Partial derivatives, gradients and Jacobians

Predict a small output change from two input changes

Consider f(x,y)=x²+xy+2y² at a=(1,−1). The inputs and output in this example are dimensionless. Its value is f(a)=2. Instead of drawing an entire surface, ask what happens when the input becomes a+h=(1+u,−1+v). Direct expansion gives f(a+h)=2+u −3v+u²+uv+2v².

The expression u −3v is linear in the displacement. It predicts the output change, while R(u,v)=u²+uv+2v² is its exact error. For h=(0.02,−0.01), the linear prediction is 2.05 and the exact value is 2.0504. The error 0.0004 is small compared with the input displacement; this comparison, rather than a surface appearing smooth, is the central idea of a derivative.

Functions of several variables 130 established domains, graphs and joint limits. Preparation also includes the derivative and linear approximation lessons, matrix multiplication 122, vector spaces 123, coordinate changes 124 and the Euclidean inner product 127. The article treats maps on open subsets of finite-dimensional real spaces. Boundary differentiation needs a separately stated domain or extension convention; do not silently take a two-sided coordinate limit through forbidden inputs.

A partial derivative follows one coordinate line

For f:U→R with U open in Rⁿ and a∈U, the i-th partial derivative is ∂ᵢf(a)=limₜ→₀ [f(a+t eᵢ)−f(a)]/t, when this finite two-sided limit exists. Here eᵢ is the i-th coordinate unit vector. Every other input coordinate is held at its value in a; it is not set to zero unless it was zero already.

For f(x,y)=x²+xy+2y², ordinary single-variable differentiation gives fₓ=2x+y and fᵧ=x+4y. At (1,−1) these are 1 and −3. Thus small positive x changes increase the output to first order, while small positive y changes decrease it. A partial derivative is a rate in a specified coordinate direction, not the change itself.

Keeping y fixed differs from following a path y=y(x). If y varies, its contribution must be included through a chain rule. A partial derivative of temperature with respect to position also holds time and every other specified input fixed. Write down those choices before attaching a physical interpretation.

One linear map must work for all small displacements

A map F:U⊂Rⁿ→Rᵐ is differentiable at a when there is a linear map L:Rⁿ→Rᵐ such that F(a+h)=F(a)+Lh+r(h), with ||r(h)||/||h||→0 as h→0 through nonzero displacements. For sufficiently small h, a+h lies in U. The derivative is L, written DF(a). This finite-dimensional total derivative is also called the Fréchet derivative.

The remainder requirement is uniform over directions: for every ε>0, a sufficiently small radius makes ||r(h)||≤ε||h|| for every nonzero h within that radius. Choosing a different radius for each fixed line is weaker. The derivative is the linear part of the output change; F(a)+DF(a)h is an affine approximation to the output value.

The linear map is unique. If L and M both work, fix a nonzero vector v and put h=tv. Subtract the two approximations, divide the resulting norm estimate by |t|, and let t→0. This gives (L−M)v=0. It holds for every v, so L=M. Differentiability also implies continuity: the linear term and the smaller remainder both tend to zero.

Control the error in every direction

For the running quadratic, propose L(u,v)=u −3v. The exact remainder is R=u²+uv+2v². Since |uv|≤(u²+v²)/2, we have |R|≤(5/2)(u²+v²)=(5/2)||h||². Therefore |R|/||h||≤(5/2)||h||→0. This proves differentiability at a without selecting paths or relying on a numerical plot.

For a unit vector d and h=td, the exact error is t²(d₁²+d₁d₂+2d₂²). Halving |t| quarters its magnitude. The error divided by ||h|| halves. These are consequences of this exact quadratic; other differentiable functions need only have a remainder that is little-o of ||h||, not necessarily a quadratic error.

An explorer may compare exact values at a finite list of displacements and make this scaling visible. Such checks test the implementation and illustrate the estimate. The uniform inequality above, which covers every sufficiently small displacement, establishes the derivative.

The Jacobian records the derivative in coordinates

If F=(F₁,…,Fₘ) is differentiable at a, its derivative matrix in the stated standard coordinates is JF(a)=[∂Fᵢ/∂xⱼ(a)]. Rows correspond to outputs and columns to inputs, so the shape is m-by-n. The j-th column is the derivative’s response to eⱼ. This follows by substituting h=t eⱼ into the total-derivative definition.

For a scalar-valued function, the derivative is the 1-by-n row [∂₁f … ∂ₙf], acting on the displacement column. For a vector-valued example F(x,y)=(x²+y,xy), the Jacobian is [[2x,1],[y,x]]. At b=(1,1), F(b)=(2,1) and JF(b)=[[2,1],[1,1]]. A displacement (u,v) predicts the output change (2u+v,u+v).

The matrix of existing partial derivatives is often called a Jacobian even before differentiability has been established. Its existence supplies a candidate linear map, not a proof that the candidate meets the uniform remainder condition. The distinction matters at irregular inputs.

For this polynomial map, the exact remainder at b is (u²,uv). Its Euclidean norm is |u|√(u²+v²)≤||h||², so division by ||h|| proves that the displayed matrix is indeed the derivative. A finite matrix and a derivative matrix become the same object only after the relevant differentiability argument.

Continuous partial derivatives give a useful sufficient condition

A standard sufficient condition is that all first partial derivatives of all components exist in a neighbourhood of a and are continuous at a. Then F is differentiable at a and DF(a) is its Jacobian. A C¹ map has continuous first partial derivatives throughout its open domain, so it is differentiable at every point there.

For two inputs and one output, split f(a₁+u,a₂+v)−f(a₁,a₂) into an x change at fixed new y and a y change at fixed old x. Apply the one-variable mean value theorem to each change. The resulting coefficients are fₓ and fᵧ at nearby intermediate points. Their continuity makes the coefficient errors tend to zero, while |u|+|v|≤√2||(u,v)|| controls the total remainder uniformly. Zero coordinate increments simply omit their corresponding changes.

Continuity of first partials is sufficient, not necessary for differentiability at a single point. For example, s(x)=x²sin(1/x) for x≠0 with s(0)=0 is differentiable at 0, but s′(x)=2x sin(1/x)−cos(1/x) does not approach s′(0)=0. The map f(x,y)=s(x) is differentiable at the origin even though its x partial is not continuous there.

A directional derivative becomes a dot product when a derivative exists

For a fixed direction v, define Dᵥf(a)=limₜ→₀ [f(a+tv)−f(a)]/t when the finite two-sided limit exists. If f is differentiable, substituting h=tv into its remainder formula gives Dᵥf(a)=Df(a)v. A unit vector u gives a rate per unit input distance. Replacing u by a nonunit v also changes the speed of the path; v=0 gives the constant path and derivative 0.

In standard Euclidean coordinates, define the gradient column ∇f(a)=(∂₁f(a),…,∂ₙf(a))ᵀ. Then Dᵤf(a)=∇f(a)·u for differentiable f. In the running example, ∇f(a)=(1,−3), so the unit direction u=(3/5,4/5) gives Dᵤf(a)=3/5−12/5=−9/5. The output initially decreases along that direction.

Cauchy–Schwarz gives |∇f·u|≤||∇f|| for ||u||=1. If ∇f≠0, the largest first-order increase is ||∇f|| in direction ∇f/||∇f||, and the greatest decrease is its negative. At a zero gradient all first-order directional rates are zero; this does not determine the second-order behaviour or classify an extremum.

Even every fixed-direction derivative can miss the problem

Define p(x,y)=x³/(x²+y²) away from the origin and p(0,0)=0. The bound |p(x,y)|≤|x| proves continuity at 0. For any nonzero v=(v₁,v₂), p(tv)/t=v₁³/(v₁²+v₂²), so every fixed-direction derivative exists. The coordinate partials are pₓ(0)=1 and pᵧ(0)=0. Nevertheless, the direction v=(1,1) gives derivative 1/2 instead of the value 1 predicted by [1,0]v. Directional rates here are not a linear function of v, so there is no total derivative.

Even continuity together with a linear assignment of all directional rates does not suffice without uniform control. Define C(x,y)=x³y/(x⁴+y²) away from 0, with C(0,0)=0. Since 2|x²y|≤x⁴+y², we have |C(x,y)|≤|x|/2, proving continuity at 0. Along a fixed line (x,y)=(ta,tb) with b≠0, C(ta,tb)/t=t a³b/(t²a⁴+b²)→0. Along a horizontal line C is identically 0. Every fixed-direction derivative therefore exists and equals 0, a linear assignment. But along h=(t,t²), C(h)=t/2, so |C(h)|/||h||=1/(2√(1+t²))→1/2. The zero candidate derivative fails its uniform remainder test.

These examples separate useful tests. Differentiability implies continuity, existing partials and a linear directional-derivative rule. The function C has all three properties at 0, yet is not differentiable there. A failed path or direction can disprove a proposed derivative; successful fixed-line tests cannot replace the uniform remainder requirement.

The chain rule multiplies the local linear maps

Let F:U⊂Rⁿ→Rᵐ be differentiable at a, and let G be defined on an open neighbourhood of F(a) and differentiable there. Then G∘F is differentiable at a and D(G∘F)(a)=DG(F(a)) DF(a). The derivative of the outer map is evaluated at the inner output. For Jacobians, the dimensions are (p-by-m)(m-by-n)=p-by-n when G takes values in Rᵖ.

The reason is that F first turns h into DF(a)h plus an error smaller than ||h||. G then applies its own linear approximation to that output displacement. Differentiability of F ensures its displacement is O(||h||), which makes G’s remainder smaller than ||h|| as well. Multiplying the two linear parts gives the derivative; it is not a component-by-component product.

Use F(x,y)=(x²+y,xy), G(u,v)=(u+v,u²−v), and b=(1,1). We have F(b)=(2,1), JF(b)=[[2,1],[1,1]], and JG(F(b))=[[1,1],[4,−1]]. Their product is [[3,2],[7,3]]. Directly differentiating (x²+y+xy,(x²+y)²−xy) gives the same matrix at b.

For a differentiable path γ(t) in U, the scalar-output version is d[f(γ(t))]/dt=∇f(γ(t))·γ′(t). A unit path direction and a path velocity are different inputs to this formula. When speed changes, the rate per unit time changes even if the path follows the same geometric curve.

A rectangular example makes the dimensions explicit. Reuse the vector map H(x,y)=(x+y,x−y,xy) from 130 and put K(u,v,w)=(u²+w,vw). At a=(1,−1), H(a)=(0,2,−1), JH(a)=[[1,1],[1,−1],[−1,1]], and JK(H(a))=[[0,0,1],[0,−1,2]]. Their 2-by-3 times 3-by-2 product is [[−1,1],[−3,3]]. Reversing the factors gives a 3-by-3 matrix, so it cannot represent the derivative of this R²→R² composition.

An approximation plane and a level-set normal describe different sets

For differentiable f:R²→R, the graph z=f(x,y) has the tangent-plane approximation z=f(a)+fₓ(a)(x−a₁)+fᵧ(a)(y−a₂). At a=(1,−1), this is z=2+(x −1)−3(y+1)=x −3y −2. Its normal vector in three-dimensional graph coordinates is (−fₓ,−fᵧ,1)=(−1,3,1). The two-dimensional gradient (1,−3) is not itself that three-dimensional normal.

A level set is instead a set of input points satisfying f(x,y)=c. If a differentiable curve γ stays in the level set, the chain rule gives ∇f(γ(t))·γ′(t)=0. If f is C¹ in a neighbourhood and its gradient is nonzero at the point, the implicit function theorem supplies a local smooth level curve, and this is its normal-direction relation. At a singular point with zero gradient, it gives no nonzero normal and does not establish that the level set is a smooth curve.

For the running example, f(a)=2 and ∇f(a)=(1,−3). The input-plane direction (3,1) is tangent to the regular level curve since its dot product with the gradient is 0. Its graph lift has zero first-order height change. Keep the input-plane level curve, the three-dimensional graph and the graph’s tangent plane distinct.

Sensitivity depends on units and the chosen geometry

The entry ∂Fᵢ/∂xⱼ has output-i units divided by input-j units. For temperature measured in degrees Celsius over spatial coordinates measured in metres, the spatial gradient has units degrees Celsius per metre. Dotting it with a velocity in metres per second yields a temperature rate in degrees Celsius per second. If temperature also depends on time, a moving observer has the additional explicit time partial: dT/dt=∂T/∂t+∇ₓT·x′(t).

A Euclidean steepest-direction statement assumes comparable input coordinates and a specified distance measure. Adding the squares of changes in seconds and kilograms without a scaling convention does not define a physically meaningful input distance. Rescaling variables changes coordinate partials and changes what a unit coordinate displacement means. State the normalisation or metric before ranking sensitivities.

If x=Bz for a fixed matrix B, the chain rule gives the derivative row D(f∘B)(z)=Df(Bz)B and the standard-coordinate gradient column ∇z(f∘B)=Bᵀ∇xf(Bz). This transpose rule describes the coefficients of the differential in z coordinates. When B is a nonsingular square basis matrix and z is a nonorthonormal coordinate description of the same Euclidean geometry, its metric matrix is BᵀB; do not confuse the raw column of partials with physical vector coordinates. The covector/vector distinction connects with tensors 129.

Second derivatives describe how the first-order map changes

When first partials can themselves be differentiated, their derivatives form second partials. Under a C² hypothesis in an open neighbourhood, mixed partial derivatives agree. The Hessian of a scalar function is the matrix of these second partials; with that hypothesis it is symmetric. Do not infer equality merely because both mixed partial values exist at one point.

For the running polynomial, H=[[2,1],[1,4]]. The exact expansion is f(a+h)=f(a)+∇f(a)·h+(1/2)hᵀHh. For a general C² function, Taylor’s formula at a has the same quadratic part and an additional remainder o(||h||²). First derivatives give slopes; the quadratic form measures how the linear approximation bends. The next lesson uses it to analyse extrema.

A concrete warning is r(x,y)=xy(x²−y²)/(x²+y²) away from 0, with r(0,0)=0. Along the y axis, ∂ₓr(0,y)=−y; along the x axis, ∂ᵧr(x,0)=x. Hence DᵧDₓr(0,0)=−1 and DₓDᵧr(0,0)=1. Writing the order of operators explicitly avoids an index-order convention hiding the difference. The hypotheses of the C² symmetry theorem fail here.

A nonsingular Jacobian supports a local inverse under regularity

The inverse function theorem says that if F is C¹ near a and its square Jacobian JF(a) is invertible, then F has a C¹ inverse between suitable neighbourhoods of a and F(a). The derivative of that inverse at F(a) is JF(a)⁻¹. This is a local conclusion; a nonsingular derivative at one point does not establish a globally one-to-one map.

For F(x,y)=(x²+y,xy) at b=(1,1), det JF(b)=1 and JF(b)⁻¹=[[1,−1],[−1,2]]. Polynomial regularity supplies the other hypothesis. The inverse output perturbation can therefore be estimated by applying that inverse matrix to a small perturbation near F(b)=(2,1). Singularity of a Jacobian does not by itself disprove ordinary invertibility: t↦t³ is one-to-one, but its inverse is not differentiable at 0.

The implicit function theorem gives a related statement for a C¹ scalar equation f(x,y)=c. If f(a)=c and fᵧ(a)≠0, it can locally be solved near a as y=φ(x), with φ′(a₁)=−fₓ(a)/fᵧ(a). For the level f=2 at (1,−1), this slope is 1/3. The theorem supplies a local function; simply dividing by fᵧ where it vanishes is invalid. Higher-dimensional versions solve for a block of variables when the corresponding derivative block is invertible.

Compare exact linear predictions before shrinking a step

In the explorer, choose a displacement direction for the quadratic and predict the sign of ∇f(a)·u. Then compare f(a+tu) with f(a)+t∇f(a)·u. Its displayed exact remainder must equal t²(u₁²+u₁u₂+2u₂²). A separate display of error divided by ||tu|| makes the derivative criterion visible. Use the stated unit directions when interpreting a rate per unit distance.

Compare the vector-map Jacobian with a directly expanded output and then follow the two matrices in the chain-rule example. The output-row/input-column convention must remain consistent in every table. Exact checks can establish the displayed finite examples and catch a transposed product; the article’s limit arguments and stated regularity hypotheses establish the general claims.

Finally compare straight paths and a curved path in the zero-directional-derivative counterexample. The point is not to search until the picture looks convincing. It is to identify why fixed-direction tests fail to supply one radius that controls every small displacement.

Worked examples

Try it yourself

1. For f(x,y)=x²y+sin y, what is fₓ(2,0)?

   Show the answer and reasoning

   0.

   Holding y fixed gives fₓ=2xy. Substitute the actual y coordinate 0; do not differentiate y as a function of x.

2. For the running quadratic at a, what is the first-order change for h=(s,2s)?

   Show the answer and reasoning

   −5s.

   [1,−3](s,2s)=s −6s. The exact quadratic remainder is 11s², which is separate from the first-order change.

3. Is f(a)+Df(a)h a linear function of h when f(a)≠0?

   Show the answer and reasoning

   It is affine, not linear.

   A linear map sends 0 to 0, whereas this approximation sends h=0 to f(a). The derivative Df(a) itself is linear.

4. Does a bound ||r(h)||≤7||h||² near 0 establish the required derivative remainder?

   Show the answer and reasoning

   Yes.

   For h≠0 the quotient is at most 7||h||, which tends to 0 uniformly over directions.

5. Does ||r(h)||≤7||h|| by itself establish differentiability with the proposed linear map?

   Show the answer and reasoning

   No.

   It only bounds the relative error by 7. Differentiability requires that relative error tend to 0.

6. What shape is the Jacobian of a differentiable map R³→R²?

   Show the answer and reasoning

   2-by-3.

   There is one row per output component and one column per input coordinate. It multiplies a 3-coordinate displacement column.

7. What is Dᵤf(a) for u=(0,1) in the running example?

   Show the answer and reasoning

   −3.

   The unit y direction gives ∇f(a)·u=(1,−3)·(0,1)=−3.

8. If a differentiable scalar function has Dᵤf(a)=4 for a unit u, what is its derivative along a+3tu at t=0?

   Show the answer and reasoning

   12.

   The path velocity is 3u, so the chain rule gives Df(a)(3u)=3Dᵤf(a).

9. If ∇f(a)=0, must a be a local minimum?

   Show the answer and reasoning

   No.

   The first-order rates vanish. For f(x,y)=x²−y² at 0, one direction increases and another decreases at second order.

10. Can a Jacobian made of existing partial derivatives fail to be the total derivative?

    Show the answer and reasoning

    Yes.

    For p=x³/(x²+y²) at 0, the partials give [1,0], but the directional rate at (1,1) is 1/2 rather than 1.

11. Why do continuity and the zero directional derivatives of C=x³y/(x⁴+y²) at 0 not prove a zero total derivative?

    Show the answer and reasoning

    They do not control the relative error uniformly over all small displacements.

    Along h=(t,t²), C(h)=t/2 and |C(h)|/||h||→1/2. Differentiability with derivative 0 would require that quotient to tend to 0.

12. F maps R²→R³ and G maps R³→R⁴. What matrix product represents D(G∘F)(a)?

    Show the answer and reasoning

    JG(F(a)) JF(a), with shape (4-by-3)(3-by-2)=4-by-2.

    The inner derivative acts first on a two-coordinate displacement. The outer derivative acts at the inner output F(a).

13. In the chain-rule example, may JG be evaluated at the original point b=(1,1) instead of F(b)=(2,1)?

    Show the answer and reasoning

    No.

    G receives F(b), so its local linear approximation must be taken at (2,1). Using the same coordinate pair by accident changes the derivative.

14. For the level f=2 at a=(1,−1), is v=(3,1) tangent to the input-plane level curve?

    Show the answer and reasoning

    Yes, as a tangent direction.

    The point is regular and (1,−3)·(3,1)=0. Normalising v changes its length, not its tangent line.

15. What is a three-dimensional normal to the graph tangent plane of the running function at a?

    Show the answer and reasoning

    (−1,3,1), or any nonzero scalar multiple.

    The graph plane has equation z=2+(x −1)−3(y+1). Its input gradient (1,−3) has only two coordinates and is not that graph normal.

16. Does existence of DᵧDₓf(a) and DₓDᵧf(a) alone force equality?

    Show the answer and reasoning

    No.

    The displayed r example has values −1 and 1. Continuous second partials in an open neighbourhood are a sufficient hypothesis for equality.

17. Does a zero derivative prove that a one-variable function is not one-to-one?

    Show the answer and reasoning

    No.

    The function t³ is one-to-one and has derivative 0 at 0. Its inverse fails to be differentiable there, so it does not contradict the C¹ inverse function theorem.

18. T is in degrees Celsius, x is in metres, and a path has x′ in metres per second. What units does Tₓx′ have?

    Show the answer and reasoning

    Degrees Celsius per second.

    The derivative contributes degrees Celsius per metre. The path velocity supplies metres per second; include any explicit time partial separately.

What to carry forward

• A partial derivative measures a coordinate-line rate with the other inputs fixed.
• A total derivative is one linear map with an error o(||h||), uniformly over sufficiently small input displacements.
• A Jacobian has output rows and input columns; existing partials alone only propose that matrix.
• Differentiability implies continuity and the directional rule Dᵥf=Df v. Even all fixed-direction derivatives can fail to establish a total derivative.
• In a stated Euclidean geometry, the gradient column represents the scalar differential by a dot product and identifies steepest first-order directions.
• Chain-rule matrices act in composition order, with the outer derivative evaluated at the inner output.
• Graph tangent planes and input-plane level sets have different ambient spaces and different normal descriptions.
• Continuous first partials, C² mixed-partial symmetry and C¹ local inversion are useful theorems with hypotheses, not automatic consequences of a displayed formula.
• Keep units, input scaling and coordinate metrics with sensitivity statements. Exact examples and numerical plots support reasoning; they do not replace uniform-limit proofs.