Tangent Planes, Linear Approximation, and the Chain Rule

Zoom in on a point of a curve whose function is differentiable, and the curve straightens into its tangent line. Zoom in on a point of a smooth surface, and it flattens into a plane. That plane is the best linear approximation to the surface, and its slopes in the two coordinate directions are the partial derivatives already in hand.

The tangent plane

Let $S$ be the surface $z = f (x, y)$ with $f$ having continuous first partials, and let $P = (x_{0}, y_{0}, z_{0})$ be a point on it. The plane $y = y_{0}$ cuts $S$ in a curve $C_{1}$ with tangent line $T_{1}$ of slope $f_{x} (x_{0}, y_{0})$ ; the plane $x = x_{0}$ cuts $S$ in $C_{2}$ with tangent $T_{2}$ of slope $f_{y} (x_{0}, y_{0})$ .

Any plane through $P$ has the form $z - z_{0} = a (x - x_{0}) + b (y - y_{0})$ . Setting $y = y_{0}$ leaves $z - z_{0} = a (x - x_{0})$ , a line of slope $a$ that must be $T_{1}$ , so $a = f_{x} (x_{0}, y_{0})$ . Setting $x = x_{0}$ forces $b = f_{y} (x_{0}, y_{0})$ .

Compare the single-variable tangent line $y - y_{0} = f^{'} (x_{0}) (x - x_{0})$ : the plane carries one slope term per input.

The tangent plane at

P

touches the surface and shares both trace tangents

T_{1}

and

T_{2}

; near

P

the surface and plane are nearly identical.

Linear approximation

Solving the plane equation for $z$ names the linear function whose graph is the tangent plane.

Differentiability

Having partial derivatives at a point is weaker than the surface being smooth there: a function can possess $f_{x}$ and $f_{y}$ at $(a, b)$ yet fail to have a tangent plane, because the partials only probe two directions. Differentiability demands that the linear approximation be genuinely good in every direction.

Write the increment $Δ z = f (a + Δ x, b + Δ y) - f (a, b)$ for the actual change in $f$ .

The leading two terms are the tangent-plane change; the $ε$ -terms are the leftover, which must vanish faster than the step. A differentiable function is one whose tangent plane approximates the surface well near the point of tangency. The definition is awkward to check directly, so the working criterion is a sufficient condition.

Every function assembled from polynomials, roots, exponentials, logarithms, and trigonometric functions by the usual operations has continuous partials on its domain, so it is differentiable there. For $f (x, y) = x e^{x y}$ , both $f_{x} = e^{x y} (1 + x y)$ and $f_{y} = x^{2} e^{x y}$ are continuous, so $f$ is differentiable everywhere; at $(1, 0)$ the linearization is $L (x, y) = x + y$ .

Differentials

For $z = f (x, y)$ the differentials $d x$ and $d y$ are independent variables, free to take any value. The total differential $d z$ is defined by

d z = f_{x} (x, y) d x + f_{y} (x, y) d y = \frac{\partial z}{\partial x} d x + \frac{\partial z}{\partial y} d y .

Setting $d x = Δ x$ and $d y = Δ y$ , the differential $d z$ is the change in height along the tangent plane, while $Δ z$ is the change in height along the surface. For a differentiable function the two agree closely when $d x, d y$ are small, and $d z$ is the estimate used in error propagation.

Δ z

is the true change along the surface;

d z

is the change along the tangent plane. Their difference is the higher-order leftover.

The chain rule

The one-variable chain rule differentiates a composition: if $y = f (x)$ and $x = g (t)$ , then $\frac{d y}{d t} = \frac{d y}{d x} \frac{d x}{d t}$ . With several intermediate variables, the rule becomes a sum with one term per intermediate route.

Tree diagram for

z = f (x, y)

with

x, y

each a function of

t

. Sum the two paths from

z

t

, each a partial of

z

times a derivative of

x

y

Here $s, t$ are the independent variables, $x, y$ the intermediate variables, and $z$ the dependent variable. Each formula has one term per intermediate variable, and each term mirrors the one-dimensional chain rule.

The tree diagram organizes the bookkeeping. Draw a branch from $z$ to each variable it depends on, then from each of those to the variables below. $\partial z / \partial s$ is the sum, over every path from $z$ down to $s$ , of the product of the derivatives along that path.

Tree diagram for

z = f (x, y)

with

x, y

functions of

s, t

. Multiply derivatives down each path from

z

s

, then add the paths.

The pattern scales without change.

Implicit differentiation

The chain rule sharpens implicit differentiation. Suppose $F (x, y) = 0$ defines $y$ as a function of $x$ . Differentiating both sides with respect to $x$ , and using $d x / d x = 1$ ,

\frac{\partial F}{\partial x} \frac{d x}{d x} + \frac{\partial F}{\partial y} \frac{d y}{d x} = 0 ⟹ \frac{d y}{d x} = - \frac{F _{x}}{F _{y}} (F_{y} \neq = 0) .

For a surface $F (x, y, z) = 0$ defining $z$ implicitly as $z = f (x, y)$ , the same argument with the three-variable chain rule gives both partials at once.

The level surface

F (x, y, z) = 0

defines

z

as a function of

x, y

wherever

F_{z} \neq = 0

; the implicit partials come from the chain rule applied to

F = 0

Existence of these functions is underwritten by the Implicit Function Theorem: if $F (a, b, c) = 0$ , $F_{z} (a, b, c) \neq = 0$ , and $F_{x}, F_{y}, F_{z}$ are continuous near $(a, b, c)$ , then $F = 0$ does define $z = f (x, y)$ near that point, with the partials above.¹

Differentiating in an arbitrary direction, rather than along the two axes, combines both partials into the gradient vector.

Stewart, §14.5 — the Implicit Function Theorem stated for $F (x, y) = 0$ and $F (x, y, z) = 0$ , giving conditions ( $F$ continuously differentiable, relevant partial nonzero) under which the implicit function exists and is differentiable. ↩