core26 min studyLesson 201 in Mathematics

Total differentiability and linear approximation

Replace separate coordinate slopes with one linear map that controls errors in every direction.

On this page

What you’ll understand

  • Test total differentiability through a normed remainder.
  • Construct the derivative matrix from partial derivatives when justified.
  • Explain why directional derivatives alone do not guarantee a linear approximation.

A temperature field depends on several coordinates, but a thermometer can move diagonally as easily as along a coordinate axis. Knowing each coordinate slope is therefore only the beginning. Total differentiability asks for a single linear rule that predicts the change for every sufficiently small displacement, with an error negligible compared with the displacement length. This uniform local statement is what makes matrix calculus reliable for geometry, optimization, and propagated measurement uncertainty.

One approximation must survive every approach

Let f map an open subset of Euclidean n-space into Euclidean m-space, and let a be an interior point. Its derivative at a is a linear map L for which subtracting f(a)+Lh leaves a remainder whose norm divided by the norm of h tends to zero. Both the direction and length of h may change during that limit. The definition is consequently stronger than checking a finite list of lines. The derivative is unique: evaluating two proposed derivatives on displacements tv forces their difference to vanish on every fixed vector v.

A differentiable function is continuous at the point because the linear part tends to zero and the remainder is smaller still. In coordinates, applying the derivative to each coordinate basis vector recovers the corresponding partial derivative. Thus the matrix of a derivative, when it exists, must be the Jacobian matrix. Computing that candidate is useful, but it does not by itself prove that the required remainder estimate holds. A proof must control arbitrary nearby displacements, either directly or through a theorem whose hypotheses have been checked.

f(a+h)=f(a)+Df(a)h+r(h),lim⁡h→0∥r(h)∥∥h∥=0f(a+h)=f(a)+Df(a)h+r(h),\qquad \lim_{h\to0}\frac{\lVert r(h)\rVert}{\lVert h\rVert}=0
The same linear map must work for all sufficiently small displacements, rather than a separate approximation for each direction.↔ Scroll wide equations horizontally.

A usable sufficient condition and a revealing failure

Continuous first partial derivatives on a neighborhood imply total differentiability there. To understand the condition, move between two nearby points one coordinate at a time and apply the one-variable mean value theorem on each segment. Continuity makes the intermediate partial derivatives close to their values at the base point, so the accumulated error is small relative to displacement. Continuity of the partials is sufficient rather than necessary: a differentiable map can have a derivative that fails to vary continuously.

There are even functions with every directional derivative at the origin but no total derivative there. Define f(x,y)=x cubed divided by x squared plus y squared away from the origin, and define its origin value as zero. Along a displacement tv, the directional derivative is v₁ cubed divided by v₁ squared plus v₂ squared for nonzero v. This dependence on v is not linear. The coordinate partials suggest the map L(h)=h₁, but on h=(t,t), f(h)=t/2 and the normalized error is a nonzero constant. The candidate therefore fails.

f(x,y)=x2+xy,Df(1,2)=(4,1),f(1+h,2+k)−3−4h−k=h2+hkf(x,y)=x^2+xy,\quad Df(1,2)=(4,1),\quad f(1+h,2+k)-3-4h-k=h^2+hk
The polynomial remainder is quadratic: its magnitude is at most two times h²+k², which is negligible relative to the displacement norm.↔ Scroll wide equations horizontally.

Accuracy, units, and an honest local promise

A derivative predicts changes, not absolute values without a base value. For a scalar output, its row matrix acts by a dot product with the gradient. In a calibrated physical model, each column carries the output units divided by the associated input units. Comparing coordinate magnitudes may require nondimensionalizing the variables; a norm that mixes meters with seconds without a declared scale hides that choice. The existence of a derivative is unchanged by fixed invertible linear rescaling, although numerical condition estimates do change.

The little-o remainder is an asymptotic promise, not a universal numerical tolerance. To certify an error at a specified displacement, derive a concrete bound, perhaps using a Hessian bound on a neighborhood. Nor does an accurate local approximation establish global invertibility or behavior far from the base point. A useful workflow separates three tasks: compute the candidate derivative, justify differentiability, and estimate the residual over the intended operating range. That separation prevents an attractive tangent picture from being mistaken for a global model.

A diagonal perturbation with a checked residual

For f(x,y)=x²+xy, estimate f(1.02,1.97) from the base point (1,2), then compute the actual error.

  1. The base value is 3. The coordinate derivatives are 2x+y and x, so the derivative row at (1,2) is (4,1). Polynomials have continuous partials, justifying its use.

  2. The displacement is (0.02,−0.03). Its linear contribution is 4·0.02−0.03=0.05, so the predicted value is 3.05.

  3. The exact remainder is h²+hk=0.0004−0.0006=−0.0002. Direct evaluation gives 1.0404+2.0094=3.0498, confirming the sign and size of that error.

The result

The estimate exceeds the actual value by 0.0002. Halving both displacement components would divide this polynomial's remainder by four, whereas the linear change would halve.

Explore the model

Interactive lab

Interactive lab · matrix

A derivative that is exactly linear

Reset, set diagonal scale a=1.5 and shear b=1, and predict the image of (1,1): (2.5,1). Play to inspect the transformed square. Reset and set both controls again for a new comparison. This selected linear map has zero approximation remainder everywhere; the interpolation is a geometric display, not an error bound for nonlinear maps.

A derivative that is exactly linearThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA derivative is one linear action on all displacement vectors; nonlinear approximation quality is an additional question.

Check your understanding

Which observation disproves a proposed total derivative L at a point?

Try it yourself

For g(x,y)=x²+y² at the origin, prove differentiability and determine the derivative. Explain why the fact that g(h) is positive for nonzero h does not contradict your answer.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compare the exact remainder with the Euclidean displacement norm rather than asking whether the remainder vanishes identically.

Compare with the solution

Take L=0. The remainder is h₁²+h₂²=‖h‖², so its ratio to ‖h‖ equals ‖h‖ and tends to zero. The derivative is the zero linear map. Positivity records a second-order increase; differentiability only requires the residual to be smaller than first order, not to be identically zero. At a stationary point, a zero derivative can coexist with substantial curvature.

Take it with you

  • Total differentiability controls one normed remainder across all directions.
  • Continuous first partials guarantee differentiability, while partials alone only identify a candidate.
  • An asymptotic derivative and a finite-distance error certificate answer different questions.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.