Differentiability and the Total Derivative

The total derivative is the starting point for the <span class="wiki-link-missing">Chain Rule in Several Variables</span>, which is only true in its clean matrix form ($D(g\circ f)(a) = Dg(f(a))\,Df(a)$) because the derivatives involved are linear maps that compose exactly like matrices — a fact that has no clean analogue if one only has partial derivatives. It is also the foundation for the <span class="wiki-link-missing">Gradient and Directional Derivatives</span>: once $Df(a)$ is known to be linear, the Riesz representation of that linear functional (for scalar-valued $f$) is the gradient vector, and every directional derivative is recovered instantly as a dot product. In differential geometry and general relativity, the total derivative generalizes to the differential of a smooth map between manifolds, the basic object underlying tangent spaces and pushforwards. In optimization and machine learning, the Jacobian and its square-matrix specialization the Hessian drive Newton's method and gradient-based training algorithms, where "differentiable" in the total-derivative sense is exactly the hypothesis needed for gradient descent to make local sense. In numerical analysis, the linear approximation given by the total derivative underlies Newton's method for systems of equations, and in control theory, linearization of nonlinear dynamical systems around an equilibrium is literally computing the total derivative (Jacobian) of the vector field.

Ask the tutor: Can you show me how the total derivative behaves under the chain rule when $f$ maps into $\mathbb{R}^m$ with $m>1$?

Think first about $m = n = 1$: a function $f$ is differentiable at $a$ exactly when there is a number $L$ — the familiar derivative $f'(a)$ — with $f(a+h) = f(a) + Lh + o(h)$. The graph of the affine function $x \mapsto f(a) + f'(a)(x-a)$ is tangent to the graph of $f$, meaning the error shrinks faster than $h$ itself. The total derivative definition is exactly this idea transplanted to $\mathbb{R}^n$: replace the number $L$ by a linear map $L : \mathbb{R}^n \to \mathbb{R}^m$, and replace "$Lh$" by evaluating that linear map at the vector $h$. The graph of $f$ near $a$ — an $n$-dimensional surface sitting in $\mathbb{R}^{n+m}$ — is approximated by a flat $n$-dimensional affine subspace (a "tangent plane" when $n=2,m=1$), and the approximation error, measured relative to $\lVert h\rVert$, must vanish no matter which direction $h \to 0$ comes from. That last clause is the crux. Partial derivatives only test what happens as $h$ approaches $0$ along the coordinate axes, one at a time. A function's graph can have a perfectly good tangent line in the $x$-direction, a perfectly good tangent line in the $y$-direction, and yet still be sharply creased, or even discontinuous, along other directions such as the diagonal. The total derivative rules this out by demanding a single linear approximation that works uniformly over all directions simultaneously — this is why differentiability is the strictly stronger and mathematically correct notion.

The picture above is the geometric content of the classic example $f(x,y) = \dfrac{xy}{x^2+y^2}$ (with $f(0,0)=0$): along each axis the function is identically $0$, so both partial derivatives at the origin exist and equal $0$. But approaching the origin along the diagonal $x = y$ gives $f(x,x) = 1/2 \neq 0$, so the function is not even continuous at the origin — utterly incompatible with any linear (indeed any reasonable) approximation. This single example already shows that "partials exist" is far too weak a hypothesis, and motivates why the definition of the total derivative insists on a two-sided, all-directions limit.

Example 1 — a genuinely asymmetric differentiability check. Let $f(x,y) = x^2 y + 3xy^2 - y^3$, and let $a = (2,-1)$. We verify differentiability at $a$ directly from partial derivatives, compute the total derivative, and use it to approximate $f$ near a nearby point. Compute partials everywhere:

Both are polynomials in $x,y$, hence continuous on all of $\mathbb{R}^2$ (a fact from single/multivariable continuity theory: polynomials are continuous everywhere). By the continuous partials theorem, $f$ is differentiable at every point of $\mathbb{R}^2$, in particular at $a = (2,-1)$. Evaluate at $a=(2,-1)$:

So the total derivative is the linear map

with Jacobian (gradient, since $m=1$) row vector $\nabla f(2,-1) = (-1,-11)$. Also compute $f(2,-1) = (2)^2(-1) + 3(2)(-1)^2 - (-1)^3 = -4 + 6 + 1 = 3$. Linear approximation near $a$: for $h = (h_1,h_2)$ small,

Test this at $h = (0.1, -0.05)$ (an asymmetric, non-special perturbation), i.e. at the point $(2.1, -1.05)$: Approximation: $3 - (0.1) - 11(-0.05) = 3 - 0.1 + 0.55 = 3.45$. Exact value: $f(2.1,-1.05) = (2.1)^2(-1.05) + 3(2.1)(-1.05)^2 - (-1.05)^3$. Compute each term: $(2.1)^2 = 4.41$, so first term $= 4.41 \times(-1.05) = -4.6305$. Second term: $(-1.05)^2 = 1.1025$, so $3(2.1)(1.1025) = 6.3(1.1025) = 6.94575$. Third term: $-(-1.05)^3 = -(-1.157625) = 1.157625$. Sum: $-4.6305 + 6.94575 + 1.157625 = 3.472875$. The linear approximation gives $3.45$, the true value is $3.472875$ — an error of about $0.0229$, which is small compared to $\lVert h \rVert = \sqrt{0.1^2+0.05^2} \approx 0.1118$ (relative error $\approx 0.205$, consistent with a genuinely quadratic remainder $O(\lVert h\rVert)$ relative size that shrinks as $h\to 0$, since $f$ is a degree-3 polynomial and the next-order terms are quadratic in $h$).

Example 2 — partials exist but differentiability fails (the boundary case made precise). Let

This is a genuinely asymmetric variant of the standard counterexample (using $x^3 - y^3$ rather than $xy$), chosen so continuity actually holds here — making the failure purely about the linear approximation, a subtler and more instructive breakdown.

Continuity at $(0,0)$: using polar coordinates $x = \rho\cos\theta, y = \rho\sin\theta$,

Since $|\cos^3\theta - \sin^3\theta| \le 2$ for all $\theta$, we get $|f(x,y)| \le 2\rho = 2\lVert (x,y)\rVert \to 0$ as $(x,y)\to(0,0)$. So $f$ is continuous at the origin, with $f(0,0)=0$.

Partial derivatives at the origin: by definition,

Both partials exist. If $f$ were differentiable at $(0,0)$, uniqueness together with the Jacobian identity would force

Testing the defining limit directly. We must check whether

Using polar coordinates again, the numerator is

so the whole quotient equals

which is independent of $\rho$ and depends only on the approach angle $\theta$. Along $\theta = \pi/2$ (i.e. approaching along the positive $y$-axis, $h_1=0$): the quotient is $(0 - 1) - (0-1) = 0$. But along $\theta = \pi/4$ (the diagonal $h_1=h_2$): $\cos\theta=\sin\theta = \tfrac{1}{\sqrt2}$, so both bracketed differences are $0$, giving quotient $0$ too — so we must probe further. Try $\theta = \pi/3$: $\cos\theta = 1/2$, $\sin\theta = \sqrt3/2 \approx 0.8660$.

Quotient $= -0.5245 - (-0.3660) = -0.1585 \neq 0$. Since the quotient takes the nonzero value $\approx -0.1585$ along rays with $\theta = \pi/3$ arbitrarily close to the origin, while the definition requires the quotient to tend to $0$ along every path, the limit does not exist (or at least is not $0$). Hence $f$ is not differentiable at $(0,0)$, even though it is continuous there and both partial derivatives exist. This shows continuity plus existence of partials is still not enough — the extra ingredient of continuity of the partials near the point (used in the sufficient condition theorem) is doing real work, since here the partials $\partial f/\partial x, \partial f/\partial y$ can be computed to be discontinuous at the origin (direct computation shows their values depend on the approach angle in the same oscillatory way).