)\), not at \(x\). That is what \(f'(g(x))\) says.
Watch one: \(y = (x^2 + 1)^5\)
Expanding is possible but miserable. Instead name the inside: \(u = x^2 + 1\), so \(y = u^5\). Then \(dy/du = 5u^4\) and \(du/dx = 2x\), and
\[\frac{dy}{dx} = 5u^4 \cdot 2x = 10x\,(x^2 + 1)^4.\]
Check at \(x = 1\): slope \(10 \cdot 16 = 160\). Numerically, \(y(1.001) - y(1) = 0.1604\), so the secant slope is about 160.4. Good. The order of the pieces matters: the outer derivative \(5u^4\) is evaluated at \(u = x^2 + 1\), never at \(x\). Writing \(5x^4 \cdot 2x\) is the classic mistake.
Your turn
- Name the inner function, then differentiate: (a) \(y = (3x + 1)^4\); (b) \(y = (2x^3 - 1)^4\).
- Differentiate \(y = \dfrac{1}{x^2 + 1}\) by writing it as \(u^{-1}\) with \(u = x^2 + 1\), using \((u^{-1})' = -u^{-2}\). Where is the slope zero, and what does the graph do there?
- Given \(g(1) = 2\), \(g'(1) = 4\), \(f(2) = 7\), \(f'(2) = 5\), \(f(1) = 3\), \(f'(1) = -2\), \(g(3) = 0\), \(g'(3) = 6\), find (a) the derivative of \(f(g(x))\) at \(x = 1\); (b) the derivative of \(g(f(x))\) at \(x = 1\). Why are they different?
- (a) Differentiate \(y = \big((x^2 + 1)^3 + x\big)^2\), a chain of three gears, and evaluate at \(x = 1\). (b) Let \(s(x) = \sqrt{x}\), so \(s(x)^2 = x\). For \(x > 0\), differentiate both sides with the chain rule and solve for \(s'(x)\). Then find the derivative of \(\sqrt{x^2 + 1}\).
Stretch: running the gears backwards
Let \(g\) be the inverse of \(f\), so \(f(g(x)) = x\). Differentiate both sides and solve for \(g'(x)\). When may you divide? Apply the formula to \(f(x) = x^3\), stating where the cube-root derivative formula applies and what happens at \(x = 0\).
Answers
- (a) \(u = 3x + 1\): \(y' = 4u^3 \cdot 3 = 12(3x + 1)^3\). (b) \(u = 2x^3 - 1\): \(y' = 4u^3 \cdot 6x^2 = 24x^2 (2x^3 - 1)^3\).
- \(y' = -u^{-2} \cdot 2x = -\dfrac{2x}{(x^2 + 1)^2}\). Zero at \(x = 0\), where the graph has its peak, \(y = 1\); it falls away on both sides.
- (a) \(f'(g(1))\,g'(1) = f'(2) \cdot 4 = 20\). (b) \(g'(f(1))\,f'(1) = g'(3) \cdot (-2) = -12\). The outer derivative is read at the inner value, and the inner values differ: \(g(1) = 2\) for one, \(f(1) = 3\) for the other. Composition is not commutative, so neither is its derivative.
- (a) \(y' = 2\big((x^2 + 1)^3 + x\big)\big(3(x^2 + 1)^2 \cdot 2x + 1\big)\). At \(x = 1\): \(2 \cdot 9 \cdot 25 = 450\). (b) \(2s(x)\,s'(x) = 1\), so \(s'(x) = \dfrac{1}{2\sqrt{x}}\) for \(x > 0\). The square root is not differentiable at \(x = 0\). Since \(x^2 + 1 > 0\), \(\big(\sqrt{x^2 + 1}\big)' = \dfrac{1}{2\sqrt{x^2 + 1}} \cdot 2x = \dfrac{x}{\sqrt{x^2 + 1}}\) for every real \(x\).
Stretch. \(f'(g(x))\,g'(x) = 1\). If \(f'(g(x)) \ne 0\), then \(g'(x) = \dfrac{1}{f'(g(x))}\): the inverse's slope is the reciprocal of the original's slope, read at the matching point. With \(f(x) = x^3\), \(f'(x) = 3x^2\) and \(g(x) = x^{1/3}\), so for \(x \ne 0\), \(g'(x) = \dfrac{1}{3(x^{1/3})^2} = \dfrac{1}{3x^{2/3}} = \tfrac13 x^{-2/3}\), exactly \(n x^{n-1}\) with \(n = \tfrac13\). At \(x = 0\), the difference quotient is \(h^{1/3}/h = 1/h^{2/3}\), which grows without bound as \(h \to 0\). The graph has a vertical tangent there, not a finite derivative. Sheet 4 uses the inverse rule to differentiate \(\ln x\).