Chain rule
Follow a change through nested functions by multiplying their local rates.
A change through two functions
A change in \(x\) first changes \(u\), which then changes \(y\). The chain rule multiplies these two local rates. For a small change, the idea is
\[\Delta u\approx g^{\prime}(x)\Delta x,\qquad \Delta y\approx f^{\prime}(u)\Delta u.\]These approximations motivate the rule; they are not its proof.
Rule
Chain rule
If \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g(x)\), then
\[\frac{d}{dx}f(g(x))=f^{\prime}(g(x))g^{\prime}(x).\]Equivalently, with \(u=g(x)\) and \(y=f(u)\),
\[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}.\]Evaluate the outer derivative at the inner output, then multiply by the inner derivative.
Proof (optional)
Fix \(x=a\) and write \(b=g(a)\). Differentiability of \(f\) means we can write
\[f(b+k)-f(b)=\bigl(f^{\prime}(b)+r(k)\bigr)k,\]where \(r(k)\to0\) as \(k\to0\). For \(k\ne0\), define \(r(k)=[f(b+k)-f(b)]/k-f^{\prime}(b)\); set \(r(0)=0\), so the identity holds at zero too.
Put \(k=g(a+h)-g(a)\). Since differentiability implies continuity, \(k\to0\) as \(h\to0\). For \(h\ne0\),
\[\frac{f(g(a+h))-f(g(a))}{h} =\bigl(f^{\prime}(b)+r(k)\bigr)\frac{g(a+h)-g(a)}h.\]Taking limits gives \(f^{\prime}(b)g^{\prime}(a)=f^{\prime}(g(a))g^{\prime}(a)\). We never divide by \(k\) in this last step, so the argument also covers nearby inputs with no inner change.
Filling a widening tank
A conical tank is filled so its water height rises steadily: \(h(t)=1+t/2\) metres, with \(t\) in seconds. The tank’s shape gives water-surface radius \(r=h/2\), so the cone-volume formula gives
\[V=\frac13\pi r^2h=\frac{\pi h^3}{12}.\]Time changes the height; height changes the volume. By the power rule, \(dh/dt=1/2\) and \(dV/dh=\pi h^2/4\). The chain rule connects them:
\[\frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt} =\frac{\pi h^2}{4}\cdot\frac12=\frac{\pi h^2}{8}.\]The height rises at the same speed throughout, but the required inflow increases as the tank widens. The factor \(dV/dh\) is the water-surface area: it converts a height rate into a volume rate.
Worked examples
Example 1
A power of an expression
For \(y=(2-x^3)^5\), choose \(u=2-x^3\), then \(y=u^5\). Differentiate each stage:
\[\frac{du}{dx}=-3x^2,\qquad \frac{dy}{du}=5u^4.\]Combine them and replace \(u\):
\[\frac{dy}{dx}=5u^4(-3x^2)=-15x^2(2-x^3)^4.\]The outer derivative alone gives a rate with respect to \(u\); multiplying by \(du/dx\) changes it to a rate with respect to \(x\).
Example 2
More than two layers
Let \(y=\sqrt{1+(2x-1)^2}\). The layers, from inside out, are \(2x-1\), its square plus \(1\), and the square root. The chain rule applies at each layer:
\[\begin{aligned} y^{\prime} &=\frac{1}{2\sqrt{1+(2x-1)^2}}\cdot2(2x-1)\cdot2\\ &=\frac{2(2x-1)}{\sqrt{1+(2x-1)^2}}. \end{aligned}\]The expression under the root is at least \(1\), so the function and derivative are defined for every real \(x\).
Example 3
Combine the product and chain rules
For \(x\ne-1/2\), rewrite
\[R(x)=\frac{x^2}{(1+2x)^3}=x^2(1+2x)^{-3}.\]The product rule handles the two factors; the chain rule handles the second factor:
\[\begin{aligned} R^{\prime}(x) &=2x(1+2x)^{-3}+x^2\bigl[-3(1+2x)^{-4}\cdot2\bigr]\\ &=\frac{2x(1+2x)-6x^2}{(1+2x)^4}\\ &=\frac{2x(1-x)}{(1+2x)^4}. \end{aligned}\]The quotient rule gives the same result if applied to the original form, using the chain rule for its denominator. The domain restriction remains.