Gradient ($\nabla f$)

In Cartesian coordinates on Euclidean space, the gradient of a scalar-valued differentiable function is the vector of its first partial derivatives. It converts a scalar field, such as temperature or a height, into a vector field describing its local change.

For a unit vector $u$, the directional derivative is $D_{u}f=u\cdot \nabla f$. The Cauchy-Schwarz inequality gives $|D_{u}f|\le |\nabla f|$. At a point where $\nabla f\ne 0$, the greatest increase per unit distance occurs along $u=\nabla f/|\nabla f|$, while the opposite direction gives the greatest decrease. These are first-order statements at the current point; they do not predict how the function behaves far along a straight path.

For $f(x,y)=x^2+2y^2$ at $(1,1)$, $\nabla f=(2,4)$. Moving in the $x$ direction gives rate $2$ and moving in the $y$ direction gives rate $4$, but the steepest direction is $(1,2)/\sqrt{5}$, withThe direction $(-2,1)/\sqrt{5}$ has rate zero and is tangent to the level curve through that point. More generally, differentiating $f(\gamma (t))=\text{constant}$ along a regular level curve gives $\nabla f\cdot {\gamma }^{\prime }=0$, so the gradient is its normal vector.

This is useful when navigating a measured scalar field or applying gradient descent: one collection of partial derivatives predicts every infinitesimal directional change. The derivative assumes a differentiable function, not merely the existence of all directional derivatives. ^[ref].

For a differentiable real-valued functional $F$ on a real Hilbert space $\mathcal{H}$, its functional gradient satisfies $DF(g)[h]=⟨{\nabla }_{\mathcal{H}}F(g),h{⟩}_{\mathcal{H}}$ for every $h\in \mathcal{H}$. A bounded linear derivative has a unique such representative by the Riesz representation theorem. The inner product matters: a formal variational derivative need not equal the functional gradient in another Hilbert space. For $F(g)=\frac{1}{2}{\sum }_i(y_i-g(x_i){)}^2$ in a Reproducing kernel Hilbert space, the reproducing property gives ${\nabla }_{\mathcal{H}}F(g)={\sum }_i(g(x_i)-y_i)k(x_i,\cdot )$. Substitution in the defining inner product proves this formula directly.

For $r=\sqrt{x_1^2+\cdots +x_d^2}>0$, the multivariable chain rule givesIndeed, $\partial r/\partial x_j=x_j/r$. Thus the gradient points radially outward when $g^{\prime }(r)>0$, inward when it is negative, and vanishes when it is zero.

At the origin the displayed quotient does not apply. If $g$ has a finite right derivative at zero, the radial function is differentiable there exactly when $g^{\prime }(0^{+})=0$, in which case its gradient is zero. Necessity follows by comparing opposite directional difference quotients: their linear parts cannot both equal a common nonzero radial slope. Sufficiency follows from $g(r)-g(0)=o(r)$, the definition of differentiability with zero linear part.

For differentiable vector fields, expand the cross products using the contraction of two Levi-Civita symbols. The directional-derivative terms cancel the corresponding terms in the two cross products, leaving ${\partial }_i(F_j{G}_j)$. This proves the displayed gradient identity by the product rule.

A gradient field is a vector field of the form $\nabla f$ for a scalar potential $f$. Its curl vanishes wherever the required second derivatives commute.