Are Directional Derivatives Scalars or Vectors?

A directional derivative describes how a function changes as you move from a point in a specified direction. The direction is represented by a vector, which is why it is easy to confuse the input with the output. The type of the derivative depends on the type of the function: a scalar-valued function has a scalar directional derivative, while a vector-valued function generally has a vector directional derivative.

The standard scalar-valued case

Suppose $f: \mathbb{R}^n \to \mathbb{R}$, so each input is a vector but each function value is a real number. At a point $\mathbf{a}$, the directional derivative in the direction of a vector $\mathbf{u}$ is defined, when the limit exists, by

$$D_{\mathbf{u}}f(\mathbf{a}) = \lim_{h \to 0} \frac{f(\mathbf{a}+h\mathbf{u})-f(\mathbf{a})}{h}.$$

The numerator is a difference of real numbers, so the limit is a real number. Thus, the directional derivative is a scalar in this common setting. The vector $\mathbf{u}$ tells you which way to move; it is not the answer.

If $f$ is differentiable, its gradient gives a convenient formula:

$$D_{\mathbf{u}}f(\mathbf{a}) = \nabla f(\mathbf{a}) \cdot \mathbf{u}.$$

This is a dot product, and therefore produces a scalar. If $\mathbf{u}$ is a unit vector, that scalar is the instantaneous rate of change per unit distance in that direction. If the vector is not a unit vector, the formula still applies, but its magnitude scales the result.

A worked example

Let

$$f(x,y)=x^2y+\sin y$$

and find the directional derivative at $(1,0)$ in the direction of $(3,4)$.

First, convert the direction into a unit vector:

$$\mathbf{u}=\frac{(3,4)}{\sqrt{3^2+4^2}}=\left(\frac35,\frac45\right).$$

Next, calculate the gradient:

$$\nabla f(x,y)=\left(2xy,\ x^2+\cos y\right).$$

At $(1,0)$, this is $\nabla f(1,0)=(0,2)$. Take its dot product with the unit direction:

$$D_{\mathbf{u}}f(1,0)=(0,2)\cdot\left(\frac35,\frac45\right)=\frac85.$$

The answer $\frac85$ is a scalar. It says that, at the given point, the function increases at an instantaneous rate of $\frac85$ per unit distance when you move in the direction $(3,4)$.

If you instead use $(3,4)$ directly in the gradient formula, you get $(0,2)\cdot(3,4)=8$. That is the directional derivative with respect to the parameter $h$ in the path $(1,0)+h(3,4)$; it is not the rate per unit distance. Normalizing the direction divides that rate by the length $5$.

What changes for a vector-valued function?

Now let $F: \mathbb{R}^n \to \mathbb{R}^m$. Its values are vectors in $\mathbb{R}^m$, so the difference quotient

$$\frac{F(\mathbf{a}+h\mathbf{u})-F(\mathbf{a})}{h}$$

is vector-valued. When its limit exists, the directional derivative is therefore a vector in $\mathbb{R}^m$.

For example, take $F(x,y)=(x^2+y,xy)$ and evaluate at $(1,2)$ in the direction $(1,0)$. Moving in that direction changes $x$ while keeping $y$ fixed. The derivative is

$$D_{(1,0)}F(1,2)=\left(2x,\ y\right)\bigg|_{(1,2)}=(2,2).$$

Here the result is a vector because $F$ itself returns a two-component vector. More generally, the derivative of $F$ at a point is a linear map, often represented by its Jacobian matrix; applying that map to the direction vector gives the directional derivative vector.

A useful way to keep the terms straight

For a differentiable function $F: \mathbb{R}^n \to \mathbb{R}^m$, the derivative at $\mathbf{a}$ is a linear map

$$DF(\mathbf{a}):\mathbb{R}^n \to \mathbb{R}^m.$$

The direction $\mathbf{u}$ is an input to that map, and $DF(\mathbf{a})[\mathbf{u}]$ is its output. For a scalar-valued function, $m=1$, so that output is a scalar. For a vector-valued function, $m$ can be greater than one, so the output is a vector. In multivariable calculus, the term “directional derivative” often refers specifically to the scalar-valued case, but the same derivative-map idea applies to vector-valued functions too.

Common mistakes

• Calling the direction the derivative: The direction is an input vector. The derivative is the resulting rate or change, whose type depends on the function’s codomain.
• Assuming every directional derivative is a scalar: That is true for real-valued functions, but not for vector-valued ones.
• Forgetting to normalize: If a question asks for change per unit distance, use a unit direction vector. If it gives a non-unit vector, check whether the intended derivative is with respect to that parameterized displacement or per unit distance.
• Confusing the gradient with the directional derivative: For a scalar-valued function, the gradient is a vector, while its dot product with a direction gives the scalar directional derivative.