The Chain Rule

Change passing through a second quantity

A composition $y=f(u)$ with $u=g(x)$ changes through the intermediate variable $u$ . Its rate with respect to $x$ is the rate with respect to $u$ multiplied by the rate of $u$ with respect to $x$ :

\frac{dy}{dx}=f'(g(x))g'(x).

The notation $dy/du\cdot du/dx$ is a useful mnemonic. The theorem requires the relevant derivatives to exist; it is not a license to cancel arbitrary symbols in every expression.

Identify the outermost operation first. A power of a polynomial is a composition; a polynomial multiplied by a trigonometric function is a product. An expression may need both rules.

Visual guide

VISUAL GUIDEThe inner change rescales the slope

sin x
sin 2x
Tangent to sin 2x at zero

Compare sin x with sin 2x. Near zero the second wave changes twice as fast, reflected by its tangent y = 2x. The derivative 2 cos 2x contains the inner derivative 2.

Worked example: retain the inside

For $y=(3x^2+1)^4$ , temporarily name $u=3x^2+1$ . The outer derivative is $4u^3$ and the inner derivative is $6x$ . Substitute back:

y'=4(3x^2+1)^3(6x)=24x(3x^2+1)^3.

Differentiating the inside too early and replacing it everywhere by $6x$ would lose the original composition. Only the extra multiplier becomes its derivative.

At $x=0$ , the inner rate is zero, so the composition's rate is zero even though the outer derivative at $u=1$ is $4$ . This is a useful way to interpret a zero in the product.

Worked example: several layers and a product

For $h(x)=x\sin(x^2)$ , first apply the product rule:

h'(x)=\sin(x^2)+x\cos(x^2)(2x).

The result is $\sin(x^2)+2x^2\cos(x^2)$ . The product creates two terms; the composition inside the second term supplies the factor $2x$ .

For $q(x)=\ln(1+e^{2x})$ , move from outside inward: logarithm, sum, exponential, then $2x$ . Thus $q'=2e^{2x}/(1+e^{2x})$ . The logarithm's argument is positive for every real input, so no additional real-domain exclusions occur.

Rates and units

If temperature $T$ depends on position $s$ and position depends on time, then $dT/dt=(dT/ds)(ds/dt)$ . A gradient along a pipe of $-3$ °C/m and a motion speed of $2$ m/s give a temperature change of $-6$ °C/s for the moving probe. The units reinforce the composition.

Practice

Differentiate $\sqrt{1+4x^2}$ .
Differentiate $e^{\sin x}$ .
A radius grows at $dr/dt=0.2$ cm/s. Find the rate of sphere volume when $r=3$ cm.

Show worked solutions

$(1/2)(1+4x^2)^{-1/2}(8x)=4x/\sqrt{1+4x^2}$ .
The exponential stays evaluated at $\sin x$ , giving $e^{\sin x}\cos x$ .
From $V=4\pi r^3/3$ , $dV/dt=4\pi r^2r'=4\pi(9)(0.2)=7.2\pi$ cm³/s.