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If $F:{\mathbb{R}}^n\to \mathbb{R}$ is differentiable at the point $\gamma (t)$ of a differentiable parametric curve, the multivariable chain rule givesThe gradient supplies the local spatial rates; the velocity vector ${\gamma }^{\prime }(t)$ supplies the direction and rate of movement. Their dot product is the change in the sampled scalar field per unit parameter. If the parameter is time, this is a rate per unit time; a directional derivative per unit distance instead uses a unit vector.

Take $F(x,y)=x^2+2y^2$ and the unit circle path $\gamma (t)=(\cos t,\sin t)$. Along the path,At $t=\pi /4$, the gradient is $(\sqrt{2},2\sqrt{2})$, the velocity vector is $(-1/\sqrt{2},1/\sqrt{2})$, and their dot product is $-1+2=1$. The sampled value is $3/2$ and increases at rate one, even though one coordinate is decreasing. This is why a moving observation must combine every coordinate's contribution.

For a level curve, $F(\gamma (t))$ is constant, so the derivative along a curve is zero and its velocity vector is orthogonal to the gradient. Along other paths, the same formula combines a field's geometry with the observer's motion. ^[ref].

The directional derivative tests a function along a chosen straight line, including lines that follow neither coordinate axis. For a fixed vector $v$ and a point $a$ in an open subset of ${\mathbb{R}}^n$, defineif the limit exists. With a unit vector $v$, the parameter $t$ measures signed distance in the input Euclidean space, so the directional derivative is change per unit distance. For a general $v$, it is change per unit parameter; the norm of $v$ also sets the rate of travel.^[ref]

For a differentiable real-valued function, the multivariable chain rule givesThus one gradient predicts the local response in every direction. This formula requires differentiability, rather than merely separate partial derivatives.^[ref]

Return to $f(x,y)=x^2+xy+2y^2$ at $a=(1,1)$. Its gradient is $(3,5)$. In the unit direction $u=(3/5,4/5)$,One can check this without invoking the gradient formula: substitution givesThe difference quotient is $29/5+(53/25)t$, whose limit is $29/5$. This calculation separates the exact change along a finite path from its first-order rate at the starting point.

Changing the travel speed changes the parameter rate. Taking $v=(3,4)=5u$ gives $D_{v}f(a)=29$, because the input travels five distance units per parameter unit. For a differentiable function, $D_{cv}f(a)=c{D}_{v}f(a)$ for any real $c$: negative $c$ reverses the orientation as well as scaling the rate. Normalizing $v$ before comparing different slopes avoids confusing direction with speed.

The Cauchy-Schwarz inequality bounds the rate in any unit direction by $\Vert \nabla f(a)\Vert$. Here the largest rate is $\sqrt{34}\approx 5.831$, attained in the direction $(3,5)/\sqrt{34}$; the chosen $(3/5,4/5)$ gives the slightly smaller rate $5.8$. The perpendicular unit direction $(-5,3)/\sqrt{34}$ gives zero. That means no first-order height change, rather than no change at all: along this direction the example has $f(a+tu)=4+(28/34)t^2$.

This is useful for following a height or temperature scalar field, choosing a local search direction in mathematical optimization, and distinguishing a coordinate sensitivity from a response along an actual path. For a curved path, derivative along a curve uses its instantaneous tangent vector. Even existence of all directional derivatives does not guarantee one common linear approximation; directional derivatives do not imply differentiability gives a concrete counterexample.

A scalar field assigns one scalar to each point of a specified space, such as Euclidean space or spacetime. For example, $F(x,y)=x^2+2y^2$ is a real-valued function on a plane, and its change along a moving observation is given by the derivative along a curve. Under a change of coordinates, a scalar field's value at the same physical point is unchanged.