Editing Gradient (section)

===Relationship with {{vanchor|Fréchet derivative}}===
Let {{math|''U''}} be an [[open set]] in {{math|'''R'''<sup>''n''</sup>}}. If the function {{math|''f'' : ''U'' → '''R'''}} is differentiable, then the differential of {{math|''f''}} is the [[Fréchet derivative]] of {{math|''f''}}. Thus {{math|∇''f''}} is a function from {{math|''U''}} to the space {{math|'''R'''<sup>''n''</sup>}} such that
<math display="block">\lim_{h\to 0} \frac{|f(x+h)-f(x) -\nabla f(x)\cdot h|}{\|h\|} = 0,</math>
where · is the dot product.

As a consequence, the usual properties of the derivative hold for the gradient, though the gradient is not a derivative itself, but rather dual to the derivative:

;[[Linearity]]
:The gradient is linear in the sense that if {{math|''f''}} and {{math|''g''}} are two real-valued functions differentiable at the point {{math|''a'' ∈ '''R'''<sup>''n''</sup>}}, and {{mvar|α}} and {{mvar|β}} are two constants, then {{math|''αf'' + ''βg''}} is differentiable at {{math|''a''}}, and moreover <math display="block">\nabla\left(\alpha f+\beta g\right)(a) = \alpha \nabla f(a) + \beta\nabla g (a).</math>
;[[Product rule]]
:If {{math|''f''}} and {{math|''g''}} are real-valued functions differentiable at a point {{math|''a'' ∈ '''R'''<sup>''n''</sup>}}, then the product rule asserts that the product {{math|''fg''}} is differentiable at {{math|''a''}}, and <math display="block">\nabla (fg)(a) = f(a)\nabla g(a) + g(a)\nabla f(a).</math>
;[[Chain rule]]
:Suppose that {{math|''f'' : ''A'' → '''R'''}} is a real-valued function defined on a subset {{math|''A''}} of {{math|'''R'''<sup>''n''</sup>}}, and that {{math|''f''}} is differentiable at a point {{math|''a''}}. There are two forms of the chain rule applying to the gradient. First, suppose that the function {{math|''g''}} is a [[parametric curve]]; that is, a function {{math|''g'' : ''I'' → '''R'''<sup>''n''</sup>}} maps a subset {{math|''I'' ⊂ '''R'''}} into {{math|'''R'''<sup>''n''</sup>}}. If {{math|''g''}} is differentiable at a point {{math|''c'' ∈ ''I''}} such that {{math|''g''(''c'') {{=}} ''a''}}, then <math display="block">(f\circ g)'(c) = \nabla f(a)\cdot g'(c),</math> where ∘ is the [[composition operator]]: {{math|1=(''f'' ∘ ''g'')(''x'') = ''f''(''g''(''x''))}}.

More generally, if instead {{math|''I'' ⊂ '''R'''<sup>''k''</sup>}}, then the following holds:
<math display="block">\nabla (f\circ g)(c) = \big(Dg(c)\big)^\mathsf{T} \big(\nabla f(a)\big),</math>
where {{math|(''Dg'')}}<sup>T</sup> denotes the transpose [[Jacobian matrix]].

For the second form of the chain rule, suppose that {{math|''h'' : ''I'' → '''R'''}} is a real valued function on a subset {{math|''I''}} of {{math|'''R'''}}, and that {{math|''h''}} is differentiable at the point {{math|''f''(''a'') ∈ ''I''}}. Then
<math display="block">\nabla (h\circ f)(a) = h'\big(f(a)\big)\nabla f(a).</math>