Multivariable analysis §

From one dimension to many §

The ε–δ definition is identical in form to the one-dimensional case (see Limits and continuity); the only change is that $|x - a|$ becomes the Euclidean norm $\|\mathbf{x} - \mathbf{a}\| = \sqrt{(x_1 - a_1)^2 + \cdots + (x_n - a_n)^2}$. The key new feature is that there are infinitely many paths to approach $\mathbf{a}$: along axes, along lines, along curves. A limit exists only if $f$ approaches $L$ along every path.

Worked example: a limit that does not exist §

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

• Along the $x$-axis ($y = 0$): $f(x, 0) = 0 \to 0$.
• Along the line $y = x$: $f(x, x) = \frac{x^2}{2x^2} = \frac{1}{2} \to \frac{1}{2}$.

Since the two paths give different limits, $\lim_{(x,y) \to (0,0)} f(x,y)$ does not exist.

Continuity of vector-valued functions §

A map $\mathbf{f}: U \subseteq \mathbb{R}^n \to \mathbb{R}^m$ is continuous at $\mathbf{a}$ if and only if each component $f_i$ is continuous at $\mathbf{a}$. This reduces the multivariable question to $m$ one-dimensional questions.

From one derivative to many §

In one variable, the derivative $f'(a)$ captures the rate of change in the only available direction. In $\mathbb{R}^n$, there are infinitely many directions. The partial derivatives $\partial f / \partial x_i$ measure the rate of change along the coordinate axes. The gradient collects all partial derivatives into a single vector that encodes the rate of change in every direction via the dot product.

Worked example §

Let $f(x, y) = x^2 y + \sin(xy)$. Then

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

At $(1, 0)$: $\nabla f(1, 0) = (0, 1)$. The steepest ascent direction is $(0, 1)$ (the positive $y$-direction), and the steepest descent direction is $(0, -1)$.

Differentiability versus partial derivatives §

Having all partial derivatives exist at a point is not sufficient for differentiability. The function $f(x,y) = \frac{xy}{\sqrt{x^2+y^2}}$ (with $f(0,0)=0$) has $\partial f/\partial x = \partial f/\partial y = 0$ at $(0,0)$, but is not differentiable there (the limit $\frac{f(h,0) - f(0,0)}{\|(h,0)\|}$ does not give a linear map). Differentiability requires that the linear approximation (given by the gradient) approximates $f$ to first order in all directions simultaneously.

The Jacobian as a linear map §

The Jacobian matrix represents the best linear approximation to $\mathbf{f}$ near $\mathbf{a}$: $\mathbf{f}(\mathbf{a} + \mathbf{h}) \approx \mathbf{f}(\mathbf{a}) + D\mathbf{f}(\mathbf{a})\,\mathbf{h}$. In the special case $m = n = 1$, the Jacobian is the $1 \times 1$ matrix $(f'(a))$, recovering the one-variable derivative.

The chain rule in matrix form §

The chain rule says that the derivative of a composition is the product of the derivatives. In one variable: $(g \circ f)'(a) = g'(f(a)) \cdot f'(a)$. In several variables, the scalars become matrices and the product is matrix multiplication. The dimensions work out: $D\mathbf{g}$ is $p \times m$ and $D\mathbf{f}$ is $m \times n$, so the product is $p \times n$.

Worked example: chain rule §

Let $f(x, y) = x^2 + y^2$ and $\mathbf{g}(t) = (\cos t, \sin t)$. Then $(f \circ \mathbf{g})(t) = \cos^2 t + \sin^2 t = 1$.

By the chain rule: $D(f \circ \mathbf{g})(t) = Df(\mathbf{g}(t)) \cdot D\mathbf{g}(t)$.

$Df(x,y) = (2x, 2y)$ (a $1 \times 2$ row), $D\mathbf{g}(t) = \begin{pmatrix} -\sin t \\ \cos t \end{pmatrix}$ (a $2 \times 1$ column).

$D(f \circ \mathbf{g})(t) = (2\cos t, 2\sin t) \cdot \begin{pmatrix} -\sin t \\ \cos t \end{pmatrix} = -2\cos t \sin t + 2\sin t \cos t = 0$. ✓

The inverse function theorem §

The condition $\det D\mathbf{f}(\mathbf{a}) \neq 0$ means the linear approximation is invertible (no direction is collapsed). This is the multivariable generalization of the one-variable condition $f'(a) \neq 0$ (see Differentiation). It guarantees that $\mathbf{f}$ is a local diffeomorphism: it maps small neighborhoods to small neighborhoods with a smooth inverse.

Why the multivariable MVT is different §

In one variable, the MVT gives a point $c$ where the derivative equals the average slope. In several variables, the gradient at a single point generally cannot capture the total change (the gradient is a vector, and the change is a vector in a different direction). The multivariable MVT for scalar-valued functions says the change equals the gradient at some intermediate point dotted with the displacement — a single scalar equation, not a vector equation.

For vector-valued functions, the MVT takes the form $\mathbf{f}(b) - \mathbf{f}(a) = D\mathbf{f}(c)(b-a)$, which is a vector equation: the change equals the Jacobian at some point applied to the displacement.

Application: Lipschitz continuity §

If $\|\nabla f(\mathbf{x})\| \leq M$ on a convex set $U$, then by the MVT,

$|f(\mathbf{b}) - f(\mathbf{a})| \leq M \|\mathbf{b} - \mathbf{a}\|.$

This is the standard proof that a function with bounded gradient is Lipschitz — the workhorse estimate in optimization and PDE theory.

From Riemann sums in two dimensions §

The double integral is defined by partitioning the rectangle into small sub-rectangles, summing $f(x_i^*, y_j^*) \Delta x \Delta y$, and taking the limit as the mesh goes to zero. Fubini’s theorem (a consequence of the Riemann/Lebesgue integration theory; see Integration) says that for “nice” functions, the double integral equals either iterated integral.

The change of variables formula §

The Jacobian determinant $|\det D\mathbf{T}|$ measures how $\mathbf{T}$ distorts area (in 2D) or volume (in 3D). A small square of area $\Delta A$ in the $\mathbf{x}$-plane is mapped to a parallelogram of area approximately $|\det D\mathbf{T}| \, \Delta A$ in the $\mathbf{y}$-plane. The change of variables formula accounts for this distortion.

Worked example: polar coordinates §

The map $\mathbf{T}(r, \theta) = (r\cos\theta, r\sin\theta)$ has Jacobian

$D\mathbf{T} = \begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix}, \qquad \det D\mathbf{T} = r.$

So $dx\, dy = r\, dr\, d\theta$. This is why the area element in polar coordinates is $r\, dr\, d\theta$ rather than $dr\, d\theta$.

Example: Compute $\iint_{x^2+y^2 \leq 1} e^{x^2+y^2}\, dx\, dy$.

In polar coordinates: $\int_0^{2\pi} \int_0^1 e^{r^2} \, r\, dr\, d\theta = 2\pi \cdot \left[-\frac{1}{2}e^{-r^2}\right]_0^1 = 2\pi \cdot \frac{1}{2}(1 - e^{-1}) = \pi(1 - e^{-1})$.

Worked example: the volume of a sphere §

In spherical coordinates $(r, \theta, \phi)$, the Jacobian is $r^2 \sin\phi$. The volume of the unit ball is

$\int_0^{2\pi} \int_0^{\pi} \int_0^1 r^2 \sin\phi \, dr\, d\phi\, d\theta = 2\pi \cdot 2 \cdot \frac{1}{3} = \frac{4\pi}{3}.$

Problem. Let $f(x, y, z) = x^2 y + y^2 z + z^2 x$.

(a) Compute $\nabla f(1, 1, 1)$.

(b) Find the directional derivative of $f$ at $(1,1,1)$ in the direction $\mathbf{v} = (1, 1, 1)/\sqrt{3}$.

(c) Use the multivariable MVT to bound $|f(2, 2, 2) - f(1, 1, 1)|$.