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Differentiable function

Definition. Let $f: \text{dom}(f) \to \mathbb{R}^m$ be a function, where its domain $\text{dom}(f) \subseteq \mathbb{R}^n$ is an open set. The function $f$ is said to be differentiable if it is differentiable at every point $\mathbf{x} \in \text{dom}(f)$.

Formally, $f$ is differentiable at a point $\mathbf{x}$ if there exists a linear map $L: \mathbb{R}^n \to \mathbb{R}^m$ (represented by an $m \times n$ matrix, the Jacobian matrix $Df(\mathbf{x})$), such that $\lim_{\mathbf{h} \to \mathbf{0}, \mathbf{h} \in \mathbb{R}^n} \frac{\| f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x}) - Df(\mathbf{x})\mathbf{h} \|}{\|\mathbf{h}\|} = 0$

Here, $\| \cdot \|$ denotes a norm (e.g., the Euclidean norm) on $\mathbb{R}^n$ and $\mathbb{R}^m$. This limit must hold for sequences $\mathbf{h} \to 0$ from any direction within $\mathbb{R}^n$.

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

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 continuous on ℝ² \ {(0, 0)}, so g is differentiable everywhere except possibly at the origin.
If g were differentiable at (0, 0): (i) the function has a minimum at (0, 0) (since g ≥ 0 and g(0,0) = 0); (ii) at a differentiable minimum, ∇g(0, 0) = (0, 0); (iii) Therefore, $D_u g(0, 0) = ∇g(0, 0) \cdot u = 0$ for all u.
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 (a sharp corner) with its vertex at the origin.
It has no tangent plane at the origin.
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 $D_{(1,0)} g = 1 ≠ 0$ ⊥, 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” with apex at the origin and sharp creases along the $x$ and $y$ axes. Partial derivatives ∂f/∂x and ∂f/∂y do not exist along the axes (sharp crease), so the linear-approximation definition fails.

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 certainly not differentiable at (0, 0).
Geometric Intuition: The graph $z = |x| + |y|$ is a pyramid. At the apex $(0,0,0)$, the surface has four distinct planar faces meeting at a point. For a function to be differentiable at a point, its graph must have a well-defined tangent plane. A tangent plane is a single flat surface that provides a good approximation to the graph in all directions.
At the pyramid’s apex, no single plane can simultaneously approximate all four faces well. Any candidate plane would only match one or two of the faces, leaving the others with a large error relative to the distance from the origin. This geometric intuition aligns perfectly with the algebraic failure of the partial derivatives.
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.
While the non-existence of partial derivatives is sufficient, examining directional derivatives provides deeper insight. 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 $D_u f(0,0) = \lim_{h\to0} \frac{f\bigl(h/\sqrt2, h/\sqrt2\bigr)}{h} = \lim_{h\to0} \frac{|h/\sqrt2|+|h/\sqrt2|}{h} = \frac{2\sqrt2\|h|}{h} = 2\sqrt2 \times \operatorname{sgn}(h).$
Again, the two one‐sided limits ($h \leadsto 0^+$) $2\sqrt2$ and ($h \leadsto 0^-$) $-2\sqrt2$ differ, so even this single directional derivative fails to exist in a consistent way.
This shows that the function is not even “linear” in a consistent way along a specific ray through the origin.

Directional Derivatives Exist but Not Linear. $f(x, y) = \begin{cases} \frac{x^2 y}{x^4 + y^2} & (x, y) \neq (0, 0) \\ 0 & (x, y) = (0, 0) \end{cases}$

All Directional Derivatives Exist at Origin. For u = (a, b) with a² + b² = 1: $D_u f(0, 0) = \lim_{t \to 0} \frac{f(ta, tb)}{t} = \lim_{t \to 0} \frac{(ta)^2(tb)}{t^4 a^4 + t^2 b^2} \cdot \frac{1}{t}$
If b ≠ 0: $= \lim_{t \to 0} \frac{t^3 a^2 b}{t(t^2 a^4 + b^2)} = \lim_{t \to 0} \frac{t^2 a^2 b}{t^2 a^4 + b^2} = \frac{0}{b^2} = 0$
If b = 0 (a² + b² = 1, so a = ±1): $D_{(\pm 1, 0)} f(0, 0) = \lim_{t \to 0} \frac{0}{t^4} = 0$.
All directional derivatives exist and equal 0!. This suggests that if f were differentiable at (0, 0), its derivative would be the zero linear map Df(0, 0) =(0, 0).
But f is NOT Differentiable at Origin.
If f were differentiable at (0, 0), with all directional derivatives = 0, then Df(0, 0) = (0, 0).
This would mean: $\lim_{(h,k) \to (0,0)} \frac{|f(h, k) - 0 - 0|}{\sqrt{h^2 + k^2}} = 0$
Along the curve y = x²: $\frac{f(h, h^2)}{\sqrt{h^2 + h^4}} = \frac{\frac{h^2 \cdot h^2}{h^4 + h^4}}{|h|\sqrt{1 + h^2}} = \frac{1/2}{|h|\sqrt{1 + h^2}}$ As $h \to 0$, this simplifies to: $\frac{1}{2|h} \to \infty$
The limit is unbounded, violating the definition of differentiability. Thus, f is not differentiable at (0, 0). ⊥

Moral: Having all directional derivatives is not enough for differentiability. The directional derivatives must combine to form a LINEAR map.
Along Axes: For u = (1, 0) or (0, 1), f is identically zero near the origin, so directional derivatives are zero.
Along y = $x^2$: The function plateaus at 1/2, creating a “ridge” that prevents linear approximation.

3D Plot Insight (trinket)

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

# Define the coordinates
x = np.linspace(-1, 1, 100)
y = np.linspace(-1, 1, 100)
X, Y = np.meshgrid(x, y)

# Avoid division by zero
Z = np.zeros_like(X)
mask = (X != 0) | (Y != 0)
# Define the function
Z[mask] = (X[mask]**2 * Y[mask]) / (X[mask]**4 + Y[mask]**2)

# Create the figure
fig = plt.figure(figsize=(10, 8))
ax = fig.add_subplot(111, projection='3d')

# Plot the surface
ax.plot_surface(X, Y, Z, cmap='viridis', alpha=0.8)

# Adding labels and title
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('f(x, y)')
plt.title('Surface Plot of f(x, y)')
plt.savefig('surface_plot.jpg')

Why Directional Derivatives Aren’t Enough

A common misconception is that if a function has directional derivatives in every direction at a point, then it must be differentiable there. This is false. Differentiability requires the existence of a single linear map that simultaneously approximates the function's behavior in all directions in a coherent, consistent way.

For a function $f: \mathbb{R}^n \to \mathbb{R}^m$ to be differentiable at a point $a$, we need a linear map $L: \mathbb{R}^n \to \mathbb{R}^m$ (the derivative $Df(a)$) such that for every direction vector $v$ (not necessarily unit), the directional derivative satisfies: $D_v f(a) = L(v) \quad \text{for all vectors } v \in \mathbb{R}^n$

This single requirement actually imposes two hidden algebraic conditions on the collection of directional derivatives:

Homogeneity (Scaling): $D_{cv} f(a) = c \cdot D_v f(a)$ for all scalars $c \in \mathbb{R}$ and all directions $v$.
Interpretation: If you double the length of the direction vector, the directional derivative should exactly double. This ensures the map scales properly.
Additivity: $D_{u+v} f(a) = D_u f(a) + D_v f(a)$ for all direction vectors $u, v \in \mathbb{R}^n$.
Interpretation: The rate of change in the direction of the sum of two vectors should equal the sum of the rates of change in each direction. This ensures the map respects vector addition.

If either of these properties fails, the function cannot be differentiable at that point —even if every single directional derivative exists.

For $f: \mathbb{R}^n \to \mathbb{R}^m$ to be differentiable at $a$, we require three things, each building on the last:

Existence. All directional derivatives $D_v f(a)$ exist for every vector $v \in \mathbb{R}^n$. This is the bare minimum; without it, differentiability is impossible.
Linearity. The map $v \mapsto D_v f(a)$ is linear. That is, it satisfies homogeneity and additivity. The derivative must be a linear transformation. If the directional derivatives don’t form a linear map, no linear approximation can exist.
Approximation. The error between $f(a+h)$ and its linear approximation $f(a) + Df(a) \cdot h$ vanishes faster than $\|h\|$ as $h \to 0$. This is the core of differentiability: the linear map actually approximates the function well near $a$.

All three requirements—existence, linearity, and approximation—are elegantly captured in the single limit definition of differentiability: $\lim_{h \to 0} \frac{\|f(a+h) - f(a) - Df(a) \cdot h\|}{\|h\|} = 0$

This definition simultaneously:

Assumes the existence of a candidate linear map $Df(a)$ (often built from partial derivatives if they exist and are continuous, but not necessarily).
Encodes the linearity requirement by using a linear map in the numerator.
Encodes the approximation requirement by demanding the error vanishes faster than $\|h\|$.

Differentiability in the Complex Plane

Definition. Let $D \subseteq \mathbb{C}$ be an open set and $f: D \to \mathbb{C}$ be a function. We say $f$ is **complex differentiable** at a point $z_0 \in D$ if the limit: $f'(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h}$ exists. The value of this limit is called the derivative of $f$ at $z_0$.. Otherwise, if this limit does not exist, then we say that f is not differentiable at $z_0$.

Key facts:

The reader should consider that h is a complex number, so it can approach 0 from infinitely many directions in the complex plane. The limit must exist and be the same regardless of the path h takes to zero.
If we write $h = re^{iθ}$ (polar form), the limit must be independent of: (i) the angle θ (direction of approach); (ii) the path (straight line, spiral, curved path, etc.)
Let $h = \Delta x + i\Delta y$. Approach along real axis ($\Delta y = 0, \Delta x \to 0$): The limit involves partial derivatives with respect to $x$.
Approach along imaginary axis ($\Delta x = 0, \Delta y \to 0$): The limit involves partial derivatives with respect to $y$, divided by $i$.
Equivalent Formulations:
Using $\Delta z = z - z_0$: $f'(z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}$.
Using differentials: $f(z_0 + h) = f(z_0) + f'(z_0)h + o(|h|) \quad \text{as } h \to 0$

Definition. A function $f$ is called analytic (or holomorphic) on an open set $D$ if it is complex differentiable at every point in $D$.

Definition. A function $f$ is called entire if it is analytic on the whole complex plane $\mathbb{C}$.

Examples of Differentiable Functions

The rules of calculation mirror real calculus because the limit definition is algebraically identical.

Constant Functions: Let $f(z) = c$ where $c \in \mathbb{C}$.

$f′(z)= \lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{c-c}{h} = lim_{h \to 0} \frac{0}{h} = lim_{h \to 0} 0 = 0.$ ∎ Constants are entire, with derivative 0.

Identity function: f(z) = z is differentiable everywhere with f’(z) = 1

$f′(z)= \lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{z+h-z}{h} = lim_{h \to 0} \frac{h}{h} = lim_{h \to 0} 1 = 1$ ∎

Power Functions (Integer $n \geq 1$): f(z) = zⁿ are differentiable everywhere.

$f′(z)= \lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{(z+h)^n-z^n}{h} =[\text{Using the binomial theorem}] \lim_{h \to 0} \frac{\sum_{k=0}^n {n \choose k}z^{n-k}h^k-z^n}{h} = \lim_{h \to 0} \sum_{k=1}^n {n \choose k}z^{n-k}h^{k-1}$

As $h \to 0$, all terms with h^k-1, k ≥ 2 vanish, leaving $f'(z)= {n \choose 1}z^{n-1} = nz^{n-1}$. Therefore, $f(z) = z^n$ is entire, with $f'(z) = nz^{n-1}$. ∎

Polynomials. Apply linearity (sum rule + constant multiple rule) and the power rule to each term: $P(z) = a_n z^n + \cdots + a_1 z + a_0$ is entire, with derivative: $P'(z) = na_n z^{n-1} + \cdots + a_1$.

Every polynomial is differentiable everywhere in ℂ. Examples: f(z) = 3z² - 2z + 5, f’(z) = 6z - 2; f(z) = z⁴ - iz³ + 2z, f’(z) = 4z³ - 3iz² + 2; f(z) = (z - 1)³, f’(z) = 3(z - 1)².

Rational Functions. Let $r(z) = \frac{p(z)}{q(z)}$ where p, q are polynomials. Derivative: $r'(z) = \frac{p'(z)q(z) - p(z)q'(z)}{[q(z)]^2}$. Domain of differentiability: ℂ \ {zeros of q}.

Example: $f(z) = \frac{3z}{z - i}$. Domain: D = ℂ \ {i}.

Derivative: $f'(z) = \frac{3(z-i) - 3z \cdot 1}{(z-i)^2} = \frac{3z - 3i - 3z}{(z-i)^2} = \frac{-3i}{(z-i)^2}$.

Let z₀ ∈ D. Then, $f'(z_0) = \lim_{h \to 0} \frac{\frac{3(z_0+h)}{z_0+h-i} - \frac{3z_0}{z_0-i}}{h}=[\text{Combining fraction}] \lim_{h \to 0} \frac{3(z_0+h)(z_0-i) - 3z_0(z_0+h-i)}{h(z_0+h-i)(z_0-i)}$

Expanding the numerator: $3(z_0+h)(z_0-i) - 3z_0(z_0+h-i) = 3z_0^2 - 3iz_0 + 3hz_0 - 3ih - 3z_0^2 - 3hz_0 + 3iz_0 = -3ih$

Therefore, $f'(z_0) = \lim_{h \to 0} \frac{-3ih}{h(z_0+h-i)(z_0-i)} = \lim_{h \to 0} \frac{-3i}{(z_0+h-i)(z_0-i)} = \frac{-3i}{(z_0-i)^2}$

This limit exists and we also know that $f'(z_0) = \frac{-3i}{(z_0-i)^2}$. Since z₀ is an arbitrary point in dom(f), f is differentiable.

Transcendental Functions

We define these functions via power series or Euler’s formula to extend them to $\mathbb{C}$.

Exponential Function. f(z) = e^z. Derivative: f’(z) = e^z

$f′(z)= \lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{e^{z+h}-e^z}{h} = e^z·\lim_{h \to 0} \frac{e^{h}-1}{h}$

Expand e^h as a Taylor series: $e^h = 1 + h + \frac{h^2}{2!} + \frac{h^3}{3!} + \cdots$. Therefore, $\frac{e^h-1}{h} = 1 + \frac{h}{2!} + \frac{h^2}{3!} + \cdots$

As $h \to 0$, all terms except the first vanish, leaving: $\lim_{h \to 0}\frac{e^h-1}{h} = 1, f′(z)= e^z·\lim_{h \to 0} \frac{e^{h}-1}{h} = e^z·1 = e^z$. Thus, $f'(z) = e^z \cdot 1 = e^z \blacksquare$.

Another approach. $e^z = e^{x+iy} = e^x(\cos y + i\sin y)$. $e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$.

Using the series definition and differentiating term-by-term (valid for power series): $\frac{d}{dz} \left(1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \right) = 0 + 1 + \frac{2z}{2!} + \frac{3z^2}{3!} + \cdots = 1 + z + \frac{z^2}{2!} + \cdots = e^z$