Differentiable Functions: A Rigorous Perspective

All things are difficult before they are easy, Thomas Fuller.

Recall

A complex number is specified by an ordered pair of real numbers (a, b) ∈ ℝ² and expressed or written in the form z = a + bi, where a and b are real numbers, and i is the imaginary unit, defined by the property i² = −1 ⇔ i = $\sqrt{-1}$, e.g., 2 + 5i, $7\pi + i\sqrt{2}.$ ℂ= { a + bi ∣a, b ∈ ℝ}.

Definition. Let D ⊆ ℂ be a set of complex numbers. A complex-valued function f of a complex variable, defined on D, is a rule that assigns to each complex number z belonging to the set D a unique complex number w, f: D ➞ ℂ.

• The set D is called the domain of the complex function f, D = Dom(f).
• The set of all actual outputs {f(z): z ∈ D} is called the range (or image) of the complex function f, also denoted f(D) = {f(z): z ∈ D}.

We often call the elements of D as points. If z = x+ iy ∈ D, then f(z) is called the image of the point z under f. f: D ➞ ℂ means that f is a complex function with domain D. We often write f(z) = u(x ,y) + iv(x, y), where u, v: ℝ² → ℝ are the real and imaginary parts.

Definition. Let D ⊆ ℂ, $f: D \rarr \Complex$ be a function and z₀ be a limit point of D (so arbitrarily close points of D lie around z₀, though possibly z₀ ∉ D). A complex number L is said to be a limit of the function f as z approaches z₀, written or expressed as $\lim_{z \to z_0} f(z)=L$, if for every epsilon ε > 0, there exist a corresponding delta δ > 0 such that |f(z) -L| < ε whenever z ∈ D and 0 < |z - z₀| < δ.

Why 0 < |z - z₀|? We exclude z = z₀ itself because the limit cares about values near z₀, not at z₀ itself. When z₀ ∉ D, you cannot evaluate f(z₀), so you only care about z approaching z₀. When z₀ ∈ D, you still want the function’s nearby behavior; this separates “limit” from “value.”

Equivalently, if ∀ε >0, ∃ δ > 0: (for every ε > 0, there exist a corresponding δ > 0) such that whenever z ∈ D ∩ B'(z₀; δ), f(z) ∈ B(L; ε) ↭ f(D ∩ B'(z₀; δ)) ⊂ B(L; ε).

If no such L exists, then we say that f(z) does not have a limit as z approaches z₀. This is exactly the same ε–δ formulation we know from real calculus, but now z and L live in the complex plane ℂ, and neighborhoods are round disks rather than intervals.

Continuity in the complex plane

Definition. Let D ⊆ ℂ. A function f: D → ℂ is said to be continuous at a point z₀ ∈ D if given any arbitrarily small ε > 0, there is a corresponding δ > 0 such that |f(z) - f(z₀)| < ε whenever z ∈ D and |z - z₀| < δ.

In words, arbitrarily small output‐changes ε can be guaranteed by restricting z to lie in a sufficiently small disk of radius δ around z₀.

Alternative (Sequential) Definition. Let D ⊆ ℂ. A function f: D → ℂ is said to be continuous at a point z₀ ∈ D if for every sequence {z_n}^∞_n=1 such that z_n ∈ D ∀n∈ℕ & z_n → z₀, we have $\lim_{z_n \to z_0} f(z_n) = f(z_0)$ .

Global Continuity

Definition. A function f: D → ℂ is said to be continuous if it is continuous at every point in its domain (∀z₀ ∈ D).

We begin by recalling the ε–δ definitions of limit and continuity in the complex plane —quite similar to the real case, but with disks instead of intervals. Then, we move to the general notion of differentiability in ℝⁿ→ℝᵐ, where the derivative becomes a matrix (the Jacobian) encoding the best linear approximation. Along the way you’ll find visual intuition, a handy “Quick Jacobian Recipe” table, and both smooth and pathological examples.

Differentiability at a point

In single-variable calculus, differentiability at a point means the familiar limit of the difference quotient exists. In ℝⁿ→ℝᵐ, it becomes the existence of a unique linear map —the Jacobian— that best approximates small changes via f(x + h) = f(x) + Df(x)h + o(∥h∥). It captures the idea that a function can be locally approximated by a linear transformation.

Figure. For f(x,y)=x²+y², the red plane at (1,1) is the Jacobian’s linear approximation.

This is a quadratic function f(x, y) = x² + y². The visualization above demonstrates several key aspects of multivariable differentiability:

1. Local Linear Approximation: At the point (x₀, y₀) = (1, 1), the red tangent plane represents the Jacobian matrix’s action as a linear map. This plane provides the best linear approximation to the function near this point.
2. Jacobian Matrix: For our example f(x, y) = x² + y², the Jacobian matrix at (1, 1) is: $Df(1, 1) = [\frac{\partial f}{\partial x} \frac{\partial f}{\partial y}]$ = [2·1 2·1] = [2 2]. This matrix defines the linear transformation that maps small changes in x and y to changes in f(x, y).
3. Approximation Property: The formula f(x + h) = f(x) + Df(x)h + o(||h||) is precisely what we’re seeing visually. The tangent plane (defined by Df(x)·h) approximates the actual surface (f(x+h)) with increasing accuracy as we approach the point of tangency.

Definition. Differentiability at a point. Let $f : ℝ^n \to ℝ^m$ be a function and let x be an interior point of the domain of f, $x \in \text{interior(dom f)} $. The function f is differentiable at x if there exists a matrix $Df(x) \in ℝ^{m \times n}$ that satisfies $\lim_{\substack{z \in \text{dom} f \\ z \neq x, z \to x}} \frac{||f(z) - f(x) - Df(x)(z-x)||_2}{||(z-x)||_2} = 0$ [*]

This matrix Df(x) is called the derivative or the Jacobian matrix of f at the point x.

Quick Jacobian Recipes

Function Type	Jacobian Df(x)
Linear: f(x) = A x	Df(x) = A
Affine: f(x) = A x + b	Df(x) = A
Quadratic: f(x) = xᵀ A x	Df(x) = (A + Aᵀ)x
Squared Norm: ||x||²	Df(x) = 2x

Differentiable function

Definition. A function f is called differentiable if its domain f (dom(f) ⊆ ℝⁿ) is open and f is differentiable at every point of its domain (∀x ∈ dom(f)).

Key Conditions and important notes

1. Open Domain. Its domain is an open set, meaning that for every point x in the domain of f, there exists a small open interval around x (open ball or a small neighborhood around x) that is also entirely contained within the domain. This condition is crucial to allow limits like $\lim_{h \to 0} \frac{||f(x+h) - f(x)||_2}{||h||_2} = 0$ to make sense for small perturbations h.
2. Pointwise Differentiability The function must be differentiable at every point in its domain: This means that the derivative of f exists at each point x in the domain or for each x ∈ dom f, there exists a Jacobian matrix Df(x) ∈ ℝ^m×n such that $\lim_{\substack{x + h \in dom f \\ h \neq 0, h \to 0}} \frac{||f(x+h) - f(x) - Df(x)h||_2}{||h||_2} = 0$. This means the best linear approximation to the function exists at each point and behaves well locally.
3. Geometric Intuition. Recall that a function is differentiable if it is “smooth” and has no sharp corners (like the absolute value function), cusps (a point on a curve where the graph sharply changes direction, like $x^{2/3}$), vertical tangents or discontinuities (i.e., the slope is undefined) within its domain. The function can be locally approximated by a linear map (i.e., its Jacobian matrix).
4. Differentiability implies continuity, but not vice versa.
5. A function may be differentiable on an open set but not on its boundary (since the definition requires openness).
6. Sufficient condition for differentiability. If all first-order partial derivatives of f exist and are continuous on an open set, then f is continuously differentiable on that set.

   If the partial derivatives exist and are continuous, it implies that the rate of change of the function with respect to each variable is well-behaved and doesn’t have sudden jumps or discontinuities. The continuity of partial derivatives ensures that the linear approximation is well-defined and consistent across the open set. This is what makes the function continuously differentiable.

Examples

• Differentiable function.

Let f(x, y) = sin(x) + y². Domain: ℝ², which is open. All partial derivatives exist and are continuous everywhere (∂f/∂x = cos(x) is continuous on ℝ² because cosine is continuous everywhere, ∂f/∂y = 2y is continuos on ℝ² as it is a linear function, hence both partial derivatives are continuous on ℝ²) ⇒ f is differentiable everywhere on its domain.

Geometric intuition. In the x-direction, the surface z = sin(x) + y² undulates like a sine wave. In the y-direction, it opens upward parabolically. Because these two behaviors blend smoothly (no corners or cusps), the surface admits a well-defined tangent plane at every point.

Polynomial Examples.

1. p(x,y) = x³ −3xy + y². Domain: ℝ². Partials: $\dfrac{\partial p}{\partial x} = 3x^2 -3y, \dfrac{\partial p}{\partial y} = -3x + 2y$ both polynomials (hence continuous everywhere), so p is C^∞.
2. Consider the polynomial function: f(x, y) = x³ + 3x²y - y³. The domain is ℝ², which is open and connected. ∂f/∂x = 3x² + 6xy, ∂f/∂y = 3x² - 3y². Both partial derivatives are polynomials, hence continuous everywhere ⇒ f is differentiable on all ℝ².
3. f(x, y)=x²y − 3xy³ + 7. Domain: ℝ² (open). All partials are polynomials (hence C¹) ⇒ differentiable everywhere.

Note

A function is said to be of class C¹ if it meets two criteria everywhere in its domain:

• All first-order partial derivatives exist.
• Those partial derivatives are continuous functions.

In other words, f ∈ C¹ means f is continuously differentiable.

• Exponential–polynomial mixture.

1. $f(x, y)=e^{x²+y²}+xy$. Domain: ℝ². f(x, y) is C¹ because the exponential is C¹ and x²+y² is C¹ and xy is C¹, too. Exponential and polynomial factors are infinitely differentiable, their sums and products remain smooth ⇒ f is differentiable on ℝ².

   Partial derivatives: ∂f/∂x = 2x·e^{x²+y²} + y, ∂f/∂y = 2y·e^{x²+y²} + x, both exist everywhere and are continuous on ℝ², so f∈C¹(ℝ²).

2. $q(x, y)=e^{x}(x^2+y)$. Domain: ℝ². Partial derivatives: ∂q/∂x = $e^{x}(x^2+y) + e^{x}·2x = e^{x}(x^2+y+2x), ∂q/∂y = e^{x}$. Again both smooth (continuously differentiable) on ℝ² ⇒ everywhere differentiable.

• Rational function away from the origin $f(x, y)=\dfrac{x}{x^{2}+y^{2}+1}$. Denominator x²+y²+1 never vanishes, f is well-defined on all of ℝ², domain = ℝ² (open).

∂f/∂x = [(x²+y²+1)·1 – x·2x] / (x²+y²+1)² = (y² – x² + 1) / (x²+y²+1)². ∂f/∂y = –x·2y / (x²+y²+1)² = –2xy / (x²+y²+1)². Both are quotients of polynomials whose denominators never hit zero, so each partial is continuous everywhere (ℝ²). By the standard theorem (continuous partials ⇒ differentiable), f is C¹ and therefore differentiable everywhere.

Standard Theorem: Let U ⊆ ℝⁿ be an open set and f : U → ℝⁿ be a function. If all the first-order partial derivatives of f exist and are continuous at a point a ∈ U, then f is differentiable at a.

Key Points

1. Hypothesis: All partial derivatives $\dfrac{∂f _i}{∂x_j}$ (i = 1,…, m; j = 1,…, n) exist and are continuous in an open neighborhood around a (and hence at a itself).
2. Conclusion: f is differentiable at a. This means the total derivative (represented by the Jacobian matrix) exists and provides a good linear approximation: $\lim_{h \to 0} \dfrac{||f(a+h)-f(a)-Df(a)·h||}{||h||} = 0$ where Df(a) is the m×n Jacobian matrix evaluated at a.
3. This theorem provides a sufficient (but not necessary) condition for differentiability. It’s often much easier to check the continuity of partial derivatives than to verify the limit definition of differentiability directly. Continuous Partial Derivatives ⟹ Differentiable.

   The converse is not true. A function can be differentiable at a point without having continuous partial derivatives at that point.

4. A function satisfying the hypothesis of this theorem is called continuously differentiable or of class C¹ at a.

• Trigonometric–hyperbolic blend, $f(x, y)=\sin(x)\cosh(y)$. Domain: ℝ². sin(x) and cosh(y) are C¹ on ℝ; products/compositions of C¹ functions are C¹ ⇒ differentiable everywhere.

Partial derivatives. At any point (x, y): ∂f/∂x = cos(x)·cosh(y), ∂f/∂y = sin(x)·sinh(y). Since sin, cos, sinh, cosh are all C^∞ on ℝ, their product (Standard Theorems for Building Smooth Functions, The product f·g of two C^∞ functions is also C^∞) is also smooth, then the original function f is differentiable everywhere.

C^∞ denotes the class of functions that are infinitely differentiable on their domain. In concrete terms, a function f is in C^∞ if for every non-negative integer k, the k-th derivative of f exists and is a continuous function.

Key Properties: 1. Every derivative of any order exists. 2. All these derivatives vary continuously (no sudden jumps or discontinuities). 3. In particular, being C^∞ implies the original function is continuous and differentiable as many times as you like.

Standard Theorems for Building Smooth Functions: If f and g are C^∞, then f + g, f – g, f·g are C^∞. If f is C^∞ on an open set U and g is C^∞ on V with g(U)⊆V, then f∘g is C^∞. For multivariable functions, having continuous partials of all orders in each variable is equivalent to being C^∞ on that domain.

• Not differentiable. Let $f(x, y)=\begin{cases} \dfrac{xy}{\sqrt{x^{2}+y^{2}}} & (x,y)\neq(0,0) \\\\ 0 & (x,y)=(0,0) \end{cases}$

1. Continuity at the origin (0, 0). To verify continuity, compute the limit as (x,y)→(0,0): $\lim_{(x, y)→(0,0)} \dfrac{xy}{\sqrt{x^{2}+y^{2}}}$. Switch to polar coordinates x = rcosθ, y = rsinθ. $\lim_{(x, y)→(0,0)} \dfrac{xy}{\sqrt{x^{2}+y^{2}}} = \lim_{(x, y)→(0,0)} \dfrac{r^2\cos(\theta)\sin(\theta)}{r} = \lim_{(x, y)→(0,0)} r\cos(\theta)\sin(\theta) = 0 = f(0, 0),$ confirming continuity at (0, 0).
2. Directional Derivatives Exist Everywhere. The directional derivative of a function f at a point in a given direction measures the instantaneous rate of change of f along that direction. It’s a generalization of the partial derivative, which measures the rate of change in the directions of the coordinate axes. Essentially, it tells you how much the function’s value will change if you move a small amount in a specific direction from a given point. $D_uf(a, b) = \lim_{h->0} \dfrac{f(a + hu, b + kv) - f(a, b)}{h}$ where u = is a unit vector.
   In our case, the directional derivative of f at (a, b) = (0, 0) in direction u = (a, b) (with $\sqrt{a^2+b^2}=1$) is: $D_uf(0, 0) = \lim_{h->0} \dfrac{f(ha, hb) - f(0, 0)}{h} = \lim_{h->0} \dfrac{(ha)(hb)}{h\sqrt{(ha)^2+(hb)^2}} = \lim_{h->0} \dfrac{h^2ab}{h^2\sqrt{a^2+b^2}} = \lim_{h->0} \dfrac{ab}{\sqrt{a^2+b^2}} = \dfrac{ab}{\sqrt{a^2+b^2}} = ab$. This limit exists for all directions u, so directional derivatives exist everywhere at (0, 0).
3. Failure of Total Differentiability. For f to be differentiable at (0,0), the directional derivatives must be linear in u and agree with the total derivative Df(0,0) ↭ A function f is differentiable at (0,0) only if there is a linear map L (the Jacobian) such that Dᵤf(0, 0) = L(u) for every unit direction u= (a, b) with a² + b² = 1.
4. The directional derivative ab is not linear (the map $\mathbf{u} \mapsto D_{\mathbf{u}}f(0,0)$ must be linear for it to represent a true derivative) in u = (a, b) with a² + b² = 1.
   Homogeneity test: Consider a scalar c∈ℝ and direction cu = (ca, cb). The unit vector in this direction is $\frac{(ca, cb)}{||cu||}=\dfrac{(ca, cb)}{|c|\sqrt{a^2+b^2}}=(\dfrac{ca}{|c|}, \dfrac{cb}{|c|})$. The directional derivative is $D_{cu}f(0, 0) = \dfrac{ca}{|c|}·\dfrac{cb}{|c|} = \dfrac{c^2ab}{|c|^2} = ab$. Linearity requires . $D_{cu}f(0, 0) = c·D_{u}f(0, 0) = c(ab)$. This holds only if ab=c(ab) for all c, which is false unless ab=0.
   For example. If u = $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), D_{u}f(0, 0) = \frac{1}{\sqrt{2}}·\frac{1}{\sqrt{2}} = \frac{1}{2}$. For c = 2, $D_{2u}f(0, 0) = \frac{1}{2} \ne 2·\frac{1}{2}=1$. Hence, Homogeneity fails.
   Additivity test. A linear map must have $L(u_1 + u_2)=L(u_1)+L(u_2),$ for u₁ = (1, 0) and u₂ = (0, 1). u₁ + u₂ = (1, 1). The unit vector is $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), D_{u_1 + u_2}f(0, 0) = \dfrac{1}{\sqrt{2}}·\dfrac{1}{\sqrt{2}} = \frac{1}{2}.$ Additivity requires $D_{u_1 + u_2}f(0, 0) = D_{u_1}f(0, 0) + D_{u_2}f(0, 0), \text{ but } \frac{1}{2} \ne 0 + 0,$ hence additivity fails, too.
5. Agreement with the Total Derivative. The directional derivatives must agree with the total derivative Df(0, 0). However,

• Partial derivatives at (0,0) are zero: $\dfrac{∂f}{∂x} = f_x(0, 0) = \lim_{h->0} \dfrac{f(h, 0) - f(0, 0)}{h} = \lim_{h->0} \dfrac{0}{h} = 0, \text{similarly} \dfrac{∂f}{∂y} = 0$. Thus, the gradient ∇f(0,0) = (0, 0). If f were differentiable at (0, 0), the total derivative would be Df_u(0, 0) = ∇f(0,0)⋅u=(0,0)⋅(a,b)=0 for all u. However, for u = $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$, Df_u(0, 0) = 1/2 ≠ 0 ⊥

• Norm function (not differentiable at the origin) g(x, y) = $\sqrt{x^2+y^2}$. Domain: ℝ²

1. For (x, y) ≠ (0, 0), one shows $\sqrt{x^2+y^2}=(x^2+y^2)^{1/2}, \dfrac{∂g}{∂x} =\dfrac{1}{2}(x^2+y^2)^{-1/2}·2x = \dfrac{x}{\sqrt{x^2+y^2}}$. Analogously, $\dfrac{∂g}{∂y} =\dfrac{y}{\sqrt{x^2+y^2}}$, both continuos on ℝ² \ {0}.

2. At the origin, the gradient would have to be ∇g(0, 0) = (0, 0).

   When a differentiable function $g\colon\R^n\to\R$ attains a local minimum at a point p, the gradient $\nabla g(p)$ has to be the zero vector. The gradient $\nabla g(p)$ points in the direction of steepest ascent. At a minimum, there is no “downhill” direction to go—every small move increases (or leaves unchanged) the value of g. Hence there cannot be any nonzero component of the gradient; otherwise you could go “downhill” in the opposite direction.

   The graph of $\sqrt{x^2+y^2}$ is the upper half of a circular cone with its vertex at the origin.

3. Directional derivatives along (1, 0) and (0, 1) give, $Dg_{(1, 0)} = \lim_{h \to 0^+} \dfrac{g(h, 0)-g(0, 0)}{h} = \lim_{h \to 0^+} \dfrac{h^2}{h}=1, Dg_{(0, 1)} = \lim_{h \to 0^+} \dfrac{g(0, h)-g(0, 0)}{h}=1$. If g were differentiable at (0, 0), the total derivative would be Dg_u(0, 0) = ∇g(0,0)⋅u=(0,0)⋅(a,b) = 0 for all u. Since the previous directional derivatives along (1, 0) and (0, 1) don’t match ∇g(0,0)⋅u = 0, g fails to be differentiable at (0, 0).

• Absolute-value ridge. f(x, y)=|x|+|y|. Failure: At (0, 0) the graph is a “pyramid.” Partial derivatives ∂f/∂x and ∂f/∂y do not exist along the axes (sharp crease), so the linear-approximation definition fails.

1. Partial derivatives do not exist at (0, 0) The partial derivative with respect to x at the origin via the limit definition: $\frac{\partial f}{\partial x}(0,0)= \lim_{h \to 0}\frac{f(h,0)-f(0,0)}{h} = \lim_{h \to 0}\frac{|h| - 0}{h}$. If h>0, the quotient (limit) is h/h = 1. If h<0, the quotient is (-h)/h = -1. Because the two one‐sided limits disagree, $\partial f/\partial x$ does not exist at (0, 0). The same computation for $\partial f/\partial y$ yields the identical failure.
   Necessary condition violation: For f to be differentiable at (0,0), all partial derivatives must exist. Since $\partial f/\partial x$ and $\partial f/\partial y$ do not exist, f is not differentiable at (0, 0).
2. Geometric Picture: The graph of z = ∣x∣ + ∣y∣ is a pyramid with a sharp point (non-smooth vertex) at (0, 0). No single tangent plane can approximate the surface well in all directions, confirming non-differentiability.
3. The hypothesis of the standard theorem (existence and continuity of partials) is not satisfied, as the partials fail to exist at (0, 0). The theorem does not apply, and the failure is consistent with the non-differentiability.
4. Directional derivatives fail along certain directions. Even if we ignore partials and try a directional derivative along the line $u=(1,1)/\sqrt2$, we get $D_u f(0,0) = \lim_{h\to0} \frac{f\bigl(h/\sqrt2, h/\sqrt2\bigr)}{h} = \lim_{h\to0} \frac{|h/\sqrt2|+|h/\sqrt2|}{h} = \frac{+\sqrt2\|h|}{h} = \sqrt2 \times \operatorname{sgn}(h).$ Again, the two one‐sided limits ($h \leadsto 0^+$) $\sqrt2$ and ($h \leadsto 0^-$) $-\sqrt2$ differ, so even this single directional derivative fails to exist in a consistent way.