Multivariable Calculus

1. Partial Derivatives

1.1 Definition

Let $f : D \subseteq \mathbb{R}^n \to \mathbb{R}$. The partial derivative of $f$ with respect to $x_i$ at $\mathbf{a} = (a_1, \ldots, a_n)$ is

$\frac{\partial f}{\partial x_i}(\mathbf{a}) = \lim_{h \to 0} \frac{f(a_1, \ldots, a_i + h, \ldots, a_n) - f(a_1, \ldots, a_n)}{h}$

provided the limit exists. This is the rate of change of $f$ in the direction of the $x_i$-axis, holding all other variables fixed.

1.2 Clairaut's Theorem

Theorem 1.1 (Clairaut's Theorem / Schwarz's Theorem). If $f_{xy}$ and $f_{yx}$ are continuous on an open set containing $(a, b)$, then

$\frac{\partial^2 f}{\partial x \partial y}(a,b) = \frac{\partial^2 f}{\partial y \partial x}(a,b)$

1.3 Differentiability

Definition. $f : D \subseteq \mathbb{R}^n \to \mathbb{R}$ is differentiable at $\mathbf{a}$ if there exists a linear map $L : \mathbb{R}^n \to \mathbb{R}$ such that

$\lim_{\mathbf{h} \to \mathbf{0}} \frac{f(\mathbf{a} + \mathbf{h}) - f(\mathbf{a}) - L(\mathbf{h})}{\lVert \mathbf{h} \rVert} = 0$

When $f$ is differentiable at $\mathbf{a}$, the linear map $L$ is given by the gradient.

1.4 The Gradient

The gradient of $f$ at $\mathbf{a}$ is

$\nabla f(\mathbf{a}) = \left(\frac{\partial f}{\partial x_1}(\mathbf{a}), \ldots, \frac{\partial f}{\partial x_n}(\mathbf{a})\right)$

The linear approximation of $f$ near $\mathbf{a}$ is

$f(\mathbf{a} + \mathbf{h}) \approx f(\mathbf{a}) + \nabla f(\mathbf{a}) \cdot \mathbf{h}$

Theorem 1.2. If all partial derivatives of $f$ exist and are continuous in a neighbourhood of $\mathbf{a}$, then $f$ is differentiable at $\mathbf{a}$.

1.5 Directional Derivatives

The directional derivative of $f$ at $\mathbf{a}$ in the direction of a unit vector $\mathbf{u}$ is

$D_{\mathbf{u}} f(\mathbf{a}) = \lim_{h \to 0} \frac{f(\mathbf{a} + h\mathbf{u}) - f(\mathbf{a})}{h}$

Theorem 1.3. If $f$ is differentiable at $\mathbf{a}$, then

$D_{\mathbf{u}} f(\mathbf{a}) = \nabla f(\mathbf{a}) \cdot \mathbf{u}$

Corollary 1.4. The gradient points in the direction of steepest ascent, and $\lVert \nabla f \rVert$ is the rate of steepest ascent.

1.6 Chain Rule

Theorem 1.5 (Multivariable Chain Rule). If $\mathbf{g} : \mathbb{R}^m \to \mathbb{R}^n$ is differentiable at $\mathbf{a}$ and $f : \mathbb{R}^n \to \mathbb{R}$ is differentiable at $\mathbf{g}(\mathbf{a})$, then

$\nabla (f \circ \mathbf{g})(\mathbf{a}) = J\mathbf{g}(\mathbf{a})^T \nabla f(\mathbf{g}(\mathbf{a}))$

where $J\mathbf{g}$ is the Jacobian matrix of $\mathbf{g}$.

1.7 Worked Example

Problem. Let $f(x, y) = x^2 y + \sin(xy)$. Compute $\nabla f$ and find the directional derivative at $(1, \pi)$ in the direction $\mathbf{u} = (1/\sqrt{2}, 1/\sqrt{2})$.

Solution.

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

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

$\nabla f(1, \pi) = (2\pi + \pi\cos(\pi), 1 + \cos(\pi)) = (2\pi - \pi, 1 - 1) = (\pi, 0)$

$D_{\mathbf{u}} f(1, \pi) = \nabla f(1, \pi) \cdot \mathbf{u} = \pi \cdot \frac{1}{\sqrt{2}} + 0 = \frac{\pi}{\sqrt{2}}$ $\blacksquare$

2. Multiple Integrals

2.1 Double Integrals

The double integral of $f$ over a rectangle $R = [a,b] \times [c,d]$ is defined as the limit of Riemann sums:

$\iint_R f(x,y)\, dA = \lim_{\lVert P \rVert \to 0} \sum_{i,j} f(x_{ij}^*, y_{ij}^*) \Delta A_{ij}$

Theorem 2.1 (Fubini's Theorem). If $f$ is continuous on $R = [a,b] \times [c,d]$, then

$\iint_R f(x,y)\, dA = \int_a^b \left(\int_c^d f(x,y)\, dy\right) dx = \int_c^d \left(\int_a^b f(x,y)\, dx\right) dy$

2.2 General Regions

For a general region $D$ in $\mathbb{R}^2$:

• Type I region: $D = \{(x,y) : a \leq x \leq b,\, g_1(x) \leq y \leq g_2(x)\}$

$\iint_D f\, dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\, dx$

• Type II region: $D = \{(x,y) : c \leq y \leq d,\, h_1(y) \leq x \leq h_2(y)\}$

$\iint_D f\, dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x,y)\, dx\, dy$

2.3 Triple Integrals

Triple integrals extend naturally to $\mathbb{R}^3$:

$\iiint_E f(x,y,z)\, dV = \iint_D \left(\int_{g_1(x,y)}^{g_2(x,y)} f(x,y,z)\, dz\right) dA$

2.4 Change of Variables

Theorem 2.2 (Change of Variables). Let $T : D \subseteq \mathbb{R}^n \to \mathbb{R}^n$ be a $C^1$ diffeomorphism with Jacobian determinant $J_T$. Then

$\int_{T(D)} f(\mathbf{u})\, d\mathbf{u} = \int_D f(T(\mathbf{x})) |J_T(\mathbf{x})|\, d\mathbf{x}$

Polar coordinates: $x = r\cos\theta$, $y = r\sin\theta$, $|J| = r$.

$\iint_D f(x,y)\, dA = \iint_{D'} f(r\cos\theta, r\sin\theta)\, r\, dr\, d\theta$

Cylindrical coordinates: $x = r\cos\theta$, $y = r\sin\theta$, $z = z$, $|J| = r$.

Spherical coordinates: $x = \rho\sin\phi\cos\theta$, $y = \rho\sin\phi\sin\theta$, $z = \rho\cos\phi$, $|J| = \rho^2 \sin\phi$.

2.5 Worked Example

Problem. Compute $\iint_D (x^2 + y^2)\, dA$ where $D$ is the region bounded by $x^2 + y^2 = 4$.

Solution. Use polar coordinates. The region $D'$ is $0 \leq r \leq 2$, $0 \leq \theta \leq 2\pi$.

$\iint_D (x^2 + y^2)\, dA = \int_0^{2\pi} \int_0^2 r^2 \cdot r\, dr\, d\theta = \int_0^{2\pi} \int_0^2 r^3\, dr\, d\theta$

$= \int_0^{2\pi} \left[\frac{r^4}{4}\right]_0^2 d\theta = \int_0^{2\pi} 4\, d\theta = 8\pi$

$\blacksquare$

3. Vector Calculus

3.1 Vector Fields

A vector field on $\mathbb{R}^n$ is a function $\mathbf{F} : D \subseteq \mathbb{R}^n \to \mathbb{R}^n$.

A vector field $\mathbf{F} = (P, Q, R)$ on $\mathbb{R}^3$ is conservative if there exists a scalar potential $\phi$ such that $\mathbf{F} = \nabla \phi$.

Theorem 3.1. $\mathbf{F}$ is conservative (on a simply connected domain) if and only if $\nabla \times \mathbf{F} = \mathbf{0}$.

3.2 Line Integrals

Definition. The line integral of a vector field $\mathbf{F}$ along a curve $C$ parameterised by $\mathbf{r}(t)$ for $a \leq t \leq b$ is

$\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\, dt$

Theorem 3.2 (Fundamental Theorem for Line Integrals). If $\mathbf{F} = \nabla \phi$ and $C$ is a piecewise smooth curve from $A$ to $B$, then

$\int_C \mathbf{F} \cdot d\mathbf{r} = \phi(B) - \phi(A)$

Corollary 3.3. The line integral of a conservative field around any closed curve is zero.

3.3 Green's Theorem

Theorem 3.4 (Green's Theorem). Let $C$ be a positively oriented, piecewise smooth, simple closed curve bounding a region $D$. If $P$ and $Q$ have continuous partial derivatives on an open set containing $D$, then

$\oint_C P\, dx + Q\, dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$

Worked Example. Evaluate $\oint_C (x^2 - y)\, dx + (y^2 + x)\, dy$ where $C$ is the unit circle traversed counterclockwise.

Solution. By Green's theorem with $P = x^2 - y$ and $Q = y^2 + x$:

$\frac{\partial Q}{\partial x} = 1, \quad \frac{\partial P}{\partial y} = -1$

$\oint_C P\, dx + Q\, dy = \iint_D (1 - (-1))\, dA = 2 \iint_D dA = 2 \cdot \pi \cdot 1^2 = 2\pi$

$\blacksquare$

3.4 Stokes' Theorem

Theorem 3.5 (Stokes' Theorem). Let $S$ be an oriented surface with piecewise smooth boundary curve $C$ (positively oriented). If $\mathbf{F}$ has continuous partial derivatives on an open set containing $S$, then

$\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$

where $d\mathbf{S} = \mathbf{n}\, dS$ is the vector surface element with unit normal $\mathbf{n}$.

3.5 Divergence Theorem

Theorem 3.6 (Divergence Theorem / Gauss's Theorem). Let $E$ be a solid region bounded by a closed surface $S$ with outward normal $\mathbf{n}$. If $\mathbf{F}$ has continuous partial derivatives on an open set containing $E$, then

$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_E \nabla \cdot \mathbf{F}\, dV$

where $\nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$ is the divergence of $\mathbf{F}$.

Worked Example. Compute the flux of $\mathbf{F} = (x^3, y^3, z^3)$ through the unit sphere $S$.

Solution. By the divergence theorem:

$\nabla \cdot \mathbf{F} = 3x^2 + 3y^2 + 3z^2 = 3(x^2 + y^2 + z^2) = 3\rho^2$

Using spherical coordinates:

$\iiint_E 3\rho^2 \cdot \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta = 3 \int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho^4 \sin\phi\, d\rho\, d\phi\, d\theta$

$= 3 \cdot 2\pi \cdot 2 \cdot \frac{1}{5} = \frac{12\pi}{5}$

$\blacksquare$

Common Pitfall

When applying Green's, Stokes', or the Divergence theorem, verify that the field has continuous partial derivatives on the region (including interior). If there are singularities inside the region, the theorems do not apply directly; the singularity must be handled separately.

4. Optimization

4.1 Local Extrema

Theorem 4.1 (First Derivative Test). If $f$ has a local extremum at an interior point $\mathbf{a}$ and $\nabla f(\mathbf{a})$ exists, then $\nabla f(\mathbf{a}) = \mathbf{0}$.

Points where $\nabla f = \mathbf{0}$ are called critical points (or stationary points).

4.2 Second Derivative Test

Theorem 4.2 (Second Derivative Test). Let $f$ have continuous second partial derivatives near a critical point $(a,b)$ with $f_x(a,b) = f_y(a,b) = 0$. Let

$D = f_{xx}(a,b) f_{yy}(a,b) - [f_{xy}(a,b)]^2$

be the Hessian determinant. Then:

• If $D > 0$ and $f_{xx}(a,b) > 0$: local minimum.
• If $D > 0$ and $f_{xx}(a,b) < 0$: local maximum.
• If $D < 0$: saddle point.
• If $D = 0$: the test is inconclusive.

4.3 Lagrange Multipliers

Theorem 4.3 (Method of Lagrange Multipliers). To find the extrema of $f(x,y,z)$ subject to the constraint $g(x,y,z) = 0$, solve the system:

$\nabla f = \lambda \nabla g, \quad g = 0$

More generally, for $k$ constraints $g_1 = 0, \ldots, g_k = 0$:

$\nabla f = \lambda_1 \nabla g_1 + \cdots + \lambda_k \nabla g_k$

4.4 Worked Example

Problem. Find the maximum of $f(x,y) = xy$ subject to $x^2 + y^2 = 1$.

Solution. Set $g(x,y) = x^2 + y^2 - 1$. The Lagrange multiplier equations:

$\nabla f = \lambda \nabla g \implies (y, x) = \lambda(2x, 2y)$

This gives $y = 2\lambda x$ and $x = 2\lambda y$. Multiplying: $xy = 4\lambda^2 xy$.

Case 1: $xy \neq 0$. Then $4\lambda^2 = 1$, so $\lambda = \pm 1/2$.

• $\lambda = 1/2$: $y = x$, and $x^2 + x^2 = 1$, so $x = \pm 1/\sqrt{2}$. Points: $(1/\sqrt{2}, 1/\sqrt{2})$ and $(-1/\sqrt{2}, -1/\sqrt{2})$ with $f = 1/2$.
• $\lambda = -1/2$: $y = -x$, and $x^2 + x^2 = 1$, so $x = \pm 1/\sqrt{2}$. Points: $(1/\sqrt{2}, -1/\sqrt{2})$ and $(-1/\sqrt{2}, 1/\sqrt{2})$ with $f = -1/2$.

Case 2: $xy = 0$. Then either $x = 0$ or $y = 0$. From the constraint: $(0, \pm 1)$ or $(\pm 1, 0)$ with $f = 0$.

Maximum: $f = 1/2$ at $(\pm 1/\sqrt{2}, \pm 1/\sqrt{2})$. Minimum: $f = -1/2$ at $(\pm 1/\sqrt{2}, \mp 1/\sqrt{2})$. $\blacksquare$

Common Pitfall

The Lagrange multiplier method finds candidates for constrained extrema but does not guarantee they are extrema. Always check which candidate gives the maximum/minimum, or use additional reasoning (e.g., compactness of the constraint set).