directional derivative

File:Directional derivative contour plot.svg of $f(x,\; y)=x^2\; +\; y^2$, showing the gradient vector in black, and the unit vector $\backslash mathbf\{u\}$ scaled by the directional derivative in the direction of $\backslash mathbf\{u\}$ in orange. The gradient vector is longer because the gradient points in the direction of greatest rate of increase of a function.]]

The directional derivative of a scalar function

$$f(\backslash mathbf\{x\})\; =\; f(x\_1,\; x\_2,\; \backslash ldots,\; x\_n)$$

along a vector

$$\backslash mathbf\{v\}\; =\; (v\_1,\; \backslash ldots,\; v\_n)$$

is the function $\backslash nabla\_\{\backslash mathbf\{v\}\}\{f\}$ defined by the limit{{cite book |author1=R. Wrede |author2=M.R. Spiegel | title=Advanced Calculus|edition=3rd| publisher=Schaum's Outline Series| year=2010 | isbn=978-0-07-162366-7}}

$$\backslash nabla\_\{\backslash mathbf\{v\}\}\{f\}(\backslash mathbf\{x\})\; =\; \backslash lim\_\{h\; \backslash to\; 0\}\{\backslash frac\{f(\backslash mathbf\{x\}\; +\; h\backslash mathbf\{v\})\; -\; f(\backslash mathbf\{x\})\}\{h\}\}\; =\; \backslash left.\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{d\}t\}f(\backslash mathbf\{x\}+t\backslash mathbf\{v\})\backslash right|\_\{t=0\}.$$

This definition is valid in a broad range of contexts, for example where the norm of a vector (and hence a unit vector) is undefined.The applicability extends to functions over spaces without a metric and to differentiable manifolds, such as in general relativity.

If the function f is differentiable at x, then the directional derivative exists along any unit vector v at x, and one has

$$\backslash nabla\_\{\backslash mathbf\{v\}\}\{f\}(\backslash mathbf\{x\})\; =\; \backslash nabla\; f(\backslash mathbf\{x\})\; \backslash cdot\; \backslash mathbf\{v\}$$

where the $\backslash nabla$ on the right denotes the gradient, $\backslash cdot$ is the dot product and v is a unit vector.If the dot product is undefined, the gradient is also undefined; however, for differentiable f, the directional derivative is still defined, and a similar relation exists with the exterior derivative. This follows from defining a path $h(t)\; =\; x\; +\; tv$ and using the definition of the derivative as a limit which can be calculated along this path to get:

$$\backslash begin\{align\}$$

0

&=\lim_{t \to 0}\frac {f(x+tv)-f(x)-tDf(x)(v)} t \\

&=\lim_{t \to 0}\frac {f(x+tv)-f(x)} t - Df(x)(v) \\

&=\nabla_v f(x)-Df(x)(v).

\end{align}

Intuitively, the directional derivative of f at a point x represents the rate of change of f, in the direction of v.

image:Geometrical interpretation of a directional derivative.svg

In a Euclidean space, some authorsThomas, George B. Jr.; and Finney, Ross L. (1979) Calculus and Analytic Geometry, Addison-Wesley Publ. Co., fifth edition, p. 593. define the directional derivative to be with respect to an arbitrary nonzero vector v after normalization, thus being independent of its magnitude and depending only on its direction.This typically assumes a Euclidean space – for example, a function of several variables typically has no definition of the magnitude of a vector, and hence of a unit vector.

This definition gives the rate of increase of {{math|f}} per unit of distance moved in the direction given by {{math|v}}. In this case, one has

$$\backslash nabla\_\{\backslash mathbf\{v\}\}\{f\}(\backslash mathbf\{x\})\; =\; \backslash lim\_\{h\; \backslash to\; 0\}\{\backslash frac\{f(\backslash mathbf\{x\}\; +\; h\backslash mathbf\{v\})\; -\; f(\backslash mathbf\{x\})\}\{h|\backslash mathbf\{v\}|\}\},$$

or in case f is differentiable at x,

$$\backslash nabla\_\{\backslash mathbf\{v\}\}\{f\}(\backslash mathbf\{x\})\; =\; \backslash nabla\; f(\backslash mathbf\{x\})\; \backslash cdot\; \backslash frac\{\backslash mathbf\{v\}\}$$

In the context of a function on a Euclidean space, some texts restrict the vector v to being a unit vector. With this restriction, both the above definitions are equivalent.{{Cite book| title=Calculus : Single and multivariable.|last1=Hughes Hallett|first1=Deborah|author1-link=Deborah Hughes Hallett|last2=McCallum|first2=William G.| author2-link=William G. McCallum|last3=Gleason|first3=Andrew M.|author3-link=Andrew M. Gleason| date=2012-01-01| publisher=John wiley|isbn=9780470888612|pages=780|oclc=828768012}}

Many of the familiar properties of the ordinary derivative hold for the directional derivative. These include, for any functions f and g defined in a neighborhood of, and differentiable at, p:

1. sum rule: $$\backslash nabla\_\{\backslash mathbf\{v\}\}\; (f\; +\; g)\; =\; \backslash nabla\_\{\backslash mathbf\{v\}\}\; f\; +\; \backslash nabla\_\{\backslash mathbf\{v\}\}\; g.$$
2. constant factor rule: For any constant c, $$\backslash nabla\_\{\backslash mathbf\{v\}\}\; (cf)\; =\; c\backslash nabla\_\{\backslash mathbf\{v\}\}\; f.$$
3. product rule (or Leibniz's rule): $$\backslash nabla\_\{\backslash mathbf\{v\}\}\; (fg)\; =\; g\backslash nabla\_\{\backslash mathbf\{v\}\}\; f\; +\; f\backslash nabla\_\{\backslash mathbf\{v\}\}\; g.$$
4. chain rule: If g is differentiable at p and h is differentiable at g(p), then $$\backslash nabla\_\{\backslash mathbf\{v\}\}(h\backslash circ\; g)(\backslash mathbf\{p\})\; =\; h\text{'}(g(\backslash mathbf\{p\}))\; \backslash nabla\_\{\backslash mathbf\{v\}\}\; g\; (\backslash mathbf\{p\}).$$

{{see also|Tangent space#Tangent vectors as directional derivatives}}

Let {{math|M}} be a differentiable manifold and {{math|p}} a point of {{math|M}}. Suppose that {{math|f}} is a function defined in a neighborhood of {{math|p}}, and differentiable at {{math|p}}. If {{math|v}} is a tangent vector to {{math|M}} at {{math|p}}, then the directional derivative of {{math|f}} along {{math|v}}, denoted variously as {{math|df(v)}} (see Exterior derivative), $\backslash nabla\_\{\backslash mathbf\{v\}\}\; f(\backslash mathbf\{p\})$ (see Covariant derivative), $L\_\{\backslash mathbf\{v\}\}\; f(\backslash mathbf\{p\})$ (see Lie derivative), or $\{\backslash mathbf\{v\}\}\_\{\backslash mathbf\{p\}\}(f)$ (see {{section link|Tangent space|Definition via derivations}}), can be defined as follows. Let {{math|γ : [−1, 1] → M}} be a differentiable curve with {{math|1=γ(0) = p}} and {{math|1=γ′(0) = v}}. Then the directional derivative is defined by

$$\backslash nabla\_\{\backslash mathbf\{v\}\}\; f(\backslash mathbf\{p\})\; =\; \backslash left.\backslash frac\{d\}\{d\backslash tau\}\; f\backslash circ\backslash gamma(\backslash tau)\backslash right|\_\{\backslash tau=0\}.$$

This definition can be proven independent of the choice of {{math|γ}}, provided {{math|γ}} is selected in the prescribed manner so that {{math|1=γ(0) = p}} and {{math|1=γ′(0) = v}}.

The Lie derivative of a vector field $W^\backslash mu(x)$ along a vector field $V^\backslash mu(x)$ is given by the difference of two directional derivatives (with vanishing torsion):

$$\backslash mathcal\{L\}\_V\; W^\backslash mu=(V\backslash cdot\backslash nabla)\; W^\backslash mu-(W\backslash cdot\backslash nabla)\; V^\backslash mu.$$

In particular, for a scalar field $\backslash phi(x)$, the Lie derivative reduces to the standard directional derivative:

$$\backslash mathcal\{L\}\_V\; \backslash phi=(V\backslash cdot\backslash nabla)\; \backslash phi.$$

Directional derivatives are often used in introductory derivations of the Riemann curvature tensor. Consider a curved rectangle with an infinitesimal vector $\backslash delta$ along one edge and $\backslash delta\text{'}$ along the other. We translate a covector $S$ along $\backslash delta$ then $\backslash delta\text{'}$ and then subtract the translation along $\backslash delta\text{'}$ and then $\backslash delta$. Instead of building the directional derivative using partial derivatives, we use the covariant derivative. The translation operator for $\backslash delta$ is thus

$$1+\backslash sum\_\backslash nu\; \backslash delta^\backslash nu\; D\_\backslash nu=1+\backslash delta\backslash cdot\; D,$$

and for $\backslash delta\text{'}$,

$$1+\backslash sum\_\backslash mu\; \backslash delta\text{'}^\backslash mu\; D\_\backslash mu=1+\backslash delta\text{'}\backslash cdot\; D.$$

The difference between the two paths is then

$$(1+\backslash delta\text{'}\backslash cdot\; D)(1+\backslash delta\backslash cdot\; D)S^\backslash rho-(1+\backslash delta\backslash cdot\; D)(1+\backslash delta\text{'}\backslash cdot\; D)S^\backslash rho=\backslash sum\_\{\backslash mu,\backslash nu\}\backslash delta\text{'}^\backslash mu\; \backslash delta^\backslash nu[D\_\backslash mu,D\_\backslash nu]S\_\backslash rho.$$

It can be argued{{cite book|last1=Zee|first1=A.|title=Einstein gravity in a nutshell|date=2013|publisher=Princeton University Press|location=Princeton|isbn=9780691145587|page=341}} that the noncommutativity of the covariant derivatives measures the curvature of the manifold:

$$[D\_\backslash mu,D\_\backslash nu]S\_\backslash rho=\backslash pm\; \backslash sum\_\backslash sigma\; R^\backslash sigma\{\}\_\{\backslash rho\backslash mu\backslash nu\}S\_\backslash sigma,$$

where $R$ is the Riemann curvature tensor and the sign depends on the sign convention of the author.

In the Poincaré algebra, we can define an infinitesimal translation operator P as

$$\backslash mathbf\{P\}=i\backslash nabla.$$

(the i ensures that P is a self-adjoint operator) For a finite displacement λ, the unitary Hilbert space representation for translations is{{cite book| last1=Weinberg|first1=Steven|title=The quantum theory of fields|date=1999|publisher=Cambridge Univ. Press| location=Cambridge [u.a.]| isbn=9780521550017|edition=Reprinted (with corr.).|url-access=registration| url=https://archive.org/details/quantumtheoryoff00stev}}

$$U(\backslash boldsymbol\{\backslash lambda\})=\backslash exp\backslash left(-i\backslash boldsymbol\{\backslash lambda\}\backslash cdot\backslash mathbf\{P\}\backslash right).$$

By using the above definition of the infinitesimal translation operator, we see that the finite translation operator is an exponentiated directional derivative:

$$U(\backslash boldsymbol\{\backslash lambda\})=\backslash exp\backslash left(\backslash boldsymbol\{\backslash lambda\}\backslash cdot\backslash nabla\backslash right).$$

This is a translation operator in the sense that it acts on multivariable functions f(x) as

$$U(\backslash boldsymbol\{\backslash lambda\})\; f(\backslash mathbf\{x\})=\backslash exp\backslash left(\backslash boldsymbol\{\backslash lambda\}\backslash cdot\backslash nabla\backslash right)\; f(\backslash mathbf\{x\})\; =\; f(\backslash mathbf\{x\}+\backslash boldsymbol\{\backslash lambda\}).$$

{{math proof|title=Proof of the last equation

|proof=

In standard single-variable calculus, the derivative of a smooth function f(x) is defined by (for small ε)

$$\backslash frac\{df\}\{dx\}\; =\; \backslash frac\{f(x+\backslash varepsilon)\; -\; f(x)\}\{\backslash varepsilon\}.$$

This can be rearranged to find f(x+ε):

$$f(x+\backslash varepsilon)=f(x)+\backslash varepsilon\; \backslash ,\backslash frac\{df\}\{dx\}=\backslash left(1+\backslash varepsilon\backslash ,\backslash frac\{d\}\{dx\}\backslash right)f(x).$$

It follows that $[1+\backslash varepsilon\backslash ,(d/dx)]$ is a translation operator. This is instantly generalized{{cite book |last1=Zee|first1=A.|title=Einstein gravity in a nutshell|date=2013|publisher=Princeton University Press| location=Princeton| isbn=9780691145587}} to multivariable functions f(x)

$$f(\backslash mathbf\{x\}+\backslash boldsymbol\{\backslash varepsilon\})\; =\; \backslash left(1+\backslash boldsymbol\{\backslash varepsilon\}\backslash cdot\backslash nabla\backslash right)\; f(\backslash mathbf\{x\}).$$

Here $\backslash boldsymbol\{\backslash varepsilon\}\backslash cdot\backslash nabla$ is the directional derivative along the infinitesimal displacement ε. We have found the infinitesimal version of the translation operator:

$$U(\backslash boldsymbol\{\backslash varepsilon\})\; =\; 1\; +\; \backslash boldsymbol\{\backslash varepsilon\}\backslash cdot\backslash nabla.$$

It is evident that the group multiplication law{{cite book | last1=Cahill |first1=Kevin Cahill | title=Physical mathematics | date=2013 | publisher=Cambridge University Press | location=Cambridge | isbn=978-1107005211 | edition=Repr.}} U(g)U(f)=U(gf) takes the form

$$U(\backslash mathbf\{a\})U(\backslash mathbf\{b\})=U(\backslash mathbf\{a+b\}).$$

So suppose that we take the finite displacement λ and divide it into N parts (N→∞ is implied everywhere), so that λ/N=ε. In other words,

$$\backslash boldsymbol\{\backslash lambda\}\; =\; N\; \backslash boldsymbol\{\backslash varepsilon\}.$$

Then by applying U(ε) N times, we can construct U(λ):

$$[U(\backslash boldsymbol\{\backslash varepsilon\})]^N\; =\; U(N\backslash boldsymbol\{\backslash varepsilon\})\; =\; U(\backslash boldsymbol\{\backslash lambda\}).$$

We can now plug in our above expression for U(ε):

$$[U(\backslash boldsymbol\{\backslash varepsilon\})]^N\; =\; \backslash left[1+\backslash boldsymbol\{\backslash varepsilon\}\backslash cdot\backslash nabla\backslash right]^N\; =\; \backslash left[1+\backslash frac\{\backslash boldsymbol\{\backslash lambda\}\backslash cdot\backslash nabla\}\{N\}\backslash right]^N.$$

Using the identity{{cite book | first1 = Ron | last1 = Larson | last2 = Edwards | first2 = Bruce H. | title = Calculus of a single variable | date = 2010 | publisher = Brooks/Cole | location =Belmont | isbn = 9780547209982 | edition = 9th}}

$$\backslash exp(x)=\backslash left[1+\backslash frac\{x\}\{N\}\backslash right]^N,$$

we have

$$U(\backslash boldsymbol\{\backslash lambda\})=\backslash exp\backslash left(\backslash boldsymbol\{\backslash lambda\}\backslash cdot\backslash nabla\backslash right).$$

And since {{math|1=U(ε)f(x) = f(x+ε)}} we have

$$[U(\backslash boldsymbol\{\backslash varepsilon\})]^N\; f(\backslash mathbf\{x\})\; =\; f(\backslash mathbf\{x\}+N\backslash boldsymbol\{\backslash varepsilon\})\; =\; f(\backslash mathbf\{x\}+\backslash boldsymbol\{\backslash lambda\})\; =\; U(\backslash boldsymbol\{\backslash lambda\})f(\backslash mathbf\{x\})\; =\; \backslash exp\backslash left(\backslash boldsymbol\{\backslash lambda\}\backslash cdot\backslash nabla\backslash right)f(\backslash mathbf\{x\}),$$

Q.E.D.

As a technical note, this procedure is only possible because the translation group forms an Abelian subgroup (Cartan subalgebra) in the Poincaré algebra. In particular, the group multiplication law U(a)U(b) = U(a+b) should not be taken for granted. We also note that Poincaré is a connected Lie group. It is a group of transformations T(ξ) that are described by a continuous set of real parameters $\backslash xi^a$. The group multiplication law takes the form

$$T(\backslash bar\{\backslash xi\})T(\backslash xi)\; =\; T(f(\backslash bar\{\backslash xi\},\backslash xi)).$$

Taking $\backslash xi^a\; =\; 0$ as the coordinates of the identity, we must have

$$f^a(\backslash xi,0)=f^a(0,\backslash xi)=\backslash xi^a.$$

The actual operators on the Hilbert space are represented by unitary operators U(T(ξ)). In the above notation we suppressed the T; we now write U(λ) as U(P(λ)). For a small neighborhood around the identity, the power series representation

$$U(T(\backslash xi))=1+i\backslash sum\_a\backslash xi^a\; t\_a+\backslash frac\{1\}\{2\}\backslash sum\_\{b,c\}\backslash xi^b\backslash xi^c\; t\_\{bc\}+\backslash cdots$$

is quite good. Suppose that U(T(ξ)) form a non-projective representation, i.e.,

$$U(T(\backslash bar\{\backslash xi\}))U(T(\backslash xi))=U(T(f(\backslash bar\{\backslash xi\},\backslash xi))).$$

The expansion of f to second power is

$$f^a(\backslash bar\{\backslash xi\},\backslash xi)=\backslash xi^a+\backslash bar\{\backslash xi\}^a+\backslash sum\_\{b,c\}f^\{abc\}\backslash bar\{\backslash xi\}^b\backslash xi^c.$$

After expanding the representation multiplication equation and equating coefficients, we have the nontrivial condition

$$t\_\{bc\}=-t\_b\; t\_c-i\backslash sum\_a\; f^\{abc\}t\_a.$$

Since $t\_\{ab\}$ is by definition symmetric in its indices, we have the standard Lie algebra commutator:

$$[t\_b,\; t\_c]=i\backslash sum\_a(-f^\{abc\}+f^\{acb\})t\_a=i\backslash sum\_a\; C^\{abc\}t\_a,$$

with C the structure constant. The generators for translations are partial derivative operators, which commute:

$$\backslash left[\backslash frac\{\backslash partial\}\{\backslash partial\; x^b\},\backslash frac\{\backslash partial\; \}\{\backslash partial\; x^c\}\backslash right]=0.$$

This implies that the structure constants vanish and thus the quadratic coefficients in the f expansion vanish as well. This means that f is simply additive:

$$f^a\_\backslash text\{abelian\}(\backslash bar\{\backslash xi\},\backslash xi)=\backslash xi^a+\backslash bar\{\backslash xi\}^a,$$

and thus for abelian groups,

$$U(T(\backslash bar\{\backslash xi\}))U(T(\backslash xi))=U(T(\backslash bar\{\backslash xi\}+\backslash xi)).$$

Q.E.D.

}}

The rotation operator also contains a directional derivative. The rotation operator for an angle θ, i.e. by an amount θ = |θ| about an axis parallel to $\backslash hat\{\backslash theta\}\; =\; \backslash boldsymbol\{\backslash theta\}/\backslash theta$ is

$$U(R(\backslash mathbf\{\backslash theta\}))=\backslash exp(-i\backslash mathbf\{\backslash theta\}\backslash cdot\backslash mathbf\{L\}).$$

Here L is the vector operator that generates SO(3):

$$\backslash mathbf\{L\}=\backslash begin\{pmatrix\}$$

0& 0 & 0\\

0& 0 & 1\\

0& -1 & 0

\end{pmatrix}\mathbf{i}+\begin{pmatrix}

0 &0 & -1\\

0& 0 &0 \\

1 & 0 & 0

\end{pmatrix}\mathbf{j}+\begin{pmatrix}

0&1 &0 \\

-1&0 &0 \\

0 & 0 & 0

\end{pmatrix}\mathbf{k}.

It may be shown geometrically that an infinitesimal right-handed rotation changes the position vector x by

$$\backslash mathbf\{x\}\backslash rightarrow\; \backslash mathbf\{x\}-\backslash delta\backslash boldsymbol\{\backslash theta\}\backslash times\backslash mathbf\{x\}.$$

So we would expect under infinitesimal rotation:

$$U(R(\backslash delta\backslash boldsymbol\{\backslash theta\}))\; f(\backslash mathbf\{x\})\; =\; f(\backslash mathbf\{x\}-\backslash delta\backslash boldsymbol\{\backslash theta\}\backslash times\backslash mathbf\{x\})=f(\backslash mathbf\{x\})-(\backslash delta\backslash boldsymbol\{\backslash theta\}\backslash times\backslash mathbf\{x\})\backslash cdot\backslash nabla\; f.$$

It follows that

$$U(R(\backslash delta\backslash mathbf\{\backslash theta\}))=1-(\backslash delta\backslash mathbf\{\backslash theta\}\backslash times\backslash mathbf\{x\})\backslash cdot\backslash nabla.$$

Following the same exponentiation procedure as above, we arrive at the rotation operator in the position basis, which is an exponentiated directional derivative:{{cite book|last1=Shankar|first1=R. | title=Principles of quantum mechanics | date=1994|publisher=Kluwer Academic / Plenum|location=New York|isbn=9780306447907|page=318|edition=2nd}}

$$U(R(\backslash mathbf\{\backslash theta\}))=\backslash exp(-(\backslash mathbf\{\backslash theta\}\backslash times\backslash mathbf\{x\})\backslash cdot\backslash nabla).$$

A normal derivative is a directional derivative taken in the direction normal (that is, orthogonal) to some surface in space, or more generally along a normal vector field orthogonal to some hypersurface. See for example Neumann boundary condition. If the normal direction is denoted by $\backslash mathbf\{n\}$, then the normal derivative of a function f is sometimes denoted as $\backslash frac\{\; \backslash partial\; f\}\{\backslash partial\; \backslash mathbf\{n\}\}$. In other notations,

$$\backslash frac\{\; \backslash partial\; f\}\{\backslash partial\; \backslash mathbf\{n\}\}\; =\; \backslash nabla\; f(\backslash mathbf\{x\})\; \backslash cdot\; \backslash mathbf\{n\}\; =\; \backslash nabla\_\{\backslash mathbf\{n\}\}\{f\}(\backslash mathbf\{x\})\; =\; \backslash frac\{\backslash partial\; f\}\{\backslash partial\; \backslash mathbf\{x\}\}\; \backslash cdot\; \backslash mathbf\{n\}\; =\; Df(\backslash mathbf\{x\})[\backslash mathbf\{n\}].$$

Several important results in continuum mechanics require the derivatives of vectors with respect to vectors and of tensors with respect to vectors and tensors.J. E. Marsden and T. J. R. Hughes, 2000, Mathematical Foundations of Elasticity, Dover. The directional directive provides a systematic way of finding these derivatives.

{{excerpt|Tensor derivative (continuum mechanics)|Derivatives with respect to vectors and second-order tensors|subsections=y}}

• {{annotated link|Del in cylindrical and spherical coordinates}}
• {{annotated link|Differential form}}
• {{annotated link|Ehresmann connection}}
• {{annotated link|Fréchet derivative}}
• {{annotated link|Gateaux derivative}}
• {{annotated link|Generalizations of the derivative}}
• {{annotated link | Semi-differentiability}}
• {{annotated link|Hadamard derivative}}
• {{annotated link|Lie derivative}}
• {{annotated link|Material derivative}}
• {{annotated link|Structure tensor}}
• {{annotated link|Tensor derivative (continuum mechanics)}}
• {{annotated link|Total derivative}}

{{reflist|2}}

• {{cite book | first=F. B. | last=Hildebrand | title=Advanced Calculus for Applications| publisher=Prentice Hall | year=1976 | isbn=0-13-011189-9 }}
• {{cite book |author1=K.F. Riley |author2=M.P. Hobson |author3=S.J. Bence | title=Mathematical methods for physics and engineering|url=https://archive.org/details/mathematicalmeth00rile |url-access=registration | publisher=Cambridge University Press| year=2010 | isbn=978-0-521-86153-3}}
• {{cite journal |first=A. |last=Shapiro |title=On concepts of directional differentiability |journal=Journal of Optimization Theory and Applications |volume=66 |issue= 3|pages=477–487 |year=1990 |doi=10.1007/BF00940933 |s2cid=120253580 }}

{{Commons category inline|Directional derivative}}

• [http://mathworld.wolfram.com/DirectionalDerivative.html Directional derivatives] at MathWorld.
• [http://planetmath.org/directionalderivative Directional derivative] at PlanetMath.

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Category:Differential calculus

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