The Chain Rule

The chain rule is the computational backbone of differentiation itself: nearly every derivative formula for elementary functions built from compositions — $\sin(x^2)$, $\ln(\cos x)$, $(3x+1)^7$ — relies on it directly. In multivariable calculus it generalizes to the Jacobian chain rule, $D(g\circ f)(a) = Dg(f(a))\, Df(a)$, a matrix product replacing the scalar product, foundational for Derivatives of Inverse Functions and the inverse and implicit function theorems. In physics, related rates and dimensional analysis in mechanics (e.g. $\frac{d}{dt}[\text{potential energy}(x(t))] = U'(x(t))\,x'(t)$, giving force times velocity as power) are direct chain-rule statements. In machine learning, the backpropagation algorithm that trains neural networks is a systematic, layer-by-layer application of the chain rule to compute gradients of a loss function through many nested compositions of affine maps and activation functions. In probability and statistics, the change-of-variables formula for probability density functions of transformed random variables uses the chain rule (via the derivative of the inverse transform) to convert one density into another. In cryptography and numerical analysis, sensitivity and error-propagation analyses — how small perturbations in an input propagate through a composed algorithm — are chain-rule estimates. The chain rule also underlies the derivation of the derivative of an inverse function, $(f^{-1})'(b) = 1/f'(f^{-1}(b))$, by differentiating the identity $f(f^{-1}(y))=y$, and it is a key ingredient in building up higher-order Taylor expansions of composite functions, connecting directly to Taylor's Theorem with Remainder.

Ask the tutor: Can you show me how the increment-based chain rule proof adapts when $a$ is a left endpoint of the domain, using one-sided derivatives?

Think of $f$ and $g$ as zoom factors. Near $a$, the function $f$ stretches small displacements by an approximate factor $f'(a)$: moving $x$ by a tiny amount $h$ moves $f(x)$ by about $f'(a)\, h$. Near $b = f(a)$, the function $g$ stretches displacements by a factor $g'(b)$. Composing the two zooms multiplies the factors: a displacement $h$ in $x$ produces about $f'(a) h$ in $y = f(x)$, which in turn produces about $g'(b) \cdot f'(a) h$ in $z = g(y)$. The overall zoom factor of $h = g\circ f$ at $a$ is the product $g'(b) f'(a)$. The delicate point is that "$f$ stretches by $f'(a)$" is only an approximate, local statement — it can fail to be exactly true for any particular nonzero increment, and in fact the increment $f(a+h)-f(a)$ might be exactly zero for some sequence of small $h$ even though $f$ is not locally constant. The rigorous proof needs a formulation of differentiability robust enough to survive this, which is exactly the increment/error-term formulation introduced below.

Why this lemma matters. The increment equation $F(c+t) = F(c) + Lt + \varepsilon(t) t$ holds identically for every small $t$, including $t = 0$ and including values of $t$ where some other quantity we might want to divide by happens to vanish. This is exactly what we need to route around the division-by-zero obstruction in the composite derivative computation.

Example 1: A composite with a non-monotonic inner function. Let

Here $f(x) = x^3 - 4x$ (which is not monotonic — it has local extrema, so $f(a+t) - f(a)$ genuinely vanishes for infinitely many small nonzero $t$ near certain points $a$, e.g. near a critical point) and $g(y) = \sin y$. We have $f'(x) = 3x^2 - 4$ and $g'(y) = \cos y$, both differentiable everywhere (polynomial and sine are differentiable on all of $\mathbb{R}$, standard facts from Algebraic Rules of Differentiation and direct computation). By the chain rule, at any $a \in \mathbb{R}$,

Check at $a = 2/\sqrt{3}$ (a critical point of $f$, since $3a^2 - 4 = 0$ there): $f'(2/\sqrt3) = 0$, so $h'(2/\sqrt3) = \cos(f(2/\sqrt3)) \cdot 0 = 0$, regardless of the cosine factor. This matches intuition: near this $a$, $f$'s increments are $O(t^2)$ rather than $O(t)$, so composing with any differentiable $g$ still gives an $h$ with zero linear term at $a$. Note also that at exactly this point, $f(a+t) - f(a)$ is not eventually nonzero for small $t \ne 0$ in general (near a local extremum of a smooth function, the difference can vanish only at isolated points typically, but the point of this example is that the proof method never needed to worry about whether it vanishes — the increment method handles it uniformly regardless). Let's also numerically sanity-check at $a = 1$: $f(1) = 1 - 4 = -3$, $f'(1) = 3 - 4 = -1$. So

Direct check via difference quotient with $t = 0.001$: $f(1.001) = (1.001)^3 - 4(1.001) = 1.003003001 - 4.004 = -3.000996999$. Then $h(1.001) = \sin(-3.000996999) \approx -0.141254$, and $h(1) = \sin(-3) \approx -0.141120$. Difference quotient $\approx (-0.141254 - (-0.141120))/0.001 = -0.000134/0.001 = -0.134$. This does not match $0.98999$ — let us recompute carefully. $\sin(-3) = -\sin(3) \approx -0.141120$. $\sin(-3.000996999) = -\sin(3.000996999)$. Using $\sin(3 + \Delta) \approx \sin 3 + \Delta\cos 3$ with $\Delta = 0.000996999$: $\sin(3.000997) \approx 0.141120 + 0.000997\times(-0.989992) \approx 0.141120 - 0.000987 = 0.140133$. So $\sin(-3.000997) \approx -0.140133$. Difference quotient: $\big(h(1.001) - h(1)\big)/0.001 = \big(-0.140133 - (-0.141120)\big)/0.001 = 0.000987/0.001 = 0.987$. This matches $h'(1) \approx 0.98999$ well (small discrepancy from linear approximation truncation), confirming the chain-rule computation. (The earlier miscomputation of $f(1.001)$ above used an arithmetic slip; the corrected value $f(1.001) \approx -3.000997$ is what was used in the second, correct pass.)

Example 2: A chain of three functions, asymmetric at every stage. Let

Write $H = f_3 \circ f_2 \circ f_1$ with

Check domains and interiority: $f_1(x) > 1 > 0$ for all real $x$, so $f_2$ is applied to a point in $(0,\infty)$, an open set, hence interior; $f_2(u) = \sqrt{u} > 1$ there, so $f_3 = \ln$ is applied to a point in $(1,\infty) \subset (0,\infty)$, again interior. All composition points are interior, satisfying the theorem's hypotheses. Derivatives:

each differentiable on the relevant domain ($f_1$ on $\mathbb R$; $f_2$ on $(0,\infty)$ since $\sqrt{u}$ is not differentiable at $u=0$ but we never land there; $f_3$ on $(0,\infty)$). By the finite-chain corollary, at any $a \in \mathbb{R}$, with $u = f_1(a) = 1+e^{2a}$ and $v = f_2(u) = \sqrt{1+e^{2a}}$:

Let's verify at $a = 0$: the formula gives $H'(0) = \dfrac{1}{1+1} = \dfrac12$. Direct check: $H(x) = \tfrac12 \ln(1+e^{2x})$ (using $\ln\sqrt{u} = \tfrac12 \ln u$ — an algebraic identity, not a differentiation shortcut, just to cross-check), so $H'(x) = \tfrac12 \cdot \dfrac{2e^{2x}}{1+e^{2x}} = \dfrac{e^{2x}}{1+e^{2x}}$, agreeing exactly with the chain-rule computation for all $x$, in particular $H'(0) = 1/2$. Numerically, $H(0.001) = \ln\sqrt{1+e^{0.002}}$. Compute $e^{0.002} \approx 1.002002$, so $1+e^{0.002} \approx 2.002002$, $\sqrt{2.002002}\approx 1.415275$, $\ln(1.415275) \approx 0.347385$. And $H(0) = \ln\sqrt2 \approx 0.346574$. Difference quotient $\approx (0.347385-0.346574)/0.001 = 0.811$ — but this should approximate $H'(0)=0.5$; the discrepancy signals an arithmetic check is needed with a smaller step or more precision, since first-order finite differences at $t=0.001$ carry $O(t)$ error only if computed with enough significant digits. Recomputing with more precision: $e^{0.002} = 1 + 0.002 + 0.000002 + \cdots \approx 1.0020020$. Then $1+e^{0.002} = 2.0020020$. $\sqrt{2.0020020}$: since $\sqrt{2}\approx 1.4142136$ and $d(\sqrt u)/du = 1/(2\sqrt u)$, $\sqrt{2.0020020} \approx 1.4142136 + \dfrac{0.0020020}{2(1.4142136)} \approx 1.4142136 + 0.0007080 = 1.4149216$. Then $\ln(1.4149216) \approx \ln(1.4142136) + \dfrac{0.0007080}{1.4142136} \approx 0.3465736 + 0.0005007 = 0.3470743$. Difference quotient: $(0.3470743 - 0.3465736)/0.001 = 0.0005007/0.001 = 0.5007$, which matches $H'(0) = 0.5$ closely (the residual $0.0007$ is the expected second-order error). This confirms the chain-rule formula.