A Continuity Criterion for Differentiability

From Partial Derivatives to Differentiability

Partial derivatives describe how a function changes when one coordinate varies and the others are held fixed. Differentiability is a stronger local property: it says that one linear function of all the coordinate changes gives a first-order approximation to the function’s total change.

The first partial derivatives provide the coefficients for that linear approximation. The question is when those coefficients genuinely describe the function’s behavior in every direction. Continuity of the partial derivatives in a neighborhood gives a useful sufficient condition.

Key idea: If the first partial derivatives exist near a point and are continuous there, then the function is differentiable at the point. The derivative is the linear map whose coefficients are the partial derivatives evaluated at that point.

Differentiability and the Linear Map

Let $f:U\to\mathbb{R}$, where $U$ is an open subset of $\mathbb{R}^n$, and let $\mathbf{a}\in U$. The function is differentiable at $\mathbf{a}$ if there is a linear map $L:\mathbb{R}^n\to\mathbb{R}$ such that

f(\mathbf{a}+\mathbf{h}) = f(\mathbf{a})+L(\mathbf{h})+r(\mathbf{h}), \qquad \lim_{\mathbf{h}\to\mathbf{0}} \frac{r(\mathbf{h})}{\|\mathbf{h}\|}=0.

Here $\|\mathbf{h}\|$ is the Euclidean length of the increment vector, and $r(\mathbf{h})$ is the remainder after subtracting the linear approximation. The condition on the remainder says that its size becomes negligible compared with $\|\mathbf{h}\|$ as $\mathbf{h}$ approaches zero.

For a differentiable function, the coefficient of the $i$th coordinate in $L$ must be the $i$th partial derivative at $\mathbf{a}$. Thus the only possible linear map is

L(\mathbf{h}) = \sum_{i=1}^{n} \frac{\partial f}{\partial x_i}(\mathbf{a})h_i, \qquad \mathbf{h}=(h_1,\ldots,h_n).

In two variables this becomes $L(h,k)=f_x(a,b)h+f_y(a,b)k$. The continuity criterion establishes that this candidate map really does approximate the function to first order.

The Continuity Criterion

Theorem: Suppose $f:U\to\mathbb{R}$ is defined on an open neighborhood $U$ of $\mathbf{a}\in\mathbb{R}^n$. If all first partial derivatives of $f$ exist in a neighborhood of $\mathbf{a}$ and are continuous at $\mathbf{a}$, then $f$ is differentiable at $\mathbf{a}$, with $$ L(\mathbf{h})= \sum_{i=1}^{n} \frac{\partial f}{\partial x_i}(\mathbf{a})h_i. $$ In particular, it is enough that the first partial derivatives exist and are continuous throughout some neighborhood of $\mathbf{a}$.

The theorem’s assumptions concern the partial derivatives near the point, not merely their values at the point. Continuity at the point is enough for the proof; requiring continuity throughout a neighborhood is a common, slightly stronger way to check the condition.

Why the Criterion Works

We prove the result by changing one coordinate at a time. This creates a sequence of one-variable changes, to each of which the one-variable Mean Value Theorem can be applied. Because each intermediate point approaches $\mathbf{a}$, continuity ensures that the partial derivative used on each short segment is close to its value at $\mathbf{a}$.

Break the total change into coordinate changes.
Starting at $\mathbf{a}$, change the first coordinate by $h_1$, then the second by $h_2$, and continue through the $n$th coordinate.

Apply the one-variable Mean Value Theorem.
For each coordinate segment, express its function change as that coordinate increment times a partial derivative at an intermediate point.

Subtract the proposed linear approximation.
Each remaining coefficient is the difference between a nearby partial derivative and the same partial derivative at $\mathbf{a}$.

Use continuity to bound the remainder.
All the intermediate points approach $\mathbf{a}$, so the coefficient differences become arbitrarily small compared with the total increment.

For a complete argument, write $\mathbf{a}=(a_1,\ldots,a_n)$ and $\mathbf{h}=(h_1,\ldots,h_n)$. Define $\mathbf{z}_0=\mathbf{a}$, and let $\mathbf{z}_i$ be the point whose first $i$ coordinates have been changed by the corresponding components of $\mathbf{h}$, while its remaining coordinates still equal those of $\mathbf{a}$. Thus $\mathbf{z}_n=\mathbf{a}+\mathbf{h}$. The open-neighborhood assumption ensures all these coordinate segments lie in $U$ when $\|\mathbf{h}\|$ is sufficiently small.

On the $i$th segment, only coordinate $i$ changes. The one-variable Mean Value Theorem gives a point $\mathbf{w}_i$ on that segment such that

f(\mathbf{z}_i)-f(\mathbf{z}_{i-1}) = h_i\frac{\partial f}{\partial x_i}(\mathbf{w}_i).

If $h_i=0$, the segment contributes zero and the same equation holds. Adding the coordinate changes gives the total change in $f$. Subtracting the proposed linear map leaves the remainder

r(\mathbf{h}) = \sum_{i=1}^{n} h_i\left( \frac{\partial f}{\partial x_i}(\mathbf{w}_i) - \frac{\partial f}{\partial x_i}(\mathbf{a}) \right).

Each $\mathbf{w}_i$ lies on a segment between points whose distance from $\mathbf{a}$ is at most $\|\mathbf{h}\|$. Consequently, $\|\mathbf{w}_i-\mathbf{a}\|\leq\|\mathbf{h}\|$, so every $\mathbf{w}_i$ approaches $\mathbf{a}$ as $\mathbf{h}$ approaches $\mathbf{0}$. Given any $\varepsilon>0$, continuity of the finitely many partial derivatives at $\mathbf{a}$ ensures that, for sufficiently small $\|\mathbf{h}\|$, every coefficient difference in the remainder has absolute value less than $\varepsilon$. Therefore

|r(\mathbf{h})| \leq \varepsilon\sum_{i=1}^{n}|h_i| \leq \varepsilon\sqrt{n}\,\|\mathbf{h}\|.

The second inequality is the Cauchy–Schwarz inequality applied to $(|h_1|,\ldots,|h_n|)$ and $(1,\ldots,1)$. Dividing by $\|\mathbf{h}\|$ shows that the remainder-to-increment ratio can be made arbitrarily small. Hence $r(\mathbf{h})/\|\mathbf{h}\|\to 0$, which is exactly differentiability at $\mathbf{a}$. This proves the criterion.

Worked Examples

Worked Example: A Polynomial at a Nonzero Point

Use the criterion to find the derivative of $f(x,y)=x^2+3xy-y^2$ at $(1,-1)$.

First, differentiate with respect to each variable while holding the other fixed:

f_x(x,y)=2x+3y, \qquad f_y(x,y)=3x-2y.

Both partial derivatives are polynomials, so they exist and are continuous throughout $\mathbb{R}^2$. The criterion applies. Evaluate the function and partial derivatives at the point:

f(1,-1)=1^2+3(1)(-1)-(-1)^2=1-3-1=-3, $$ $$ f_x(1,-1)=2(1)+3(-1)=2-3=-1, \qquad f_y(1,-1)=3(1)-2(-1)=3+2=5.

Thus the derivative is $L(h,k)=-h+5k$, and the corresponding first-order expression is $-3-h+5k$. To check the remainder directly, substitute $x=1+h$ and $y=-1+k$:

\begin{aligned} f(1+h,-1+k) &=(1+h)^2+3(1+h)(-1+k)-(-1+k)^2\\ &=-3-h+5k+h^2+3hk-k^2. \end{aligned}

The remainder is $h^2+3hk-k^2$. Its absolute value is at most $h^2+3|hk|+k^2$, which is at most $\frac{5}{2}(h^2+k^2)$, since $2|hk|\leq h^2+k^2$. Dividing by $\sqrt{h^2+k^2}$ gives a bound tending to zero. This confirms the derivative found using the criterion.

Worked Example: Exponential and Trigonometric Terms

Find the derivative of $g(x,y)=e^{xy}+\sin(x+y)$ at $(0,0)$, and check the first-order remainder.

By the chain and product rules,

g_x(x,y)=ye^{xy}+\cos(x+y), \qquad g_y(x,y)=xe^{xy}+\cos(x+y).

These partial derivatives are continuous everywhere. At the origin,

g(0,0)=e^0+\sin 0=1, \qquad g_x(0,0)=0+1=1, \qquad g_y(0,0)=0+1=1.

Therefore $L(h,k)=h+k$, and the first-order expression is $1+h+k$. To check it, substitute $(h,k)$ into $g$ and subtract this expression:

g(h,k)-(1+h+k) = (e^{hk}-1)+\bigl(\sin(h+k)-(h+k)\bigr).

Let $r=\sqrt{h^2+k^2}$. Then $|hk|\leq r^2/2$, and $|h+k|\leq\sqrt{2}\,r$. For small $r$, the standard estimates $|e^u-1|\leq e^{|u|}|u|$ and $|\sin s-s|\leq |s|^3/6$ show that the absolute value of the remainder is bounded by a constant times $r^2$ plus a constant times $r^3$. After division by $r$, both terms tend to zero. This verifies the linear approximation as well as the theorem’s conclusion.

Worked Example: A Logarithmic Function

Find the derivative of $q(x,y)=\ln(4+x^2+y^2)$ at $(1,-1)$.

The logarithm is defined throughout $\mathbb{R}^2$ here because $4+x^2+y^2\geq 4>0$. Differentiation gives

q_x(x,y)=\frac{2x}{4+x^2+y^2}, \qquad q_y(x,y)=\frac{2y}{4+x^2+y^2}.

The denominators never vanish, so both partial derivatives are continuous in a neighborhood of $(1,-1)$. The criterion applies. Substitution gives

q(1,-1)=\ln(4+1+1)=\ln 6, $$ $$ q_x(1,-1)=\frac{2}{6}=\frac{1}{3}, \qquad q_y(1,-1)=\frac{-2}{6}=-\frac{1}{3}.

Thus $L(h,k)=\frac{1}{3}h-\frac{1}{3}k$, and the first-order expression at $(1,-1)$ is $\ln 6+\frac{h-k}{3}$. Substituting the point into the derivative formulas confirms both coefficients, and the continuous-partials criterion guarantees that the difference between the function and this expression, divided by $\sqrt{h^2+k^2}$, tends to zero.

A Common Mistake: Checking Partial Derivatives Only at the Point

The theorem does not say that partial derivatives existing at a point are enough. For a concrete warning, define

p(x,y)= \begin{cases} \dfrac{xy}{x^2+y^2},&(x,y)\ne(0,0),\\[4pt] 0,&(x,y)=(0,0). \end{cases}

Along either coordinate axis, the function is zero. Hence the partial derivatives at the origin exist and are both zero:

p_x(0,0)=\lim_{t\to 0}\frac{p(t,0)-p(0,0)}{t}=0, \qquad p_y(0,0)=\lim_{t\to 0}\frac{p(0,t)-p(0,0)}{t}=0.

A mistaken conclusion would be: “The partial derivatives are zero, so the derivative is the zero map.” That conclusion assumes differentiability without checking the required local behavior. Along the line $y=x=t$, for every $t\ne0$,

p(t,t)=\frac{t^2}{2t^2}=\frac{1}{2}.

This does not approach $p(0,0)=0$. The function is not even continuous at the origin, so it cannot be differentiable there. The continuous-partials condition is a sufficient test; merely calculating partial derivatives at the point does not meet its assumptions.

When the Criterion Is Useful

In applications, the criterion often turns a difficult question about all possible directions into a manageable check of first partial derivatives. For a formula built from familiar differentiable functions, one can often establish continuity of its partial derivatives directly from the usual differentiation rules. Once that local continuity is verified, differentiability follows without testing every possible path of increments.

The assumptions still matter. The function must be defined on an open neighborhood so that small coordinate changes are available, and the partial derivatives must exist nearby and be continuous at the point. When these conditions hold, the derivative’s coefficients are the partial derivatives at the point, and the remainder is negligible compared with the length of the increment.

Check Your Understanding

Use the theorem and its assumptions to answer the following questions.

State the continuity criterion for differentiability of a function $f:U\to\mathbb{R}$ at a point in $\mathbb{R}^n$.
For $f(x,y)=x^2y$, what linear map does the criterion give at $(1,2)$?
Why does the coordinate-by-coordinate proof use the one-variable Mean Value Theorem?
In the proof, why do the partial derivative values at the intermediate points approach their values at $\mathbf{a}$?
What is the error in concluding that a function is differentiable solely because both partial derivatives exist at the point?