Differentiability and Continuous Differentiability

If you’re going through hell, keep on going, Winston Churchill

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 $\lim_{\mathbf{h} \to \mathbf{0}, \mathbf{h} \in \mathbb{R}^n} \frac{\| f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x}) - Df(\mathbf{x})\mathbf{h} \|}{\|\mathbf{h}\|} = 0$

The Connection to Real Differentiability

Any complex function f: ℂ → ℂ can be written as f(z) = u(x, y) + iv(x, y), which defines a map F: ℝ² → ℝ²: $F(x, y) = (u(x, y), v(x, y))$.

The Jacobian matrix of F is: $J_F = \begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix}$

Theorem. A function f(z) = u(x,y) + iv(x,y) is complex differentiable at z₀ = x₀ + iy₀ if and only if:

1. f is real-differentiable at (x₀, y₀) (as F: ℝ² → ℝ²)
2. The Cauchy-Riemann equations hold at (x₀, y₀): $u_x = v_y \quad \text{and} \quad u_y = -v_x$

When these hold, the complex derivative is: $f'(z_0) = u_x(x_0, y_0) + i v_x(x_0, y_0) = v_y(x_0, y_0) - i u_y(x_0, y_0)$

The Jacobian has the special form $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ if and only if the Cauchy-Riemann equations hold. This matrix represents multiplication by the complex number f’(z) = a + ib!

Function	Jacobian	Cauchy Riemman holds?	Complex differentiable
f(z) = z²	$\begin{pmatrix} 2x & -2y \\ 2y & 2x \end{pmatrix}$	✓	✓ everywhere
f(z) = eᶻ	$\begin{pmatrix} e^x\cos y & -e^x\sin y \\ e^x\sin y & e^x\cos y \end{pmatrix}$	✓	✓ everywhere
f(z) = z̄	$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$	✗	✗ anywhere
f(z) = |z|²	$\begin{pmatrix} 2x & 2y \\ 0 & 0 \end{pmatrix}$	only at (0,0)	only at z = 0
f(z) = Re(z)	$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$	✗	✗ anywhere

Definition. A function f: D → ℂ is holomorphic on D if f is complex differentiable at every point of the open set D.

Definition. A function f: ℂ → ℂ is entire if it is holomorphic on all of ℂ.

Examples of entire functions: all polynomials, eᶻ, sin(z), cos(z), sinh(z), cosh(z).

Non-examples: 1/z (not defined at 0); tan(z) (poles at π/2 + kπ); log(z) (branch cut)

Rules of differentiation for complex functions

Complex differentiation mirrors real differentiation formally, but the underlying requirements are much stricter. These rules apply to holomorphic (complex differentiable) functions — those satisfying the Cauchy-Riemann equations.

1. Linearity Rule: If f(z) and g(z) are differentiable functions, and a and b are complex constants, then the function af(z) + bg(z) is also differentiable, and its derivative is given by: $\frac{d}{dz}$[af(z) + bg(z)] = af’(z) + bg’(z).
   Proof via Limit Definition:
   $\frac{d}{dz}[f(z) + g(z)] = \lim_{h \to 0} \frac{f(z+h)+g(z+h)-f(z)-g(z)}{h} = \lim_{h \to 0} (\frac{f(z+h)-f(z)}{h} + \frac{g(z+h)-g(z)}{h}) = f'(z) + g'(z).$
   $\frac{d}{dz}[cf(z)] = \lim_{h \to 0} \frac{cf(z+h)-cf(z)}{h} = c·\lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = c·f'(z).$

2. Product Rule: If f(z) and g(z) are differentiable functions, then their product f(z)g(z) is also differentiable, and its derivative is given by: $\frac{d}{dz}$[f(z) · g(z)] = f’(z)g(z) + f(z)g’ (z).
   

   $$\begin{aligned} &\frac{f(z+h)g(z+h) - f(z)g(z)}{h} \\ &= \frac{f(z+h)g(z+h) - f(z+h)g(z) + f(z+h)g(z) - f(z)g(z)}{h} \\ &= f(z+h)\frac{g(z+h)-g(z)}{h} + g(z)\frac{f(z+h)-f(z)}{h} \end{aligned}$$

   
   As $h \to 0$: $f(z+h) \to f(z)$ by continuity of differentiable functions, yielding $f(z)g'(z) + g(z)f'(z)$.

3. Quotient Rule: If f(z) and g(z) are differentiable functions, and g(z) ≠ 0, then their quotient $\frac{f(z)}{g(z)}$ is also differentiable, and its derivative is given by: $\frac{d}{dz} \frac{f(z)}{g(z)} = \frac{f'(z)g(z) - f(z)g'(z)}{(g(z)^2)}$
   Apply the product rule to $f(z) \cdot [g(z)]^{-1}$, using the chain rule for the reciprocal.

4. Chain Rule: If f(z) is a differentiable function at z and g(w) is a differentiable function of w, then the composition g(f(z)) is also differentiable, and its derivative is given by: $\frac{d}{dz}$[g(f(z))] = g’(f(z))·f’(z).
   It underlies change of variables in contour integrals and differentiation of inverse functions.

5. Power Rule: If n is an integer, then the function zⁿ is differentiable, and its derivative is given by: $\frac{d}{dz}[z^n] = nz^{n-1}$.
   n ≥ 0, the domain is the complex plane. It is an entire function. Proof by induction using the product rule, or directly $\frac{(z+h)^n-z^n}{h} = \frac{\sum_{k=0}^n {n \choose k}z^{n-k}h^k-z^n}{h} = nz^{n-1}+O(h) \to nz^{n-1}$
   n < 0, Domain = $\mathbb{C} \setminus \{ 0 \}$. Pole at origin, derivative has higher order pole.

6. Exponential Rule: The exponential function e^z is entire, and its derivative is given by: $\frac{d}{dz}[e^z] = e^{z}$. In other words, $e^z$ is its own derivative.
   $\frac{e^{z+h} - e^z}{h} = e^z \cdot \frac{e^h - 1}{h} = e^z \cdot \frac{1 + h + \frac{h^2}{2!} + \cdots - 1}{h} = e^z\left(1 + \frac{h}{2!} + \cdots\right) \to e^z$

7. Logarithmic Rule: On any simply connected domain not containing 0 with the principal branch of the logarithmic function log(z) selected, log(z) is differentiable, and its derivative is given by: $\frac{d}{dz}[log(z)] = \frac{1}{z}$.
   Let $w = \log z$, so $z = e^w$. The complex logarithm w = log(z) is defined as the inverse of the exponential function $z = e^w$. Since the derivative of the exponential function is never zero: $\frac{dz}{dw}= \frac{d}{dw}e^w = e^w \ne 0$.
   The Inverse Function Theorem for holomorphic functions guarantees that w = log(z) is holomorphic at any point where $e^w$ has a non-zero derivate.
   By differentiating both sides of the identity $e^{w(z)} = z, \frac{d}{dz}(e^w) = \frac{d}{dz}(z), e^w\cdot \frac{dw}{dz} =1 \implies[\text{Solve for the derivative}] \frac{dw}{dz} = \frac{1}{e^w}= \frac{1}{z}$.
   $\log z$ is not a single-valued holomorphic function on $\mathbb{C} \setminus \{0\}$. We must (i) choose a branch cut (e.g., negative real axis); (2) define the principal branch $\text{Log } z = \ln|z| + i\text{Arg } z$ with $-\pi < \text{Arg } z \leq \pi$
   Conclusion: The derivative is valid everywhere except on the branch cut and at the branch point (z = 0).

8. Trigonometric Rules: The complex sine and cosine are entire with $\frac{d}{dz}[\sin z] = \cos z, \qquad \frac{d}{dz}[\cos z] = -\sin z$
   Proof via Euler’s formula: $\sin z = \frac{e^{iz} - e^{-iz}}{2i} \implies \frac{d}{dz}\sin z = \frac{ie^{iz} + ie^{-iz}}{2i} = \frac{e^{iz} + e^{-iz}}{2} = \cos z$
   Complex behavior: Unlike their real counterparts, $\sin z$ and $\cos z$ are unbounded on $\mathbb{C}$ (e.g., $\cos(iy) = \cosh y \to \infty$ as $y \to \infty$).

Summary Table

Rule	Formula	Domain Requirements
Linearity	$(af + bg)' = af' + bg'$	$f, g$ differentiable at $z$
Product	$(fg)' = f'g + fg'$	$f, g$ differentiable at $z$
Quotient	$(f/g)' = (f'g - fg')/g^2$	$g(z) \neq 0$
Chain	$(g \circ f)' = (g' \circ f) \cdot f'$	$f$ at $z$, $g$ at $f(z)$
Power	$(z^n)' = nz^{n-1}$	$n \in \mathbb{Z}$; $z \neq 0$ if $n < 0$
Exponential	$(e^z)' = e^z$	Entire
Logarithm	$(\log z)' = 1/z$	Branch cut required; $z \neq 0$
Trigonometric	$(\sin z)' = \cos z$, $(\cos z)' = -\sin z$	Entire

Continuous differentiability

In calculus, differentiability tells us that a function has a well-defined rate of change at a point. But differentiability alone can be surprisingly weak —a function can be differentiable everywhere yet have a derivative that jumps wildly.

Continuous differentiability (being $C^1$) is the natural strengthening: not only does the derivative exist, but it varies smoothly. This condition guarantees predictable, non-erratic behavior; provides the “right” notion of smoothness for most theorems and applications.

In single-variable calculus, a function f: ℝ→ℝ is defined as continuously differentiable ($f \in C^1(a, b)$) on an open interval (a,b) if its derivative f′(x) exists for every point x∈(a,b) and this derivative function f’:$(a, b) \to \mathbb{R}$ is itself a continuous function. Geometrically, it means that not only does the function have a well-defined tangent at every point within the interval, but the slope of this tangent changes smoothly as you move along the curve, without any abrupt jumps, corners, or breaks.

Many common elementary functions are continuously differentiable ($C^1$) on their entire domains: polynomial functions (x², 8x⁴ + 3x² -7x + 2), trigonometrical and inverse trigonometrical functions (sin(x) and cos(x) on ℝ, tan(x) on ℝ \ {π/2 + kπ}, arcsin(x) and arccos(x) on (−1, 1), arctan(x) on ℝ), power and logarithmic functions (e^x for all real x, a^x for any base a > 0, ln(x) on (0, ∞), $log_a(x)=\frac{ln(x)}{ln(a)}, \forall x \gt 0$), hyperbolic functions (sinh(x), cosh(x), tanh(x) on ℝ) etc.

The requirement that the interval (a, b) is open ensures that we are dealing with points where the function is well defined in a neighborhood, allowing for the proper definition of limits and continuity of the derivative.

The derivative of a function, f′(x), represents its instantaneous rate of change. If f′(x) is itself a continuous function, it means that this rate of change does not jump suddenly from one value to another, but varies continuously. Therefore, small changes in x lead to small changes in f′(x).

This is a stronger condition than mere differentiability. A function can be differentiable at a point, meaning it has a defined rate of change there, but this rate of change might not be a continuous function, e.g.,

However, $f(x) = \begin{cases} x^2 \sin(1/x) & x \neq 0 \\ 0 & x = 0 \end{cases}$ is differentiable everywhere, but not $C^1$.
At x ≠ 0: Using standard rules: $f'(x) = 2x \sin(1/x) + x^2 \cdot \cos(1/x) \cdot (-1/x^2) = 2x \sin(1/x) - \cos(1/x)$
At x = 0: Using the limit definition: $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin(1/h)}{h} = \lim_{h \to 0} h \sin(1/h)$
Since $0 \leq |h \sin(1/h)| \leq |h| \to 0$ (squeeze theorem), we have f’(0) = 0.
As x → 0: The term 2x sin(1/x) → 0 (squeeze theorem), but -cos(1/x) oscillates wildly between -1 and +1, so f’ is not continuous at 0.

Definition (Multivariable Case). Let $\mathbb{f}:U \subseteq \mathbb{R}^n \to \mathbb{R}$ be a function defined on an open set. If all the partial derivatives $\frac{\partial f}{\partial x_1}, \ldots, \frac{\partial f}{\partial x_n}$ exist and are continuous on U (at every point x ∈ U), then f is said to be continuously differentiable on U. If the domain S = dom(f) itself is an open set and f is continuously differentiable on S, then f is said to be continuously differentiable.

Specifying $U \subseteq S$ as an open set is crucial, as differentiability is typically defined in open sets where points have neighborhoods entirely contained in the domain, allowing partial derivatives to be evaluated in all directions.

A function is continuously differentiable if its derivative (in single-variable calculus) or all its partial derivatives (in multivariable calculus) exist and are themselves continuous functions. This means the function’s rate of change varies smoothly, without abrupt jumps. This concept is crucial because it ensures predictable behavior (they can’t suddenly change slope; it rules out “corners” and makes graphs visually and analytically well-behaved), enables key theorems and results, etc.

Theorem (C¹ implies Differentiable). If $U \subset \mathbb{R}^n$ is open and $f \in C^1(U, \mathbb{R}^m)$, then $f$ is (totally) differentiable at every point of $U$.

This means:

1. The total derivative (Jacobian matrix) $Df(a)$ exists
2. The linear approximation is valid: $f(x) \approx f(a) + Df(a)(x - a)$
3. The error vanishes faster than the displacement: $\lim_{h \to 0} \frac{\|f(a+h) - f(a) - Df(a)h\|}{\|h\|} = 0$

Proof:

Step 1. Setup and Reduction to the Scalar Case.

Let $U \subset \mathbb{R}^n$ be open and $f \in C^1(U, \mathbb{R}^m)$.

Fix a point $a \in U$. We want to show that $f$ is differentiable at $a$, i.e., there exists a linear map $Df(a): \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that: $f(a+h) = f(a) + Df(a)h + o(\|h\|) \quad \text{as } h \rightarrow 0$

We can express $f$ in terms of its scalar components as $f = (f^1, \dots, f^m)$, where each $f^k: U \rightarrow \mathbb{R}$. If we can prove the theorem for scalar-valued functions, then for each $k \in \{1, \dots, m\}$ we will have: $f^k(a+h) = f^k(a) + Df^k(a)h + o(\|h\|)$

Stacking these scalar equations component-wise yields the general vector version. The linear map $Df(a)$ will simply be the $m \times n$ Jacobian matrix of partial derivatives at $a$, acting on the vector $h$. Because limits in $\mathbb{R}^m$ can be evaluated component by component, it suffices to prove the result for $m=1$.

Step 2. Defining the Linear Map for m = 1. From now on, assume $f: U \rightarrow \mathbb{R}$ is a scalar function with $f \in C^1(U)$.

Let $a = (a_1, \dots, a_n)$. Since $f \in C^1$, all partial derivatives $\frac{\partial f}{\partial x_j}$ exist and are continuous on $U$. We define our candidate for the total derivative as the linear map $L: \mathbb{R}^n \rightarrow \mathbb{R}$ given by the dot product of the gradient at $a$ with $h$: $L(h) = \sum_{j=1}^n \frac{\partial f}{\partial x_j}(a)h_j, \quad \text{where } h = (h_1, \dots, h_n)$

Our goal is to show that $\lim_{h \to 0} \frac{|f(a+h) - f(a) - L(h)|}{\|h\|} = 0$.

Step 3. The Telescoping Sum (Moving along coordinate axes).

Since $U$ is open, for a sufficiently small vector $h$, the entire box containing $a$ and $a+h$ lies within $U$. To evaluate $f(a+h) - f(a)$, we move from $a$ to $a+h$ one coordinate at a time.

Define intermediate points $p_0, p_1, \dots, p_n$ such that we change coordinates one by one: $p_0 = a, p_1 = a + (h_1, 0, \dots, 0), p_2 = a + (h_1, h_2, 0, \dots, 0), \cdots, p_n = a + (h_1, h_2, \dots, h_n) = a + h$

We can write the total change in $f$ as a telescoping sum: $f(a+h) - f(a) = \sum_{j=1}^n [f(p_j) - f(p_{j-1})]$

Step 4: Applying the Mean Value Theorem.

Notice that the points $p_j$ and $p_{j-1}$ differ only in their $j$-th coordinate (by the amount $h_j$). Because $f$ is differentiable with respect to each variable, we can apply the one variable Mean Value Theorem to $f$ along the line segment connecting $p_{j-1}$ and $p_j$.

There exists some point $c_j$ on the line segment between $p_{j-1}$ and $p_j$ such that: $f(p_j) - f(p_{j-1}) = \frac{\partial f}{\partial x_j}(c_j)h_j$

Substituting this into our telescoping sum gives: $f(a+h) - f(a) = \sum_{j=1}^n \frac{\partial f}{\partial x_j}(c_j)h_j$

Step 5: Bounding the Error using Continuity.

Now, let’s look at the approximation error by comparing this to our linear map $L(h)$:

$$f(a+h) - f(a) - L(h) = \sum_{j=1}^n \left[ \frac{\partial f}{\partial x_j}(c_j) - \frac{\partial f}{\partial x_j}(a) \right] h_j$$

Take the absolute value and apply the triangle inequality: $|f(a+h) - f(a) - L(h)| \le \sum_{j=1}^n \left| \frac{\partial f}{\partial x_j}(c_j) - \frac{\partial f}{\partial x_j}(a) \right| |h_j|$

Since $|h_j| \le \|h\|$ for all $j$, we can divide both sides by $\|h\|$ to isolate the limit term: $\frac{|f(a+h) - f(a) - L(h)|}{\|h\|} \le \sum_{j=1}^n \left| \frac{\partial f}{\partial x_j}(c_j) - \frac{\partial f}{\partial x_j}(a) \right|$

Now, consider what happens as $h \rightarrow 0$. As the vector $h$ shrinks to zero, the point $p_n = a+h$ approaches $a$. Consequently, all intermediate points $p_j$ approach $a$, and therefore the points $c_j$ (which are themselves squeezed between $p_{j-1}$ and $p_j$) must also approach $a$.

Because $f \in C^1$, the partial derivatives are continuous at $a$. Therefore, as $c_j \rightarrow a$, we have: $\lim_{h \to 0} \frac{\partial f}{\partial x_j}(c_j) = \frac{\partial f}{\partial x_j}(a)$

This means that for every $j$, the term $\left| \frac{\partial f}{\partial x_j}(c_j) - \frac{\partial f}{\partial x_j}(a) \right|$ goes to $0$.

Therefore, the entire sum bounds to $0$, proving that: $\lim_{h \to 0} \frac{|f(a+h) - f(a) - L(h)|}{\|h\|} = 0$. This establishes that $f$ is differentiable at $a$, completing the proof. $\blacksquare$

Examples:

• Polinomials: $x^³ –5x + 2$ is $C^\infty$ —its derivative 3x²–5 is continuous everywhere. f(x, y) = x³ - 3xy + y², Partial derivatives: $\frac{\partial f}{\partial x} = 3x^2 - 3y, \quad \frac{\partial f}{\partial y} = -3x + 2y$; Both are polynomials in x, y, hence continuous everywhere, $f \in C^{\infty}(ℝ²).$
• Exponential Combinations: f(x, y) = $e^{x+y} + xy$. Partial derivatives: $\frac{\partial f}{\partial x} = e^{x+y} + y, \quad \frac{\partial f}{\partial y} = e^{x+y} + x$. Both of these partial derivatives are continuous functions because they are composed of products, compositions, and sums of continuous functions (polynomial, trigonometric, and exponential functions), $f \in C^{\infty}(ℝ²)$.
• Trigonometric Functions, f(x, y) = sin(xy), Partial derivatives: $\frac{\partial f}{\partial x} = y \cos(xy), \quad \frac{\partial f}{\partial y} = x \cos(xy)$. Both are continuous on ℝ², $f \in C^{\infty}(ℝ²)$.
• Rational Functions, $f(x, y) = \frac{x}{x^2 + y^2 + 1}$. Domain: ℝ² (denominator is always ≥ 1). Partial derivatives: $\frac{\partial f}{\partial x} = \frac{(x^2 + y^2 + 1) - x(2x)}{(x^2 + y^2 + 1)^2} = \frac{y^2 - x^2 + 1}{(x^2 + y^2 + 1)^2}, \frac{\partial f}{\partial y} = \frac{-2xy}{(x^2 + y^2 + 1)^2}$. Both are continuous on ℝ² (rational functions with non-vanishing denominators), $f \in C^{\infty}(ℝ²)$.

This definition is really key in multivariable calculus since the existence of partial derivatives alone is not sufficient to guarantee that a function is differentiable in the full sense. However, continuous partial derivatives ensure the function is differentiable in the full sense (i.e., has a linear approximation via the Jacobian).

• $f(x, y) = \begin{cases} \frac{xy}{x^2 + y^2} & (x, y) \neq (0, 0) \\ 0 & (x, y) = (0, 0) \end{cases}$

Partial derivatives at origin: $\frac{\partial f}{\partial x}(0, 0) = \lim_{h \to 0} \frac{f(h, 0) - f(0, 0)}{h} = \lim_{h \to 0} \frac{0}{h} = 0$. Similarly, $\frac{\partial f}{\partial y}(0, 0) = 0$.

However, f is NOT differentiable at (0, 0).

If f were differentiable with Df(0,0) = (0, 0), then: $\lim_{(h,k) \to (0,0)} \frac{|f(h, k) - 0 - 0|}{\sqrt{h^2 + k^2}} = 0$

Along h = k = t: $\frac{|f(t, t)|}{\sqrt{2}|t|} = \frac{|t^2/(2t^2)|}{\sqrt{2}|t|} = \frac{1/2}{\sqrt{2}|t|} \to \infty$. The limit is not 0, hence f is not differentiable at (0, 0).

Why did this happen? The partial derivatives exist at (0, 0), but they are NOT continuous there.

Definition. Let F: U → ℝᵐ where U ⊆ ℝⁿ is open, with component functions F = (F₁, F₂, …, Fₘ) where each $F_i: U \to \mathbb{R}$. We say that $\mathbf{F}$ is continuously differentiable on U, denoted $\mathbf{F} \in C^1(U, \mathbb{R}^m)$ or simply $\mathbf{F} \in C^1(U)$, if and only if each of its component functions $F_i$ has continuous first-order partial derivatives on $U$.

Equivalently: : For each $i = 1, \dots, m$ and $j = 1, \dots, n$, the partial derivative $\frac{\partial F_i}{\partial x_j}(\mathbf{x})$ exists for all $\mathbf{x} \in U$, and the resulting function $\frac{\partial F_i}{\partial x_j}: U \to \mathbb{R}$ is continuous.

Key Properties and Theorems

• If $\mathbf{F} \in C^1(U)$, then at every point $\mathbf{x} \in U$, the total derivative (or Jacobian matrix) $D\mathbf{F}(\mathbf{x})$ exists. It is an $m \times n$ matrix whose entries are the partial derivatives, $[DF(x)]_{ij} = \frac{\partial F_i}{\partial x_j}(x)$. $J_F(x) = \begin{pmatrix} \frac{\partial F_1}{\partial x_1} & \cdots & \frac{\partial F_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial F_m}{\partial x_1} & \cdots & \frac{\partial F_m}{\partial x_n} \end{pmatrix}$ has entries that are continuous functions of x.
• $C^1$ (all partial derivatives exist and are continuous) implies Differentiability. A function being $C^1$ is a stronger condition than simply being differentiable.
• Algebraic Properties. If $f, g \in C^1(U)$ , then, $af + bg \in C^1(U)$ for any constants $a, b \in \mathbb{R}$, $f \cdot g \in C^1(U)$. If $g(\mathbf{x}) \neq 0$ for all $\mathbf{x} \in U$, then the quotient $f/g \in C^1(U)$.
• The Chain Rule. Let $U \subseteq \mathbb{R}^n$ and $V \subseteq \mathbb{R}^m$ be open sets. Suppose $\mathbf{F}: U \to V$ is a $C^1$ map and $\mathbf{G}: V \to \mathbb{R}^p$ is a $C^1$ map. Then the composition $\mathbf{G} \circ \mathbf{F}: U \to \mathbb{R}^p$ is also in $C^1(U)$.

Moreover, the derivative (Jacobian) at a point $\mathbf{x} \in U$ is the matrix product of the individual derivatives: $D(\mathbf{G}\circ \mathbf{F})(x)=D\mathbf{G}(\mathbf{F}(x))\cdot D\mathbf{F}(x).$. The continuity of the partial derivatives of $\mathbf{F}$ and $\mathbf{G}$ ensures that the resulting composition is continuously differentiable.

• Inverse Function Theorem. Let $\mathbf{F}: U \to \mathbb{R}^n$ be a $C^1$ map (here, $m=n$, so the Jacobian is a square matrix). If at some point $\mathbf{a} \in U$, the Jacobian matrix $D\mathbf{F}(\mathbf{a})$ is invertible (i.e., $\det D\mathbf{F}(\mathbf{a}) \neq 0$), then:
  (i) Local Invertibility. There exists an open neighborhood $U_0 \subseteq U$ of $\mathbf{a}$ and an open neighborhood $V_0$ of $\mathbf{b} = \mathbf{F}(\mathbf{a})$ such that $\mathbf{F}: U_0 \to V_0$ is bijective.
  (ii) $C^1$ Inverse. Its inverse function $\mathbf{F}^{-1}: V_0 \to U_0$ is also of class $C^1$.
  The derivative of the inverse is given by $D\mathbf{F}^{-1}(\mathbf{y}) = [D\mathbf{F}(\mathbf{F}^{-1}(\mathbf{y}))]^{-1}$.
• Equality of Mixed Partials (Clairaut’s Theorem).If a function $f: U \to \mathbb{R}$ is of class $C^2(U)$ (meaning all its second-order partial derivatives exist and are continuous), then the order of differentiation does not matter: $\frac{\partial^2 f}{\partial x \partial y}(x, y) = \frac{\partial^2 f}{\partial y \partial x}(x, y) \forall (x, y) \in U$.
• Higher-Order Classes ($C^k$ and $C^\infty$). The concept of $C^1$ can be extended naturally. A function is $C^k(U)$ if all partial derivatives of order up to and including $k$ exist and are continuous on $U$. A function is $C^\infty(U)$ or smooth if it has continuous partial derivatives of all orders on $U$.
• There is a hierarchical relationship: $C^\infty \subset \dots \subset C^{k+1} \subset C^k \subset \dots \subset C^1 \subset C^0$, where $C^0$ denotes the set of continuous functions.

Enhanced Explanation of Multivariable Differentiability

A function $f: \mathbb{R}^n \to \mathbb{R}$ is differentiable at a point $x_0 \in U$ (where $U \subseteq \mathbb{R}^n$ is open) if there exists a linear map (represented by the Jacobian matrix $Df(x_0)$) such that: $f(x_0 + h) = f(x_0) + Df(x_0)·h + o(||h||)$ where $h = (h_1, \cdots, h_n)$ and o(||h||) is a remainder term that vanishes faster than o||h|| as $h \to 0$, i.e., satisfying $\lim_{h \to 0}\frac{o(||h||)}{||h||} = 0$

Continuity of partial derivatives ensures the function’s behavior is smooth enough for the linear approximation to hold, that is, the linear approximation $f(x_0) + Df(x_0)·h$ becomes arbitrarily accurate as $h \to 0$.

The existence of partial derivatives at $x_0$ is necessary but not sufficient for differentiability. For f to be differentiable at $x_0$, the partial derivatives must be continuous in a neighborhood of $x_0$.

It ensures ensures that the rate of change of f in all coordinate directions varies smoothly. The partial derivatives $\frac{\partial f}{\partial x_i}$ represent the instantaneous slope in each direction, and their continuity prevents abrupt jumps or discontinuities.

Geometric Interpretation: Tangent Hyperplane

In $\mathbb{R}^n$, the tangent hyperplane at $x_0$ is defined as $T_{x_0} = f(x_0) + Df(x_0)\cdot (x - x_0)$, where $x = (x_1, \cdots, x_n)$. The Jacobian matrix $Df(x_0)$ encodes the slopes of f in each coordinate direction, and the tangent hyperplane is the best linear approximation to f near $x_0$.

Example: f(x, y) = x² + y². Partial derivatives: $\frac{\partial f}{\partial x} = 2x, \frac{\partial f}{\partial y} = 2y$. Both 2x and 2y are continuous everywhere (as they are linear functions), so f is continuously differentiable on $\mathbb{R}^2$, an open set. The Jacobian is $\mathbb{Df}(1, 1) = [2, 2]$.

Note: Continuous differentiability ensures a well-defined tangent plane. In higher dimension (ℝⁿ), the tangent plane becomes a tangent hyperplane, defined similarly using the Jacobian.

In ℝ², for a function f(x, y), the tangent plane at a point $(x_0, y_0)$ is: $z = f(x_0, y_0) + \frac{\partial f}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial f}{\partial y}(x_0, y_0)(y - y_0).$ In our example, f(1, 1) = 1² + 1² = 2, and the partial derivatives are $\frac{\partial f}{\partial x}(x_0, y_0) = 2·x|_{(1, 1)} = 2, \frac{\partial f}{\partial y}(x_0, y_0) = 2·y|_{(1, 1)} = 2$. The tangent plane is: z = 2 + 2(x - 1) + 2(y - 1) = 2x + 2y - 2.