Directional Derivatives, the Gradient, and the Chain Rule

The derivative $D f (x_{0})$ is a full linear transformation. Two narrower questions extract less from it and are often what one needs: the rate of change of $f$ along a single direction, and, for a scalar field, the direction of fastest increase. The first answer is the directional derivative, a single number; the second is the gradient, the vector that assembles the directional derivatives.

The directional derivative

Fix an open neighborhood $U$ of $x_{0} \in R^{n}$ and a scalar field $f : U \to R$ . To measure the rate of change of $f$ as we leave $x_{0}$ along a unit vector $e$ , restrict $f$ to the line $t \mapsto x_{0} + t e$ and differentiate the resulting one-variable function at $t = 0$ .

When $f$ is differentiable at $x_{0}$ , the directional derivative is not a new computation — it is the derivative read along $e$ . Taking $x = x_{0} + t e$ in the definition of $D f (x_{0})$ gives, for any $ε > 0$ and small $∣ t ∣$ ,

\frac{f ( x _{0} + t e ) - f ( x _{0} )}{t} - D f (x_{0}) e \leq ε,

so the limit exists and

t \to 0 lim \frac{f ( x _{0} + t e ) - f ( x _{0} )}{t} = D f (x_{0}) e .

The partial derivatives are the special case $e = e_{i}$ : $\partial f / \partial x_{i}$ is the directional derivative along the $i$ -th coordinate axis.

The directional derivative is the slope of

f

restricted to the line through

x_{0}

along

e

: the rise of the graph per unit step in direction

e

Directional derivatives do not imply differentiability. As with partials, the existence of every directional derivative at a point is weaker than differentiability — it does not even force continuity.

This is stronger than the axis counterexample for partial derivatives: sampling $f$ along every straight line still misses its behavior along curves, and only differentiability controls all approaches at once.

Tangent plane. For $f : R^{2} \to R$ , the directional derivatives assemble the tangent plane to the graph at $(x_{0}, f (x_{0}))$ :

z = f (x_{0}) + D f (x_{0}) (x - x_{0}) .

For $f (x, y) = x^{2} + y$ at $(1, 2)$ , $D f = (2 x, 1)$ so $D f (1, 2) = (2, 1)$ , and the tangent plane is $z = 3 + 2 (x - 1) + (y - 2)$ , i.e. $2 x + y - z = 1$ .

The chain rule

Derivatives of composites are computed by matrix multiplication of the Jacobians, in the order the maps are applied.

The dimensions line up: $D f (x_{0}) : R^{n} \to R^{m}$ and $D g (f (x_{0})) : R^{m} \to R^{p}$ , so their product is a well-defined linear map $R^{n} \to R^{p}$ . In coordinates, with $h = g \circ f$ ,

\frac{\partial h _{i}}{\partial x _{k}} (x) = j = 1 \sum m \frac{\partial g _{i}}{\partial y _{j}} (f (x)) \frac{\partial f _{j}}{\partial x _{k}} (x),

which reproduces the row-times-column rule of matrix multiplication.

The chain rule composes derivatives by multiplying Jacobians in the order the maps are applied, each a linear map between the intermediate spaces.

The other differentiation rules also hold. The derivative operator $D$ is linear — $D (u + v) = D u + D v$ and $D (α u) = α D u$ — and there is a product rule for a scalar times a vector field.

Abbreviated, $D (g f) = g D f + D g f$ , with the understanding that $g$ scales the vector $f$ and $D g w$ is a scalar scaling it too.

The geometry of the gradient

For a scalar field $f : A \subseteq R^{n} \to R$ , the derivative is the single row vector

grad f (x) = \nabla f (x) = (\frac{\partial f}{\partial x _{1}}, \dots, \frac{\partial f}{\partial x _{n}}),

and the directional derivative becomes an inner product, $D f (x_{0}) e = \nabla f (x_{0}) \cdot e$ . Two geometric facts follow from that one identity.

Steepest ascent. Writing the inner product with the angle $θ$ between $\nabla f (x_{0})$ and the unit direction $e$ ,

\nabla f (x_{0}) \cdot e = ∥ \nabla f (x_{0})∥ cos θ .

This is largest when $θ = 0$ , i.e. when $e$ points along $\nabla f (x_{0})$ . The gradient points in the direction of fastest increase of $f$ , and its length is that maximal rate. The opposite direction, $- \nabla f$ , is the direction of steepest descent.

The directional rate

\nabla f \cdot e = ∥ \nabla f ∥ cos θ

is the projection of the gradient onto

e

, largest when

e

aligns with the gradient.

Orthogonality to level sets. Let $S = {x : f (x) = c}$ be a level set. If $t \mapsto γ (t)$ is a differentiable curve lying in $S$ with $γ (0) = x_{0}$ , then $f (γ (t)) = c$ is constant, and differentiating with the chain rule,

\nabla f (x_{0}) \cdot γ^{'} (0) = 0.

Since $γ^{'} (0)$ can be any tangent vector to $S$ at $x_{0}$ , the gradient is orthogonal to $S$ . The tangent plane to $S$ at $x_{0}$ is therefore $\nabla f (x_{0}) \cdot (x - x_{0}) = 0$ .

The gradient at a point is orthogonal to the level curve through it and points toward higher values; a step along a level curve leaves

f

unchanged.

The mean value theorem in several variables

The one-variable mean value theorem — $f (b) - f (a) = f^{'} (c) (b - a)$ for some interior $c$ — does not carry over intact to vector-valued maps.

For a scalar field the theorem does hold, along the segment joining two points. Call $c$ between $x$ and $y$ if $c = (1 - λ) x + λ y$ for some $λ \in [0, 1]$ .

The segment must actually lie inside $A$ , which is guaranteed when $A$ is convex: every segment between two of its points stays in the set. A useful consequence: on a convex open set, $D f = 0$ everywhere forces $f$ to be constant, because each component satisfies $f_{j} (y) - f_{j} (x) = D f_{j} (c_{j}) (y - x) = 0$ . This is the several-variable version of zero derivative means constant, and it needs the convexity hypothesis to connect the points.

The theorem needs the whole segment from

x

y

inside the domain; a convex set contains it, and on a non-convex set the segment can escape.

For vector-valued $f : R^{n} \to R^{m}$ the theorem holds in componentwise form: there are points $c_{1}, \dots, c_{m}$ on the segment (one per component, not a single shared $c$ ) with $f_{j} (y) - f_{j} (x) = D f_{j} (c_{j}) (y - x)$ .

Statement	Scalar field $f : R^{n} \to R$	Vector map $f : R^{n} \to R^{m}$
one shared point $c$	yes, on the segment	no (the $(x^{2}, x^{3})$ example)
componentwise, one $c_{j}$ each	trivially (single component)	yes
$D f = 0$ on convex open set $\Rightarrow$ constant	yes	yes

The vanishing of $\nabla f$ at a critical point is classified by the second derivative, through the Hessian and Taylor's theorem.¹

Shkoller, MAT125B Lecture Notes, §2.6 (directional derivative, Definition 2.23, and Example 2.24), §2.7 (chain rule Theorem 2.27 and product rule Theorem 2.31), §2.8 (geometry of the gradient), and §2.9 (mean value theorem for scalar fields, Theorem 2.37, and the vector-valued failure). ↩