Partial derivatives, differentiability, chain rule

In one sentence

In one variable there is one notion of derivative; in two there are four, they are not equivalent, and the exam is built almost entirely on the gaps between them.

Why the exam asks it

Because "differentiable" is the word candidates carry over from one variable unexamined. In one variable, the derivative existing is the whole story — it forces continuity and everything else follows. In two, partial derivatives can exist at a point where the function is not even continuous, and that single sentence generates more wrong options in this topic than any other fact. The paper knows it, and sets it every year.

The idea

Write $f$ for a real-valued function of $(x,y)$ near a point $a$.

Partial derivatives are one-variable derivatives along the two axes. $f_x(a)$ freezes $y$ and differentiates in $x$. That is two directions out of infinitely many, which is the source of all the trouble.

Directional derivative in the direction $u$: the ordinary derivative at $0$ of $t \mapsto f(a + tu)$. The partials are the special cases $u = (1,0)$ and $u = (0,1)$.

Differentiable at $a$ means something stronger than any collection of directional statements. It asks for a linear map $L$ with

$f(a + h) = f(a) + L(h) + o(|h|) \quad \text{as } h \to 0.$

The error must be small compared with $|h|$ however $h$ approaches zero — not along each line separately, but uniformly in direction. When such an $L$ exists it is unique, it is given by the partials, $L(h) = f_x(a)h_1 + f_y(a)h_2$, and $f$ is continuous at $a$.

The ladder is the topic. Read down and every arrow is a theorem; read up and every arrow is a counterexample, and each has a standard witness.

Partials exist, function not continuous

$f(x,y) = \frac{xy}{x^2 + y^2}, \qquad f(0,0) = 0.$

Along the axes $f$ is identically $0$, so $f_x(0,0) = f_y(0,0) = 0$: both partials exist and are as well-behaved as they could be. But along the diagonal $y = x$,

$f(x,x) = \frac{x^2}{2x^2} = \frac12$

for every $x \ne 0$. So $f$ takes the value $\frac12$ arbitrarily close to the origin where it is $0$: not continuous, let alone differentiable.

The partials looked at two directions. The function misbehaves in a third.

Every directional derivative exists, function still not continuous

The natural repair is to demand all directions, not just two. It is not enough.

$f(x,y) = \frac{x^2 y}{x^4 + y^2}, \qquad f(0,0) = 0.$

Along the line $y = mx$,

$f(x, mx) = \frac{m x^3}{x^4 + m^2x^2} = \frac{mx}{x^2 + m^2} \longrightarrow 0,$

and along the $y$-axis $f$ is $0$ outright. So every directional derivative at the origin exists and equals $0$ — the function looks, from every straight line, exactly like the zero function.

Now approach along the parabola $y = x^2$:

$f(x, x^2) = \frac{x^2 \cdot x^2}{x^4 + x^4} = \frac12 .$

Constantly one half, all the way in. The limit does not exist, so $f$ is not continuous at the origin, and certainly not differentiable there.

This is the example to remember, because it closes the obvious escape route. Straight lines are a one-parameter family of paths; the plane has more paths than that.

What actually guarantees differentiability

The usable sufficient condition:

If $f_x$ and $f_y$ exist in a neighbourhood of $a$ and are continuous at $a$, then $f$ is differentiable at $a$.

This is the workhorse. In practice every function the paper hands you built from polynomials, exponentials, sines and logarithms is $C^1$ away from its obvious bad points, so it is differentiable there and the only question is what happens at the bad point.

Note that continuity of the partials is sufficient and not necessary. A function can be differentiable at a point where its partials are discontinuous — the one-variable $x^2\sin(1/x)$ dressed up in two variables does it. So $C^1$ is a strictly smaller class than differentiable, and an option asserting the reverse is wrong.

Equality of mixed partials

Schwarz/Clairaut: if $f_{xy}$ and $f_{yx}$ exist near $a$ and are continuous at $a$, they are equal there. The continuity hypothesis is real, and dropping it is a standard question.

$f(x,y) = \frac{xy(x^2 - y^2)}{x^2 + y^2}, \qquad f(0,0) = 0.$

Working from first principles along the axes gives $f_x(0,y) = -y$ and $f_y(x,0) = x$. Differentiating those,

$f_{xy}(0,0) = -1, \qquad f_{yx}(0,0) = +1 .$

Both mixed partials exist at the origin and they are different. Nothing is wrong with Schwarz — its hypothesis simply is not met, because $f_{xy}$ is not continuous at the origin.

The chain rule

For $f$ differentiable at $a$ and a curve $\gamma$ differentiable at $t_0$ with $\gamma(t_0) = a$,

$\frac{d}{dt}f(\gamma(t)) \Big|_{t_0} = f_x(a)\,\gamma_1'(t_0) + f_y(a)\,\gamma_2'(t_0).$

The hypothesis that matters is differentiability of $f$, not existence of its partials. With partials alone the formula is false, and $xy/(x^2+y^2)$ is again the witness: its partials at the origin are $0$, the formula would predict derivative $0$ along $\gamma(t) = (t,t)$, and in fact $f(\gamma(t)) = \frac12$ is a constant that does not even agree with $f(0,0) = 0$.

Where intuition breaks

"Both partials exist at $a$, so $f$ is continuous at $a$." $xy/(x^2+y^2)$. Tempting because in one variable differentiability does force continuity, and the partials look like the same statement. They are not: each partial sees a single line, and two lines do not control a neighbourhood. This is the topic's central hypothesis dropped.

"All directional derivatives exist, so $f$ is differentiable." $x^2y/(x^4+y^2)$. This is the more sophisticated version of the same error, and it is the one that catches candidates who have already learned the first. Differentiability is not a statement about each direction separately; it is a statement about all directions at once, with a single linear approximation and a uniform error bound.

"$f$ is differentiable, so $f_x$ and $f_y$ are continuous." A converse assumed: continuity of the partials is sufficient for differentiability, not necessary. $C^1 \subsetneq$ differentiable, in that direction.

"$f_{xy} = f_{yx}$ always." $xy(x^2-y^2)/(x^2+y^2)$ gives $-1$ and $+1$ at the origin. Symmetry of second derivatives is a theorem with a continuity hypothesis, not an identity.

"$f$ is continuous and the partials exist, so $f$ is differentiable." Still false, and worth separating because it feels like it has patched the hole. Take $f(x,y) = \dfrac{x^3}{x^2+y^2}$ with $f(0,0) = 0$. It is continuous at the origin (the quotient is bounded by $|x|$), both partials exist there, and it is not differentiable: the candidate linear map would be $h \mapsto h_1$, and $f(h) - h_1$ is not $o(|h|)$ along $h_1 = h_2$.

"$f$ restricted to every line through $a$ is continuous, so $f$ is continuous at $a$." The same $x^2y/(x^4+y^2)$ again. Restriction to lines is a strictly weaker condition than continuity, in every one of these examples.

The exam's angle

1. Identify which rung the option is on. Partials, all directions, differentiable, $C^1$ — these are four different claims, and the wrong option almost always swaps two of them.
2. When asked about a limit at the origin, try lines first, then a parabola. Lines are quick and often settle it. If every line gives the same value, do not conclude the limit exists — try $y = x^2$, then $y = x^3$ or $x = y^2$ as the powers suggest.
3. Match the degrees. In $x^ay^b/(x^c + y^d)$, the path to try is the one making numerator and denominator the same order: for $x^2y/(x^4+y^2)$ that is $y \sim x^2$. This is a reliable way to find the bad path rather than guessing.
4. For "is it differentiable at the origin", compute the candidate map, then test the error. The candidate is forced: $L(h) = f_x(0)h_1 + f_y(0)h_2$. Then check whether $(f(h) - f(0) - L(h))/|h| \to 0$. If it fails along any path, you are done.
5. Away from the bad point, say $C^1$ and move on. Elementary formulas are $C^1$ wherever the denominator does not vanish, so the whole question lives at the one exceptional point.

The night before

• Partials exist $\Leftarrow$ all directional derivatives exist $\Leftarrow$ differentiable $\Leftarrow$ $C^1$. Every arrow one way.
• Differentiable $\Rightarrow$ continuous. Partials existing does not $\Rightarrow$ continuous.
• $xy/(x^2+y^2)$: partials exist at $0$, not continuous there. Value $\frac12$ on $y = x$.
• $x^2y/(x^4+y^2)$: every directional derivative exists at $0$, not continuous. Value $\frac12$ on $y = x^2$.
• $x^3/(x^2+y^2)$: continuous at $0$, partials exist, still not differentiable.
• $xy(x^2-y^2)/(x^2+y^2)$: $f_{xy}(0,0) = -1$, $f_{yx}(0,0) = +1$. Schwarz needs continuity of the mixed partial.
• Sufficient condition to reach for: partials continuous near $a$ $\Rightarrow$ differentiable at $a$. Not necessary.
• To find a bad path in $x^ay^b/(x^c+y^d)$, balance the degrees.