MATH-283 (Calculus III) Curriculum

Session 11 - Chain Rule and Higher Derivatives

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

By the end of this session, you will be able to:

Apply the multivariable chain rule $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$ and its vector dot product form $\frac{dz}{dt} = \nabla z \cdot \vec{v}(t)$ to compute instantaneous rates of change experienced along parametric motion trajectories and dynamic physical paths.
Evaluate partial derivatives of composite multivariable functions with multiple independent parameters ( $\frac{\partial z}{\partial u}, \frac{\partial z}{\partial v}$ ) using dependency tree diagrams, and compute derivatives of implicitly defined curves and surfaces using multivariable partial formulas ( $\frac{dy}{dx} = -\frac{F_x}{F_y}$ and $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$ , $\frac{\partial z}{\partial y} = -\frac{F_y}{F_z}$ ).
Calculate pure and mixed second-order partial derivatives ( $f_{xx}, f_{yy}, f_{xy}, f_{yx}$ ), verify Clairaut’s Theorem on the symmetry of mixed partials, compute higher-order partial derivatives ( $n \ge 3$ ), and assemble Hessian matrices to evaluate local surface curvature, mechanical strain energy, and structural stability.

Micro-Lecture

1. Dynamic Tracking on Spatial Landscapes (Chain Rule Case 1)

Mathematical Foundation: In single-variable calculus, when $y = f(u)$ and $u = g(x)$ , the chain rule multiplies rates: $\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$ . In multivariable environments, a surface $z = f(x, y)$ represents an environmental field (such as temperature, elevation, or chemical concentration). When an observer, vehicle, or sensor travels along a curved path parameterized by time $\vec{r}(t) = \langle x(t), y(t) \rangle$ , both spatial inputs change concurrently. As established in Multivariable Chain Rule, the total rate of change experienced by the moving observer is the sum of changes along all parallel pathways: $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$ . In vector calculus, this equals the dot product of the spatial slope gradient and the travel velocity vector: $\frac{dz}{dt} = \nabla z \cdot \vec{v}(t)$ .
Engineering Context: Flight navigation computers on autonomous quadcopters continuously evaluate the multivariable chain rule to track atmospheric pressure and altitude changes as the craft maneuvers through 3D wind gusts.

2. Surface Parameterizations and Implicit Differentiation (Chain Rule Case 2)

Mathematical Foundation: When a surface is mapped into alternative curvilinear coordinates (such as Polar, Cylindrical, or Finite Element parametric coordinates $u$ and $v$ ), every coordinate is driven by multiple independent inputs: $x = g(u, v)$ and $y = h(u, v)$ . The multivariable chain rule generalizes using dependency tree diagrams, multiplying derivatives along connected branches and summing across parallel paths: $\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$ . Furthermore, as explored in Multivariable Chain Rule, applying the chain rule to level sets $F(x, y) = 0$ and $F(x, y, z) = 0$ yields the rapid implicit differentiation formulas $\frac{dy}{dx} = -\frac{F_x}{F_y}$ and $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$ , bypassing algebraic isolate-and-differentiate steps.
Engineering Context: Environmental engineers convert Cartesian contaminant dispersion equations into radial polar coordinates to monitor aquifer filtration rings, while electrical power engineers use implicit differentiation to compute dynamic AC impedances on non-linear semiconductor diodes.

3. Surface Curvature, Clairaut's Symmetry, and the Hessian Matrix

Mathematical Foundation: While first-order partial derivatives ( $f_x, f_y$ ) establish the tangent plane to a surface, second-order partial derivatives describe how the surface curves and twists in 3D space. As established in Second-Order Partial Derivatives, pure partials $f_{xx}$ and $f_{yy}$ quantify concavity along coordinate slices, while mixed partials $f_{xy}$ and $f_{yx}$ measure cross-directional twisting and coupling shear. By Clairaut's Theorem, for all smooth functions ( $C^2$ ), the order of differentiation is strictly commutative: $f_{xy} = f_{yx}$ . Organizing these four partial derivatives into the symmetric Hessian Matrix $\mathbf{H}_f$ provides the fundamental mathematical engine for multivariable optimization and stability classification.
Engineering Context: Civil and mechanical engineers inspect the Hessian determinant of structural strain energy potentials to verify that loaded bridge columns remain in a stable, convex energy trough rather than buckling into an unstable saddle collapse.

Skill Block 1

Worked Example A: Environmental Pollutant Plume Under Dynamic Wind Advection

The concentration of a chemical pollutant in a municipal reservoir is modeled by the Gaussian distribution:

C(x, y) = 40e^{-0.1(x^2 + y^2)}

where

x

and

y

are spatial coordinates in kilometers, and

C

is in parts per million (

\text{ppm}

). A localized weather front shifts the pollutant plume across a stationary water intake sensor such that the relative coordinates of the plume center vary with time

t

(in hours) according to

x(t) = 3 + t

and

y(t) = 2t

Find the instantaneous rate of change of the measured pollutant concentration, $\frac{dC}{dt}$ , at the arrival time $t = 0\text{ hours}$ .

Worked Solution:

State the Governing Multivariable Chain Rule Formula: State the Multivariable Chain Rule for a Single Parameter (defined in Multivariable Chain Rule):
$\frac{dC}{dt} = \frac{\partial C}{\partial x}\frac{dx}{dt} + \frac{\partial C}{\partial y}\frac{dy}{dt}$
Compute Spatial Partial Derivatives of Concentration: Apply the single-variable exponential chain rule, treating the alternate variable as a constant:
$\frac{\partial C}{\partial x} = 40e^{-0.1(x^2 + y^2)} \cdot \frac{\partial}{\partial x}[-0.1(x^2 + y^2)] = 40e^{-0.1(x^2 + y^2)}(-0.2x) = -8x e^{-0.1(x^2 + y^2)}$ $\frac{\partial C}{\partial y} = 40e^{-0.1(x^2 + y^2)} \cdot \frac{\partial}{\partial y}[-0.1(x^2 + y^2)] = 40e^{-0.1(x^2 + y^2)}(-0.2y) = -8y e^{-0.1(x^2 + y^2)}$
Compute Parametric Time Derivatives of the Trajectory:
$\frac{dx}{dt} = \frac{d}{dt}[3 + t] = 1\text{ km/h}$ $\frac{dy}{dt} = \frac{d}{dt}[2t] = 2\text{ km/h}$
Evaluate Positions and Derivatives at the Target Time $t = 0$ :
- Spatial coordinates at $t = 0$ : $x(0) = 3 + 0 = 3\text{ km}$ $y(0) = 2(0) = 0\text{ km}$
- Radial distance argument: $x^2 + y^2 = 3^2 + 0^2 = 9$
- Evaluate spatial partials at $(3, 0)$ : $\frac{\partial C}{\partial x}(3, 0) = -8(3)e^{-0.1(9)} = -24e^{-0.9}\text{ ppm/km}$ $\frac{\partial C}{\partial y}(3, 0) = -8(0)e^{-0.1(9)} = 0\text{ ppm/km}$
Synthesize Total Time Derivative: Substitute all evaluated terms into the chain rule formula:
$\frac{dC}{dt} = \left(-24e^{-0.9}\right)(1) + (0)(2) = -24e^{-0.9}\text{ ppm/h}$

$e^{-0.9} \approx 0.40657 \implies \frac{dC}{dt} \approx -24(0.40657) \approx -9.76\text{ ppm/h}$
- Physical Interpretation: At $t = 0$ , the sensor records an instantaneous concentration drop of approximately $9.76\text{ ppm}$ per hour as the plume core moves away from the intake.

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Explain why $\frac{dz}{dt}$ is written with an ordinary Roman $d$ , whereas $\frac{\partial z}{\partial x}$ is written with the curly partial symbol $\partial$ .
(b) In Worked Example A, explain why the $y$ -velocity component ( $\frac{dy}{dt} = 2$ ) contributed zero to the instantaneous rate $\frac{dC}{dt}$ at $t = 0$ .
(c) In Bridge Truss Gusset Plate Fatigue (Mechanical & Civil Engineering), how does dynamic crack growth rate $\frac{da}{dt}$ amplify the stress intensity factor rate $\frac{dK}{dt}$ ?

Solution Key:

(a) As detailed in Multivariable Chain Rule, the curly symbol $\partial$ indicates that a function depends directly on multiple independent variables, meaning other variables are artificially held constant during differentiation. Once the intermediate spatial coordinates are parameterized by a single driving variable $t$ (via $x = x(t)$ and $y = y(t)$ ), the composite output $z(t) = f(x(t), y(t))$ depends strictly on the single parameter $t$ . Therefore, $\frac{dz}{dt}$ represents a total derivative with respect to that single variable, written with standard Roman $d$ .
(b) In Worked Example A, the spatial partial derivative with respect to $y$ is $\frac{\partial C}{\partial y} = -8ye^{-0.1(x^2+y^2)}$ . At the initial arrival instant $t = 0$ , the sensor is positioned on the $x$ -axis where $y(0) = 0$ . Consequently, $\frac{\partial C}{\partial y}(3, 0) = 0$ . Even though the plume is moving in the $y$ -direction at a speed of $\frac{dy}{dt} = 2\text{ km/h}$ , the local concentration contour is flat tangent to the $y$ -direction at that exact point, producing zero instantaneous change along that component.
(c) In Bridge Truss Gusset Plate Fatigue (Mechanical & Civil Engineering), the stress intensity factor is $K(\sigma, a) = 1.5\sigma\sqrt{a}$ . By the multivariable chain rule, $\frac{dK}{dt} = \frac{\partial K}{\partial \sigma}\frac{d\sigma}{dt} + \frac{\partial K}{\partial a}\frac{da}{dt}$ . The sensitivity coefficient for crack propagation is $\frac{\partial K}{\partial a} = \frac{0.75\sigma}{\sqrt{a}}$ . Because $\sqrt{a}$ sits in the denominator, any non-zero crack growth rate $\frac{da}{dt}$ is multiplied by a massive sensitivity coefficient when the crack is small ( $a = 0.044\text{ m}$ ), contributing significantly to the risk of sudden brittle fracture.

Show your steps and calculations here.

Practice Problem 2

A thin metal inspection plate has a steady-state temperature field $T(x, y) = 5x^2y + 2y^3$ , where $x$ and $y$ are spatial coordinates in centimeters, and $T$ is in degrees Celsius ( $^\circ\text{C}$ ). A micro-sensor moves across the plate following the parametric path $x(t) = \sin(2t)$ and $y(t) = e^t$ , where $t$ is time in seconds.

Find the rate at which the sensor's measured temperature is changing, $\frac{dT}{dt}$ , when $t = 0\text{ seconds}$ .

Solution Key:

State the Governing Multivariable Chain Rule Formula: State the Multivariable Chain Rule (defined in Multivariable Chain Rule):
$\frac{dT}{dt} = \frac{\partial T}{\partial x}\frac{dx}{dt} + \frac{\partial T}{\partial y}\frac{dy}{dt}$
Compute Spatial Partial Derivatives:
- With respect to $x$ (treating $y$ as a constant): $\frac{\partial T}{\partial x} = \frac{\partial}{\partial x}[5x^2y + 2y^3] = 10xy$
- With respect to $y$ (treating $x$ as a constant): $\frac{\partial T}{\partial y} = \frac{\partial}{\partial y}[5x^2y + 2y^3] = 5x^2 + 6y^2$
Compute Velocity Derivatives of the Path:
- Differentiate $x(t)$ using the single-variable chain rule: $\frac{dx}{dt} = \frac{d}{dt}[\sin(2t)] = 2\cos(2t)$
- Differentiate $y(t)$ : $\frac{dy}{dt} = \frac{d}{dt}[e^t] = e^t$
Evaluate Path Coordinates and Velocities at $t = 0$ :
- Coordinates: $x(0) = \sin(0) = 0\text{ cm}$ $y(0) = e^0 = 1\text{ cm}$
- Velocities: $\frac{dx}{dt}\Big|_{t=0} = 2\cos(0) = 2(1) = 2\text{ cm/s}$ $\frac{dy}{dt}\Big|_{t=0} = e^0 = 1\text{ cm/s}$
Evaluate Spatial Partials at $(0, 1)$ :
$\frac{\partial T}{\partial x}(0, 1) = 10(0)(1) = 0^\circ\text{C/cm}$ $\frac{\partial T}{\partial y}(0, 1) = 5(0)^2 + 6(1)^2 = 6^\circ\text{C/cm}$
Assemble Total Rate of Change:
$\frac{dT}{dt} = (0)(2) + (6)(1) = 0 + 6 = 6^\circ\text{C/s}$

Show your steps and calculations here.

Practice Problem 3

Given the multivariable function $z = 3x^2y + \sin(y)$ , where the spatial variables follow the parametric trajectories $x = e^t$ and $y = \cos(t)$ , determine the total derivative $\frac{dz}{dt}$ expressed entirely in terms of the independent parameter $t$ .

Solution Key:

State the Governing Multivariable Chain Rule Formula:
$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$ (Defined in Multivariable Chain Rule.)
Compute Partial Derivatives of $z(x, y)$ :
$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}[3x^2y + \sin(y)] = 6xy$ $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}[3x^2y + \sin(y)] = 3x^2 + \cos(y)$
Compute Parametric Derivatives with Respect to $t$ :
$\frac{dx}{dt} = \frac{d}{dt}[e^t] = e^t$ $\frac{dy}{dt} = \frac{d}{dt}[\cos(t)] = -\sin(t)$
Substitute Components into the Chain Rule Formula:
$\frac{dz}{dt} = (6xy)(e^t) + (3x^2 + \cos(y))(-\sin(t))$ $\frac{dz}{dt} = 6xy e^t - (3x^2 + \cos(y))\sin(t)$
Eliminate Intermediate Variables $x$ and $y$ : Substitute $x = e^t$ and $y = \cos(t)$ :
$\frac{dz}{dt} = 6(e^t)(\cos(t))(e^t) - \left(3(e^t)^2 + \cos(\cos(t))\right)\sin(t)$ Combine exponential powers ( $e^t \cdot e^t = e^{2t}$ and $(e^t)^2 = e^{2t}$ ): $\frac{dz}{dt} = 6e^{2t}\cos(t) - 3e^{2t}\sin(t) - \sin(t)\cos(\cos(t))$

Show your steps and calculations here.

Practice Problem 4

An internal combustion engine piston's vertical clearance height is governed by the kinematic model $h(r, \theta) = r^2\sin(\theta)$ , where $r$ is the crankshaft throw radius and $\theta$ is the crank rotation angle. During engine acceleration, the parameters vary dynamically with time $t$ (in seconds) according to $r(t) = 2t$ and $\theta(t) = \pi t$ .

Find the instantaneous rate of change of the clearance height, $\frac{dh}{dt}$ , at time $t = 0.25\text{ seconds}$ .

Solution Key:

State the Governing Multivariable Chain Rule Formula:
$\frac{dh}{dt} = \frac{\partial h}{\partial r}\frac{dr}{dt} + \frac{\partial h}{\partial \theta}\frac{d\theta}{dt}$ (Defined in Multivariable Chain Rule.)
Compute Partial Derivatives of $h(r, \theta)$ :
$\frac{\partial h}{\partial r} = \frac{\partial}{\partial r}[r^2\sin(\theta)] = 2r\sin(\theta)$ $\frac{\partial h}{\partial \theta} = \frac{\partial}{\partial \theta}[r^2\sin(\theta)] = r^2\cos(\theta)$
Compute Time Derivatives of the Mechanism:
$\frac{dr}{dt} = \frac{d}{dt}[2t] = 2$ $\frac{d\theta}{dt} = \frac{d}{dt}[\pi t] = \pi$
Evaluate Variables at $t = 0.25\text{ s}$ :
- $r(0.25) = 2(0.25) = 0.5$
- $\theta(0.25) = \pi(0.25) = \frac{\pi}{4}\text{ rad}$
- Evaluate trigonometric terms: $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \qquad \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
- Evaluate partial derivatives: $\frac{\partial h}{\partial r} = 2(0.5)\left(\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{2}$ $\frac{\partial h}{\partial \theta} = (0.5)^2\left(\frac{\sqrt{2}}{2}\right) = (0.25)\left(\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{8}$
Assemble Total Derivative:
$\frac{dh}{dt} = \left(\frac{\sqrt{2}}{2}\right)(2) + \left(\frac{\sqrt{2}}{8}\right)(\pi) = \sqrt{2} + \frac{\pi\sqrt{2}}{8}$ Factoring out $\sqrt{2}$ : $\frac{dh}{dt} = \sqrt{2}\left(1 + \frac{\pi}{8}\right) \approx 1.4142(1 + 0.3927) \approx 1.9696 \approx 1.97$

Show your steps and calculations here.

Regroup 1

Key Takeaway: When traversing a multivariable landscape $z = f(x, y)$ , total change is the sum of changes experienced along all coordinate directions simultaneously: $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$ .
Vector Insight: The total derivative is identical to the dot product $\nabla z \cdot \vec{v}(t)$ . If the velocity vector is perpendicular to the spatial gradient (moving tangent to a level contour), the total rate of change is identically zero.
Common Pitfall: Never confuse partial derivative $\frac{\partial z}{\partial x}$ (which holds $y$ constant) with total derivative $\frac{dz}{dt}$ (where $x$ and $y$ evolve concurrently).

Skill Block 2

Worked Example A: Multivariable Chain Rule with Two Independent Parameters

Let the function $z = x^2 y - y^3$ be defined over intermediate parameters $x(u, v) = u^2 v$ and $y(u, v) = 2u - v$ . Use the Multivariable Chain Rule for two independent variables to compute $\frac{\partial z}{\partial u}$ and $\frac{\partial z}{\partial v}$ , and evaluate both partial rates at the parameter coordinate $(u, v) = (2, 1)$ .

Worked Solution:

State the Governing Multivariable Chain Rule Formulas (Case 2): State the Multivariable Chain Rule for Two Independent Variables (defined in Multivariable Chain Rule):
$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$ $\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial v}$
Compute Intermediate Partial Derivatives of $z(x, y)$ :
$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}[x^2 y - y^3] = 2xy$ $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}[x^2 y - y^3] = x^2 - 3y^2$
Compute Parameter Partial Derivatives:
- With respect to $u$ : $\frac{\partial x}{\partial u} = \frac{\partial}{\partial u}[u^2 v] = 2uv, \qquad \frac{\partial y}{\partial u} = \frac{\partial}{\partial u}[2u - v] = 2$
- With respect to $v$ : $\frac{\partial x}{\partial v} = \frac{\partial}{\partial v}[u^2 v] = u^2, \qquad \frac{\partial y}{\partial v} = \frac{\partial}{\partial v}[2u - v] = -1$
Evaluate Intermediate Values at $(u, v) = (2, 1)$ :
- Intermediate coordinates: $x(2, 1) = (2)^2(1) = 4$ $y(2, 1) = 2(2) - 1 = 3$
- Intermediate partials at $(4, 3)$ : $\frac{\partial z}{\partial x}(4, 3) = 2(4)(3) = 24$ $\frac{\partial z}{\partial y}(4, 3) = (4)^2 - 3(3)^2 = 16 - 27 = -11$
- Parameter partials at $(2, 1)$ : $\frac{\partial x}{\partial u} = 2(2)(1) = 4, \qquad \frac{\partial y}{\partial u} = 2$ $\frac{\partial x}{\partial v} = (2)^2 = 4, \qquad \frac{\partial y}{\partial v} = -1$
Assemble and Evaluate Both Partial Rates:
- Rate with respect to $u$ : $\frac{\partial z}{\partial u} = (24)(4) + (-11)(2) = 96 - 22 = 74$
- Rate with respect to $v$ : $\frac{\partial z}{\partial v} = (24)(4) + (-11)(-1) = 96 + 11 = 107$

Practice Problem 5: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) If a function $w = f(x, y, z)$ depends on three intermediate variables, and each intermediate variable depends on two independent parameters $(u, v)$ , how many total derivative products must be summed to compute $\frac{\partial w}{\partial u}$ ?
(b) In the geometric derivation of the implicit curve differentiation formula $\frac{dy}{dx} = -\frac{F_x}{F_y}$ , explain algebraically where the negative sign comes from.
(c) In Groundwater Contaminant Plume in Radial Aquifer (Environmental Engineering), why does the angular derivative $\frac{\partial C}{\partial \theta}$ equal zero for all radii?

Solution Key:

(a) According to the tree diagram rules in Multivariable Chain Rule, there is one primary output variable $w$ at the top level, three intermediate variables ( $x, y, z$ ) in the middle tier, and two independent parameters ( $u, v$ ) at the bottom level. To find $\frac{\partial w}{\partial u}$ , we track every downward path connecting $w$ to $u$ . Since there are three distinct paths passing through $x$ , $y$ , and $z$ , the formula sums exactly three branch products: $\frac{\partial w}{\partial u} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial u} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial u}$
(b) Along an implicit level curve $F(x, y) = c$ , the function value is constant, so its total differential vanishes identically: $dF = F_x dx + F_y dy = 0$ . Dividing through by the differential $dx$ gives $F_x + F_y \frac{dy}{dx} = 0$ . Subtracting $F_x$ from both sides yields $F_y \frac{dy}{dx} = -F_x$ . Dividing by $F_y$ (assuming $F_y \neq 0$ ) introduces the negative sign: $\frac{dy}{dx} = -\frac{F_x}{F_y}$ .
(c) In Groundwater Contaminant Plume in Radial Aquifer (Environmental Engineering), the concentration is governed by $C(x, y) = 100 e^{-0.02(x^2+y^2)}$ . Because $x^2 + y^2 = r^2$ , the plume depends purely on radial distance from the origin ( $C(r) = 100e^{-0.02r^2}$ ) and is completely independent of the polar angle $\theta$ . Applying the chain rule yields $\frac{\partial C}{\partial \theta} = (-4x e^{-0.02r^2})(-r\sin\theta) + (-4y e^{-0.02r^2})(r\cos\theta) = 4r^2 e^{-0.02r^2}(\cos\theta\sin\theta - \sin\theta\cos\theta) = 0$ .

Show your steps and calculations here.

Practice Problem 6

A multivariable surface is represented in Cartesian coordinates by $z = 4x^2 + y^2$ . Engineers convert this model into polar coordinates $(r, \theta)$ via the transformation $x(r, \theta) = r\cos(\theta)$ and $y(r, \theta) = r\sin(\theta)$ .

State the Multivariable Chain Rule formula expressing $\frac{\partial z}{\partial r}$ in terms of Cartesian partials and polar angle $\theta$ .
Evaluate $\frac{\partial z}{\partial r}$ at the polar coordinate point $(r, \theta) = (3, \frac{\pi}{6})$ .

Solution Key:

State the Governing Multivariable Chain Rule Formula:
$\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$
(Defined in Multivariable Chain Rule.)
- Compute transformation derivatives: $\frac{\partial x}{\partial r} = \frac{\partial}{\partial r}[r\cos\theta] = \cos\theta$ $\frac{\partial y}{\partial r} = \frac{\partial}{\partial r}[r\sin\theta] = \sin\theta$
- Substitute into the chain rule: $\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\cos\theta + \frac{\partial z}{\partial y}\sin\theta$
- Compute Cartesian partials for $z = 4x^2 + y^2$ : $\frac{\partial z}{\partial x} = 8x = 8(r\cos\theta) = 8r\cos\theta$ $\frac{\partial z}{\partial y} = 2y = 2(r\sin\theta) = 2r\sin\theta$
- Combine: $\frac{\partial z}{\partial r} = (8r\cos\theta)(\cos\theta) + (2r\sin\theta)(\sin\theta) = 8r\cos^2\theta + 2r\sin^2\theta$
Evaluate at $(r, \theta) = (3, \frac{\pi}{6})$ :
- Trigonometric values at $\theta = \frac{\pi}{6}$ : $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \implies \cos^2\left(\frac{\pi}{6}\right) = \frac{3}{4}$ $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \implies \sin^2\left(\frac{\pi}{6}\right) = \frac{1}{4}$
- Substitute $r = 3$ : $\frac{\partial z}{\partial r} = 8(3)\left(\frac{3}{4}\right) + 2(3)\left(\frac{1}{4}\right) = 24\left(\frac{3}{4}\right) + 6\left(\frac{1}{4}\right) = 18 + 1.5 = 19.5 = \frac{39}{2}$

Show your steps and calculations here.

Practice Problem 7

Use the multivariable implicit differentiation formula $\frac{dy}{dx} = -\frac{F_x}{F_y}$ to determine the exact slope of the tangent line to the Folium of Descartes curve:

x^3 + y^3 - 6xy = 0

at the non-trivial coordinate point

(3, 3)

Solution Key:

Define the Implicit Function and State Governing Formula: Define $F(x, y) = x^3 + y^3 - 6xy = 0$ . State the Implicit Differentiation Formula for Curves (defined in Multivariable Chain Rule):
$\frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$
Compute First-Order Partial Derivatives of $F(x, y)$ :
- With respect to $x$ (treating $y$ as a constant): $F_x = \frac{\partial}{\partial x}[x^3 + y^3 - 6xy] = 3x^2 - 6y$
- With respect to $y$ (treating $x$ as a constant): $F_y = \frac{\partial}{\partial y}[x^3 + y^3 - 6xy] = 3y^2 - 6x$
Assemble the Symbolic Derivative:
$\frac{dy}{dx} = -\frac{3x^2 - 6y}{3y^2 - 6x} = -\frac{3(x^2 - 2y)}{3(y^2 - 2x)} = -\frac{x^2 - 2y}{y^2 - 2x}$
Evaluate at the Point $(3, 3)$ :
$\frac{dy}{dx}\Big|_{(3, 3)} = -\frac{(3)^2 - 2(3)}{(3)^2 - 2(3)} = -\frac{9 - 6}{9 - 6} = -\frac{3}{3} = -1$
- Geometric Insight: At the apex loop point $(3, 3)$ , the curve has a tangent slope of $m = -1$ , confirming symmetry across the line $y = x$ .

Show your steps and calculations here.

Practice Problem 8

An implicit curved aerodynamic shell surface is modeled by the equation:

x^2 y + y^2 z + z^2 x = 7

Use multivariable implicit differentiation to compute the surface elevation gradients

\frac{\partial z}{\partial x}

and

\frac{\partial z}{\partial y}

at the point

P_0(1, 1, 2)

Solution Key:

Define the Level Surface Function and State Governing Formulas: Define $F(x, y, z) = x^2 y + y^2 z + z^2 x - 7 = 0$ . State the Implicit Surface Differentiation Formulas (defined in Multivariable Chain Rule):
$\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \qquad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z}$
Compute All Three First-Order Partial Derivatives of $F(x, y, z)$ :
$F_x = \frac{\partial}{\partial x}[x^2 y + y^2 z + z^2 x - 7] = 2xy + z^2$ $F_y = \frac{\partial}{\partial y}[x^2 y + y^2 z + z^2 x - 7] = x^2 + 2yz$ $F_z = \frac{\partial}{\partial z}[x^2 y + y^2 z + z^2 x - 7] = y^2 + 2zx$
Evaluate the Partial Derivatives at $P_0(1, 1, 2)$ :
- $F_x(1, 1, 2) = 2(1)(1) + (2)^2 = 2 + 4 = 6$
- $F_y(1, 1, 2) = (1)^2 + 2(1)(2) = 1 + 4 = 5$
- $F_z(1, 1, 2) = (1)^2 + 2(2)(1) = 1 + 4 = 5$
Compute the Surface Derivatives:
$\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{6}{5}$ $\frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = -\frac{5}{5} = -1$

Show your steps and calculations here.

Regroup 2

Key Takeaway: Multi-parameter chain rules multiply derivatives along vertical tree branches and sum across parallel branches. The bottom level always determines how many partial derivative equations must be written.
Implicit Differentiation Power: Never solve for $y$ or $z$ algebraically. Applying $\frac{dy}{dx} = -\frac{F_x}{F_y}$ or $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$ computes exact slopes and surface gradients in seconds.
Common Pitfall: Forgetting the minus sign in $-\frac{F_x}{F_y}$ or inverting the numerator and denominator (remember: the differentiated variable appears in the denominator of the ratio!).

Skill Block 3

Worked Example A: Second-Order Partials on an Exponential-Polynomial Surface

Compute all second-order partial derivatives ( $f_{xx}, f_{yy}, f_{xy}, f_{yx}$ ) for the multivariable exponential surface:

f(x, y) = e^{x^2 y}

and verify that Clairaut's Theorem holds.

Worked Solution:

State the Governing Second-Order Derivative Formulas: State the Second-Order Partial Derivative Definitions (see Second-Order Partial Derivatives):
$f_{xx} = \frac{\partial}{\partial x}(f_x), \quad f_{yy} = \frac{\partial}{\partial y}(f_y), \quad f_{xy} = \frac{\partial}{\partial y}(f_x), \quad f_{yx} = \frac{\partial}{\partial x}(f_y)$
Compute First-Order Partial Derivatives: Apply the single-variable exponential chain rule $\frac{d}{du}[e^u] = e^u u'$ :
- With respect to $x$ (treating $y$ as a constant): $f_x = e^{x^2 y} \cdot \frac{\partial}{\partial x}[x^2 y] = e^{x^2 y}(2xy) = 2xye^{x^2 y}$
- With respect to $y$ (treating $x$ as a constant): $f_y = e^{x^2 y} \cdot \frac{\partial}{\partial y}[x^2 y] = e^{x^2 y}(x^2) = x^2 e^{x^2 y}$
Compute Pure Second Partial Derivatives:
- Compute $f_{xx}$ (requires the product rule on $u = 2xy$ and $v = e^{x^2 y}$ ): $f_{xx} = \frac{\partial}{\partial x}[2xye^{x^2 y}] = \frac{\partial}{\partial x}[2xy] e^{x^2 y} + 2xy \frac{\partial}{\partial x}[e^{x^2 y}]$ $f_{xx} = 2y e^{x^2 y} + 2xy(2xy e^{x^2 y}) = 2y e^{x^2 y} + 4x^2 y^2 e^{x^2 y} = 2y(1 + 2x^2 y)e^{x^2 y}$
- Compute $f_{yy}$ (treating $x^2$ as a constant multiplier): $f_{yy} = \frac{\partial}{\partial y}[x^2 e^{x^2 y}] = x^2 \frac{\partial}{\partial y}[e^{x^2 y}] = x^2 (x^2 e^{x^2 y}) = x^4 e^{x^2 y}$
Compute and Verify Mixed Partial Derivatives:
- Compute $f_{xy} = \frac{\partial}{\partial y}(f_x)$ (using product rule on $u = 2xy$ and $v = e^{x^2 y}$ with respect to $y$ ): $f_{xy} = \frac{\partial}{\partial y}[2xy] e^{x^2 y} + 2xy \frac{\partial}{\partial y}[e^{x^2 y}]$ $f_{xy} = 2x e^{x^2 y} + 2xy (x^2 e^{x^2 y}) = 2x e^{x^2 y} + 2x^3 y e^{x^2 y} = 2x(1 + x^2 y)e^{x^2 y}$
- Compute $f_{yx} = \frac{\partial}{\partial x}(f_y)$ (using product rule on $u = x^2$ and $v = e^{x^2 y}$ with respect to $x$ ): $f_{yx} = \frac{\partial}{\partial x}[x^2] e^{x^2 y} + x^2 \frac{\partial}{\partial x}[e^{x^2 y}]$ $f_{yx} = 2x e^{x^2 y} + x^2 (2xy e^{x^2 y}) = 2x e^{x^2 y} + 2x^3 y e^{x^2 y} = 2x(1 + x^2 y)e^{x^2 y}$
Verify Clairaut's Theorem:
$f_{xy} = 2x(1 + x^2 y)e^{x^2 y} = f_{yx}$ Because $f(x, y)$ is a smooth composition of exponential and polynomial functions, its partial derivatives of all orders are continuous on $\mathbb{R}^2$ , confirming Clairaut's Theorem.

Practice Problem 9: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Under what mathematical condition does Clairaut's Theorem guarantee that $f_{xy} = f_{yx}$ ?
(b) In Worked Example A, explain why the product rule was necessary when differentiating $f_x = 2xy e^{x^2 y}$ with respect to both $x$ and $y$ .
(c) In Structural Column Strain Energy Stability (Civil & Mechanical Engineering), explain the physical meaning of the mixed partial derivative $U_{xy}$ and why a positive Hessian determinant ( $D > 0$ ) guarantees structural stability.

Solution Key:

(a) As established in Clairaut's Theorem, the equality of mixed second-order partial derivatives ( $f_{xy} = f_{yx}$ ) requires that the function $f(x, y)$ and both of its mixed second-order partial derivatives $f_{xy}$ and $f_{yx}$ be continuous on an open disk $D$ centered at the point $(a, b)$ . All elementary functions satisfy this continuity requirement everywhere on their natural domains.
(b) In Worked Example A, the first partial derivative is $f_x(x, y) = 2xye^{x^2 y}$ . When differentiating with respect to $x$ , the variable $x$ appears as a linear factor in $2xy$ and as a quadratic term in the exponent $x^2 y$ , requiring the product rule ( $u = 2xy, v = e^{x^2 y}$ ). When differentiating with respect to $y$ , the variable $y$ likewise appears in both the prefactor $2xy$ and in the exponent $x^2 y$ , requiring the product rule with respect to $y$ as well.
(c) In Structural Column Strain Energy Stability (Civil & Mechanical Engineering), the internal strain energy is $U(x, y) = 0.5x^2 + 0.04x^2 y + 3y^2$ , where $x$ is axial compressive load and $y$ is temperature rise. The mixed partial $U_{xy} = 0.08x$ represents the thermal-mechanical coupling stiffness—how temperature changes alter the column's resistance to axial deflection. A positive Hessian determinant ( $D = \det(\mathbf{H}_U) = 17.36 > 0$ ) combined with positive pure concavity ( $U_{xx} = 3.0 > 0$ ) confirms that the energy surface curves upward in every direction, proving the column is in a stable equilibrium state immune to sudden elastic buckling.

Show your steps and calculations here.

Practice Problem 10

Let the multivariable function be defined by:

f(x, y) = x^2 y^3 + 4xy + \sin(xy)

Compute the first-order partial derivatives $f_x$ and $f_y$ .
Compute the pure second-order partial derivatives $f_{xx}$ and $f_{yy}$ .
Compute the mixed second-order partial derivatives $f_{xy}$ and $f_{yx}$ and verify Clairaut's Theorem.

Solution Key:

Compute First-Order Partial Derivatives:
$\frac{\partial}{\partial x}[\sin(xy)] = \cos(xy) \cdot \frac{\partial}{\partial x}[xy] = y\cos(xy)$ $f_x = \frac{\partial}{\partial x}[x^2 y^3 + 4xy + \sin(xy)] = 2xy^3 + 4y + y\cos(xy)$ $\frac{\partial}{\partial y}[\sin(xy)] = \cos(xy) \cdot \frac{\partial}{\partial y}[xy] = x\cos(xy)$ $f_y = \frac{\partial}{\partial y}[x^2 y^3 + 4xy + \sin(xy)] = 3x^2 y^2 + 4x + x\cos(xy)$
Compute Pure Second-Order Partial Derivatives:
- Compute $f_{xx}$ : $f_{xx} = \frac{\partial}{\partial x}[2xy^3 + 4y + y\cos(xy)] = 2y^3 + 0 + y\left(-\sin(xy) \cdot y\right) = 2y^3 - y^2\sin(xy)$
- Compute $f_{yy}$ : $f_{yy} = \frac{\partial}{\partial y}[3x^2 y^2 + 4x + x\cos(xy)] = 6x^2 y + 0 + x\left(-\sin(xy) \cdot x\right) = 6x^2 y - x^2\sin(xy)$
Compute Mixed Partial Derivatives and Verify Clairaut's Theorem:
- Compute $f_{xy} = \frac{\partial}{\partial y}(f_x)$ (apply product rule to $y\cos(xy)$ with respect to $y$ ): $f_{xy} = \frac{\partial}{\partial y}[2xy^3 + 4y + y\cos(xy)]$ $f_{xy} = 6xy^2 + 4 + \left[(1)\cos(xy) + y(-\sin(xy) \cdot x)\right]$ $f_{xy} = 6xy^2 + 4 + \cos(xy) - xy\sin(xy)$
- Compute $f_{yx} = \frac{\partial}{\partial x}(f_y)$ (apply product rule to $x\cos(xy)$ with respect to $x$ ): $f_{yx} = \frac{\partial}{\partial x}[3x^2 y^2 + 4x + x\cos(xy)]$ $f_{yx} = 6xy^2 + 4 + \left[(1)\cos(xy) + x(-\sin(xy) \cdot y)\right]$ $f_{yx} = 6xy^2 + 4 + \cos(xy) - xy\sin(xy)$ Both expressions are identical, successfully verifying Clairaut's Theorem.

Show your steps and calculations here.

Practice Problem 11

Compute all four second-order partial derivatives ( $f_{xx}, f_{yy}, f_{xy}, f_{yx}$ ) for the polynomial surface:

f(x, y) = 2x^3 y^2 + 5xy

Solution Key:

Compute First-Order Partial Derivatives:
$f_x = \frac{\partial}{\partial x}[2x^3 y^2 + 5xy] = 6x^2 y^2 + 5y$ $f_y = \frac{\partial}{\partial y}[2x^3 y^2 + 5xy] = 4x^3 y + 5x$
Compute Pure Second-Order Partial Derivatives:
- Differentiate $f_x$ with respect to $x$ : $f_{xx} = \frac{\partial}{\partial x}[6x^2 y^2 + 5y] = 12xy^2 + 0 = 12xy^2$
- Differentiate $f_y$ with respect to $y$ : $f_{yy} = \frac{\partial}{\partial y}[4x^3 y + 5x] = 4x^3(1) + 0 = 4x^3$
Compute Mixed Second-Order Partial Derivatives:
- Differentiate $f_x$ with respect to $y$ : $f_{xy} = \frac{\partial}{\partial y}[6x^2 y^2 + 5y] = 6x^2(2y) + 5 = 12x^2 y + 5$
- Differentiate $f_y$ with respect to $x$ : $f_{yx} = \frac{\partial}{\partial x}[4x^3 y + 5x] = 4(3x^2)y + 5 = 12x^2 y + 5$ As guaranteed by Clairaut's Theorem, $f_{xy} = f_{yx} = 12x^2 y + 5$ .

Show your steps and calculations here.

Practice Problem 12

For the multivariable function $f(x, y) = x^4 y^2 - 2x^2 y^3 + 5x$ :

Construct the complete $2 \times 2$ Hessian matrix $\mathbf{H}_f(x, y)$ .
Compute the third-order mixed partial derivatives $f_{xxy}$ and $f_{xyx}$ , and verify that they are equal.

Solution Key:

Compute First-Order Partials:
$f_x = \frac{\partial}{\partial x}[x^4 y^2 - 2x^2 y^3 + 5x] = 4x^3 y^2 - 4xy^3 + 5$ $f_y = \frac{\partial}{\partial y}[x^4 y^2 - 2x^2 y^3 + 5x] = 2x^4 y - 6x^2 y^2$
Compute Second-Order Partials:
- Pure partials: $f_{xx} = \frac{\partial}{\partial x}[4x^3 y^2 - 4xy^3 + 5] = 12x^2 y^2 - 4y^3$ $f_{yy} = \frac{\partial}{\partial y}[2x^4 y - 6x^2 y^2] = 2x^4 - 12x^2 y$
- Mixed partials: $f_{xy} = \frac{\partial}{\partial y}[4x^3 y^2 - 4xy^3 + 5] = 8x^3 y - 12xy^2$ $f_{yx} = \frac{\partial}{\partial x}[2x^4 y - 6x^2 y^2] = 8x^3 y - 12xy^2$
Construct the Hessian Matrix: State the Hessian Matrix Definition (see Second-Order Partial Derivatives):
$\mathbf{H}_f(x, y) = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} 12x^2 y^2 - 4y^3 & 8x^3 y - 12xy^2 \\ 8x^3 y - 12xy^2 & 2x^4 - 12x^2 y \end{bmatrix}$
Compute and Verify Third-Order Mixed Partials:
- Differentiate $f_{xx}$ with respect to $y$ : $f_{xxy} = \frac{\partial}{\partial y}[f_{xx}] = \frac{\partial}{\partial y}[12x^2 y^2 - 4y^3] = 12x^2(2y) - 4(3y^2) = 24x^2 y - 12y^2$
- Differentiate $f_{xy}$ with respect to $x$ : $f_{xyx} = \frac{\partial}{\partial x}[f_{xy}] = \frac{\partial}{\partial x}[8x^3 y - 12xy^2] = 8(3x^2)y - 12(1)y^2 = 24x^2 y - 12y^2$ Both third-order partial derivatives match identically ( $f_{xxy} = f_{xyx} = 24x^2 y - 12y^2$ ), confirming that Clairaut's Theorem extends to higher-order differentiation.

Show your steps and calculations here.

Regroup 3

Key Takeaway: Pure second partials ( $f_{xx}, f_{yy}$ ) describe concavity along the coordinate axes; mixed partials ( $f_{xy}, f_{yx}$ ) describe twisting and cross-coupling shear across variables.
Clairaut Symmetry: For any smooth function ( $C^2$ ), mixed partials are commutative: $f_{xy} = f_{yx}$ . This symmetry guarantees that the Hessian matrix is symmetric ( $\mathbf{H}_f^T = \mathbf{H}_f$ ), saving computational effort in finite element analysis and optimization algorithms.
Preview of Optimization: In Session 13, the determinant of the Hessian matrix ( $D = f_{xx}f_{yy} - (f_{xy})^2$ ) will serve as the discriminant in the Second Derivative Test to classify local minima, local maxima, and saddle points.

Synthesis Wrap-up

Core Takeaways from Session 11:
1. Chain Rule Case 1: The total rate along a parameter path $\vec{r}(t)$ is $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} = \nabla z \cdot \vec{v}(t)$ .
2. Chain Rule Case 2 & Implicit Differentiation: Tree diagrams map multi-parameter systems; implicit formulas $\frac{dy}{dx} = -\frac{F_x}{F_y}$ and $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$ compute slopes without isolating variables.
3. Higher Partials & Clairaut: Smooth functions have symmetric mixed partials ( $f_{xy} = f_{yx}$ ), populating the symmetric Hessian curvature matrix $\mathbf{H}_f$ .
Homework Assignment: Complete the Knewton Alta assignment: Second Partial Derivative and Chain Rule.
Looking Ahead to Session 12: Next session, we combine spatial partial derivatives into the full vector operator—the Gradient Vector $\nabla f$ —and use it to calculate Directional Derivatives along arbitrary spatial headings and identify paths of steepest ascent.\n