Session 11 - Chain Rule and Higher Derivatives

📅 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.

Segment 1 of 8

Skill Block 1

Real-Life Engineering Applications

🏗️

Civil Engineering

Bridge Truss Gusset Plate Fatigue (Mechanical & Civil Engineering)

During a structural health inspection of a steel bridge truss, engineers monitor the stress intensity factor $K$ (in $\text{MPa}\cdot\sqrt{\text{m}}$ ) around a critical micro-crack on a primary gusset plate. The intensity factor depends on the applied tension stress $\sigma$ (in $\text{MPa}$ ) and the physical length of the crack $a$ (in meters), modeled by:

K(\sigma, a) = 1.5\sigma \sqrt{a} = 1.5\sigma a^{0.5}

As heavy morning commuter traffic moves onto the bridge, the load-induced stress peaks. Simultaneously, due to structural fatigue, the crack lengthens dynamically over time $t$ (in minutes) according to the parameters:

$\sigma(t) = 200 + 10t^2$
$a(t) = 0.04 + 0.002t$

Compute the baseline tensile stress $\sigma(2)$ and crack length $a(2)$ at $t = 2\text{ minutes}$ .
State the multivariable chain rule formula for the total time rate of change $\frac{dK}{dt}$ .
Calculate the instantaneous rate of change of the stress intensity factor $\frac{dK}{dt}$ at $t = 2\text{ minutes}$ .
In fracture mechanics, crack propagation becomes unstable when $K \ge K_{Ic}$ (fracture toughness). Explain why structural health monitoring systems track the rate $\frac{dK}{dt}$ rather than just the instantaneous magnitude $K$ .

💡 Show Worked Derivation

During a structural health inspection of a steel bridge truss, engineers monitor the stress intensity factor $K$ (in $\text{MPa}\cdot\sqrt{\text{m}}$ ) around a critical micro-crack on a primary gusset plate. The intensity factor depends on the applied tension stress $\sigma$ (in $\text{MPa}$ ) and the physical length of the crack $a$ (in meters), modeled by:

K(\sigma, a) = 1.5\sigma \sqrt{a} = 1.5\sigma a^{0.5}

As heavy morning commuter traffic moves onto the bridge, the load-induced stress peaks. Simultaneously, due to structural fatigue, the crack lengthens dynamically over time $t$ (in minutes) according to the parameters:

$\sigma(t) = 200 + 10t^2$
$a(t) = 0.04 + 0.002t$

Compute the baseline tensile stress $\sigma(2)$ and crack length $a(2)$ at $t = 2\text{ minutes}$ .
State the multivariable chain rule formula for the total time rate of change $\frac{dK}{dt}$ .
Calculate the instantaneous rate of change of the stress intensity factor $\frac{dK}{dt}$ at $t = 2\text{ minutes}$ .
In fracture mechanics, crack propagation becomes unstable when $K \ge K_{Ic}$ (fracture toughness). Explain why structural health monitoring systems track the rate $\frac{dK}{dt}$ rather than just the instantaneous magnitude $K$ .

Step-by-Step Solution:

Evaluate Baseline Parameters at $t = 2\text{ min}$ :
$\sigma(2) = 200 + 10(2)^2 = 200 + 40 = 240\text{ MPa}$ $a(2) = 0.04 + 0.002(2) = 0.04 + 0.004 = 0.044\text{ m}$ $\sqrt{a(2)} = \sqrt{0.044} \approx 0.20976\text{ m}^{0.5}$
State the Governing Multivariable Chain Rule Formula: Because $K$ depends on intermediate variables $\sigma$ and $a$ , which both vary with independent parameter $t$ :
$\frac{dK}{dt} = \frac{\partial K}{\partial \sigma}\frac{d\sigma}{dt} + \frac{\partial K}{\partial a}\frac{da}{dt}$ (Defined in Multivariable Chain Rule.)
Compute Partial Derivatives and Time Derivatives:
- Partial Derivatives: $\frac{\partial K}{\partial \sigma} = 1.5 a^{0.5} = 1.5 \sqrt{a}$ $\frac{\partial K}{\partial a} = 1.5\sigma \left(0.5 a^{-0.5}\right) = \frac{0.75\sigma}{\sqrt{a}}$
- Time Derivatives: $\frac{d\sigma}{dt} = \frac{d}{dt}[200 + 10t^2] = 20t$ $\frac{da}{dt} = \frac{d}{dt}[0.04 + 0.002t] = 0.002\text{ m/min}$
Evaluate Individual Derivatives at $t = 2\text{ min}$ :
- $\frac{\partial K}{\partial \sigma} = 1.5\sqrt{0.044} \approx 1.5(0.20976) = 0.31464\text{ m}^{0.5}$
- $\frac{\partial K}{\partial a} = \frac{0.75(240)}{\sqrt{0.044}} = \frac{180}{0.20976} \approx 858.11\text{ MPa}/\sqrt{\text{m}}$
- $\frac{d\sigma}{dt} = 20(2) = 40\text{ MPa/min}$
- $\frac{da}{dt} = 0.002\text{ m/min}$
Synthesize Total Derivative:
$\frac{dK}{dt} = (0.31464)(40) + (858.11)(0.002)$ $\frac{dK}{dt} = 12.5857 + 1.7162 = 14.30\text{ MPa}\cdot\sqrt{\text{m}}/\text{min}$
Engineering Interpretation: Tracking $\frac{dK}{dt}$ warns civil inspection engineers of rapid acceleration in structural damage before catastrophic brittle failure occurs. Even if the current intensity $K$ is below the material fracture toughness threshold $K_{Ic}$ (typically $\sim 50\text{ MPa}\cdot\sqrt{\text{m}}$ for structural steel), a large positive derivative $\frac{dK}{dt} = 14.30$ indicates that the bridge joint is rapidly deteriorating under live commuter load and requires immediate lane restrictions.

⚡

Electrical Engineering

Processor Thermal Power Dissipation (Computer & Electrical Engineering)

In modern computer microchip architecture, the local dynamic thermal power dissipation $P$ (in milliwatts, $\text{mW}$ ) of a processing core transistor sub-grid depends on its clock frequency $f$ (in gigahertz, $\text{GHz}$ ) and logic supply voltage $V$ (in volts, $\text{V}$ ), governed by:

P(f, V) = 12fV^2

To handle a demanding computational workload spike, an embedded Dynamic Voltage and Frequency Scaling (DVFS) microcontroller algorithm adjusts both operational parameters concurrently over run-time $t$ (in milliseconds, $\text{ms}$ ):

$f(t) = 2 + \ln(t + 1)$
$V(t) = 1.0 + 0.05t^2$

Compute the operating frequency $f(1)$ and voltage $V(1)$ at $t = 1\text{ ms}$ .
State the multivariable chain rule formula for the total time derivative $\frac{dP}{dt}$ .
Calculate the instantaneous rate of change of power dissipation $\frac{dP}{dt}$ at $t = 1\text{ ms}$ .
Explain how modern thermal management firmware uses the partial sensitivities $\frac{\partial P}{\partial f}$ and $\frac{\partial P}{\partial V}$ to prevent processor thermal throttling.

💡 Show Worked Derivation

In modern computer microchip architecture, the local dynamic thermal power dissipation $P$ (in milliwatts, $\text{mW}$ ) of a processing core transistor sub-grid depends on its clock frequency $f$ (in gigahertz, $\text{GHz}$ ) and logic supply voltage $V$ (in volts, $\text{V}$ ), governed by:

P(f, V) = 12fV^2

To handle a demanding computational workload spike, an embedded Dynamic Voltage and Frequency Scaling (DVFS) microcontroller algorithm adjusts both operational parameters concurrently over run-time $t$ (in milliseconds, $\text{ms}$ ):

$f(t) = 2 + \ln(t + 1)$
$V(t) = 1.0 + 0.05t^2$

Compute the operating frequency $f(1)$ and voltage $V(1)$ at $t = 1\text{ ms}$ .
State the multivariable chain rule formula for the total time derivative $\frac{dP}{dt}$ .
Calculate the instantaneous rate of change of power dissipation $\frac{dP}{dt}$ at $t = 1\text{ ms}$ .
Explain how modern thermal management firmware uses the partial sensitivities $\frac{\partial P}{\partial f}$ and $\frac{\partial P}{\partial V}$ to prevent processor thermal throttling.

Step-by-Step Solution:

Evaluate Operating State at $t = 1\text{ ms}$ :
- Frequency: $f(1) = 2 + \ln(1 + 1) = 2 + \ln(2) \approx 2 + 0.69315 = 2.69315\text{ GHz}$
- Voltage: $V(1) = 1.0 + 0.05(1)^2 = 1.05\text{ V}$
- Baseline Power: $P = 12(2.69315)(1.05)^2 \approx 12(2.69315)(1.1025) \approx 35.63\text{ mW}$
State Governing Multivariable Chain Rule Formula: Since $P$ depends on intermediate parameters $f$ and $V$ , which are parameterized by run-time $t$ :
$\frac{dP}{dt} = \frac{\partial P}{\partial f}\frac{df}{dt} + \frac{\partial P}{\partial V}\frac{dV}{dt}$ (Defined in Multivariable Chain Rule.)
Compute Partial and Parametric Derivatives:
- Partial Derivatives: $\frac{\partial P}{\partial f} = 12V^2$ $\frac{\partial P}{\partial V} = 24fV$
- Time Derivatives: $\frac{df}{dt} = \frac{d}{dt}[2 + \ln(t + 1)] = \frac{1}{t + 1}$ $\frac{dV}{dt} = \frac{d}{dt}[1.0 + 0.05t^2] = 0.10t$
Evaluate Individual Derivatives at $t = 1\text{ ms}$ :
- $\frac{\partial P}{\partial f} = 12(1.05)^2 = 12(1.1025) = 13.23\text{ mW/GHz}$
- $\frac{\partial P}{\partial V} = 24(2.69315)(1.05) \approx 67.867\text{ mW/V}$
- $\frac{df}{dt} = \frac{1}{1 + 1} = 0.50\text{ GHz/ms}$
- $\frac{dV}{dt} = 0.10(1) = 0.10\text{ V/ms}$
Assemble Total Rate of Change:
$\frac{dP}{dt} = (13.23)(0.50) + (67.867)(0.10)$ $\frac{dP}{dt} = 6.615 + 6.787 = 13.40\text{ mW/ms}$
Firmware Sensitivity Analysis: Notice that $\frac{\partial P}{\partial V} = 67.87\text{ mW/V}$ is more than 5 times larger than $\frac{\partial P}{\partial f} = 13.23\text{ mW/GHz}$ . Because dynamic power scales quadratically with voltage ( $V^2$ ) but only linearly with frequency ( $f$ ), DVFS algorithms prioritize reducing voltage first during thermal emergencies to maximize temperature drops while minimizing clock frequency penalties.

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Multivariable Chain Rule (Single Parameter Paths & Trajectories)

General Overview
Skill Block 1 covers the formulation and evaluation of the multivariable chain rule when intermediate spatial variables depend on a single driving parameter (typically time $t$ ). Students learn how to construct dependency trees, distinguish partial sensitivities from total time rates, express the chain rule as a vector dot product between spatial gradients and path velocities, and evaluate dynamic physical rates in mechanical, thermal, and environmental systems.

Conceptual Framework

The Total Time Derivative (Case 1): Given $z = f(x, y)$ where $x = x(t)$ and $y = y(t)$ : $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$ (Defined in Multivariable Chain Rule.)
Extension to Three Spatial Variables: Given $w = f(x, y, z)$ along trajectory path $\vec{r}(t) = \langle x(t), y(t), z(t) \rangle$ : $\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt}$
Vector Dot Product Interpretation: $\frac{dz}{dt} = \langle z_x, z_y \rangle \cdot \left\langle \frac{dx}{dt}, \frac{dy}{dt} \right\rangle = \nabla z \cdot \vec{v}(t)$ The rate of change experienced by a moving sensor equals the projection of the spatial gradient along its velocity vector.

Key Equations & Structural Rules

Branch Multiplication Rule: Multiply derivative rates along each branch from the top output variable down to the bottom driving parameter.
Path Addition Rule: Add the products of all parallel branches terminating at the same independent parameter.
Partial vs. Total Distinction: $\frac{\partial z}{\partial x}$ is the static spatial slope holding $y$ fixed; $\frac{dz}{dt}$ is the dynamic time rate of change experienced along the path.

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 $t = 0\text{ hours}$ .

💡 Reveal 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 $\frac{dK}{dt}$ ?

Practice Space Solve this problem on paper or tablet.

Review the Conceptual Framework and the solutions to Worked Example A and the linked bridge gusset application. (Defined in Multivariable Chain Rule.)

Answer: * (a) Because $z(t) = f(x(t), y(t))$ is ultimately a function of a single independent variable $t$ , making its derivative an ordinary derivative, whereas $f(x, y)$ depends on multiple independent spatial coordinates.

(b) Because at $t = 0$ , $y = 0$ , which causes the spatial concentration gradient in the $y$ -direction ( $\frac{\partial C}{\partial y}$ ) to vanish identically.
(c) Because the crack length appears under a square root in the denominator of $\frac{\partial K}{\partial a} = \frac{0.75\sigma}{\sqrt{a}}$ , making the sensitivity to crack growth exceptionally high when cracks are small.

Step-by-Step Execution

(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.

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 $t = 0\text{ seconds}$ .

Practice Space Solve this problem on paper or tablet.

State the Multivariable Chain Rule for a Single Parameter: $\frac{dT}{dt} = \frac{\partial T}{\partial x}\frac{dx}{dt} + \frac{\partial T}{\partial y}\frac{dy}{dt}$ . Differentiate $T(x, y)$ partially, differentiate the path equations with respect to $t$ , evaluate at $t=0$ , and synthesize. (Defined in Multivariable Chain Rule.)

Answer:

\frac{dT}{dt}\Big|_{t=0} = 6^\circ\text{C/s}

Step-by-Step Execution

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:
$\frac{dT}{dt} = (0)(2) + (6)(1) = 0 + 6 = 6^\circ\text{C/s}$

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 $t$ .

Practice Space Solve this problem on paper or tablet.

State the Multivariable Chain Rule: $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$ . Compute partials with respect to $x$ and $y$ , differentiate $x(t)$ and $y(t)$ , and substitute $x = e^t, y = \cos(t)$ . (Defined in Multivariable Chain Rule.)

Answer:

\frac{dz}{dt} = 6e^{2t}\cos(t) - 3e^{2t}\sin(t) - \sin(t)\cos(\cos(t))

Step-by-Step Execution

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 $(e^t)^2 = e^{2t}$ ): $\frac{dz}{dt} = 6e^{2t}\cos(t) - 3e^{2t}\sin(t) - \sin(t)\cos(\cos(t))$

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 $t = 0.25\text{ seconds}$ .

Practice Space Solve this problem on paper or tablet.

State the Multivariable Chain Rule: $\frac{dh}{dt} = \frac{\partial h}{\partial r}\frac{dr}{dt} + \frac{\partial h}{\partial \theta}\frac{d\theta}{dt}$ . Evaluate $r$ and $\theta$ at $t = 0.25$ , compute partial and time derivatives, and evaluate. (Defined in Multivariable Chain Rule.)

Answer:

\frac{dh}{dt}\Big|_{t=0.25} = \sqrt{2}\left(1 + \frac{\pi}{8}\right) \approx 1.97

Step-by-Step Execution

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 $\sqrt{2}$ : $\frac{dh}{dt} = \sqrt{2}\left(1 + \frac{\pi}{8}\right) \approx 1.4142(1 + 0.3927) \approx 1.9696 \approx 1.97$

Segment 2 of 8

📅 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).

Segment 3 of 8

Skill Block 2

Real-Life Engineering Applications

⚙️

General Engineering

Groundwater Contaminant Plume in Radial Aquifer (Environmental Engineering)

An environmental engineer is modeling the chemical contaminant concentration $C(x, y)$ (in milligrams per liter, $\text{mg/L}$ ) leaking from an underground storage tank into a uniform, porous sand aquifer:

C(x, y) = 100 e^{-0.02(x^2 + y^2)}

where $x$ and $y$ represent Cartesian coordinates in meters measured relative to the source leak.

To monitor groundwater remediation and design circular extraction well barriers, the engineering team transforms the analytical model from Cartesian coordinates to polar coordinates ( $r, \theta$ ) via:

x(r, \theta) = r\cos\theta, \qquad y(r, \theta) = r\sin\theta

State the Case 2 Multivariable Chain Rule formulas for $\frac{\partial C}{\partial r}$ and $\frac{\partial C}{\partial \theta}$ .
Compute the Cartesian partial derivatives $\frac{\partial C}{\partial x}$ and $\frac{\partial C}{\partial y}$ .
Evaluate the radial concentration gradient $\frac{\partial C}{\partial r}$ at a monitoring radius of $r = 10\text{ m}$ .
Show analytically that $\frac{\partial C}{\partial \theta} = 0$ , and interpret the environmental significance of this result for circular barrier placement.

💡 Show Worked Derivation

An environmental engineer is modeling the chemical contaminant concentration $C(x, y)$ (in milligrams per liter, $\text{mg/L}$ ) leaking from an underground storage tank into a uniform, porous sand aquifer:

C(x, y) = 100 e^{-0.02(x^2 + y^2)}

where $x$ and $y$ represent Cartesian coordinates in meters measured relative to the source leak.

To monitor groundwater remediation and design circular extraction well barriers, the engineering team transforms the analytical model from Cartesian coordinates to polar coordinates ( $r, \theta$ ) via:

x(r, \theta) = r\cos\theta, \qquad y(r, \theta) = r\sin\theta

State the Case 2 Multivariable Chain Rule formulas for $\frac{\partial C}{\partial r}$ and $\frac{\partial C}{\partial \theta}$ .
Compute the Cartesian partial derivatives $\frac{\partial C}{\partial x}$ and $\frac{\partial C}{\partial y}$ .
Evaluate the radial concentration gradient $\frac{\partial C}{\partial r}$ at a monitoring radius of $r = 10\text{ m}$ .
Show analytically that $\frac{\partial C}{\partial \theta} = 0$ , and interpret the environmental significance of this result for circular barrier placement.

Step-by-Step Solution:

State the Governing Multivariable Chain Rule (Case 2): Because $C$ depends on intermediate variables $x$ and $y$ , which each depend on independent polar coordinates $r$ and $\theta$ :
$\frac{\partial C}{\partial r} = \frac{\partial C}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial C}{\partial y}\frac{\partial y}{\partial r}$ $\frac{\partial C}{\partial \theta} = \frac{\partial C}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial C}{\partial y}\frac{\partial y}{\partial \theta}$ (Defined in Multivariable Chain Rule.)
Compute Cartesian Partial Derivatives: Using the single-variable exponential chain rule:
$\frac{\partial C}{\partial x} = 100 e^{-0.02(x^2+y^2)} \cdot \frac{\partial}{\partial x}[-0.02(x^2+y^2)] = 100 e^{-0.02(x^2+y^2)} (-0.04x) = -4x e^{-0.02(x^2+y^2)}$ $\frac{\partial C}{\partial y} = 100 e^{-0.02(x^2+y^2)} \cdot \frac{\partial}{\partial y}[-0.02(x^2+y^2)] = 100 e^{-0.02(x^2+y^2)} (-0.04y) = -4y e^{-0.02(x^2+y^2)}$
Compute Polar Intermediate Derivatives:
- With respect to $r$ : $\frac{\partial x}{\partial r} = \cos\theta, \qquad \frac{\partial y}{\partial r} = \sin\theta$
- With respect to $\theta$ : $\frac{\partial x}{\partial \theta} = -r\sin\theta, \qquad \frac{\partial y}{\partial \theta} = r\cos\theta$
Assemble and Simplify $\frac{\partial C}{\partial r}$ :
$\frac{\partial C}{\partial r} = \left(-4x e^{-0.02(x^2+y^2)}\right)(\cos\theta) + \left(-4y e^{-0.02(x^2+y^2)}\right)(\sin\theta)$
Factor out $-4 e^{-0.02(x^2+y^2)}$ :
$\frac{\partial C}{\partial r} = -4 e^{-0.02(x^2+y^2)} (x\cos\theta + y\sin\theta)$
Substitute $x = r\cos\theta$ , $y = r\sin\theta$ , and $x^2 + y^2 = r^2$ :
$x\cos\theta + y\sin\theta = r\cos^2\theta + r\sin^2\theta = r(\cos^2\theta + \sin^2\theta) = r$

$\frac{\partial C}{\partial r} = -4r e^{-0.02r^2}$
- Evaluate at $r = 10\text{ m}$ : $\frac{\partial C}{\partial r}\Big|_{r=10} = -4(10) e^{-0.02(10)^2} = -40 e^{-2.0} \approx -40(0.1353) = -5.41\text{ (mg/L)/m}$ The concentration decreases at $5.41\text{ mg/L}$ per meter as one moves outward from the source.
Assemble and Simplify $\frac{\partial C}{\partial \theta}$ :
$\frac{\partial C}{\partial \theta} = \left(-4x e^{-0.02(x^2+y^2)}\right)(-r\sin\theta) + \left(-4y e^{-0.02(x^2+y^2)}\right)(r\cos\theta)$ Factor out $-4 e^{-0.02 r^2}$ : $\frac{\partial C}{\partial \theta} = -4 e^{-0.02 r^2} \left( -x r\sin\theta + y r\cos\theta \right)$ Substitute $x = r\cos\theta$ and $y = r\sin\theta$ : $-x r\sin\theta + y r\cos\theta = -(r\cos\theta)(r\sin\theta) + (r\sin\theta)(r\cos\theta) = -r^2\sin\theta\cos\theta + r^2\sin\theta\cos\theta = 0$ $\frac{\partial C}{\partial \theta} = 0$
Environmental Engineering Interpretation: Because $\frac{\partial C}{\partial \theta} = 0$ , the plume exhibits perfect radial symmetry (isotropic dispersion). All monitoring wells drilled at radius $r = 10\text{ m}$ will measure identical pollutant concentrations ( $C(10) = 100e^{-2} \approx 13.53\text{ mg/L}$ ) regardless of angular orientation $\theta$ , allowing engineers to build concentric extraction rings that treat the entire perimeter equally.

⚡

Electrical Engineering

Photovoltaic Solar Cell Dynamic Impedance (Electrical & Energy Engineering)

An electrical energy systems engineer is designing a Maximum Power Point Tracking (MPPT) inverter for a utility-scale photovoltaic solar array. The non-linear current-voltage ( $I\text{-}V$ ) relationship of the solar cell is defined implicitly by the diode equation:

F(I, V) = I - I_{sc} + I_0 \left(e^{\beta V} - 1\right) = 0

where:

$I$ is output current in Amperes ( $\text{A}$ ),
$V$ is terminal voltage in Volts ( $\text{V}$ ),
$I_{sc} = 8.0\text{ A}$ is the short-circuit illumination current,
$I_0 = 1.0 \times 10^{-6}\text{ A}$ is the reverse saturation dark current,
$\beta = \frac{q}{n k T} = 25.0\text{ V}^{-1}$ is the thermal voltage scaling coefficient.

At the target operating point, the cell operates at voltage $V = 0.60\text{ V}$ , which yields a current of $I \approx 4.73\text{ A}$ .

State the multivariable implicit differentiation formula for the incremental voltage-current derivative $\frac{dV}{dI}$ .
Compute the partial derivatives $F_I = \frac{\partial F}{\partial I}$ and $F_V = \frac{\partial F}{\partial V}$ .
Calculate the small-signal dynamic output resistance $R_d = -\frac{dV}{dI}$ of the solar cell at $V = 0.60\text{ V}$ .
Explain how MPPT control loops use the ratio $\frac{dV}{dI}$ to verify that the array is operating at peak power ( $P_{\max}$ ).

💡 Show Worked Derivation

An electrical energy systems engineer is designing a Maximum Power Point Tracking (MPPT) inverter for a utility-scale photovoltaic solar array. The non-linear current-voltage ( $I\text{-}V$ ) relationship of the solar cell is defined implicitly by the diode equation:

F(I, V) = I - I_{sc} + I_0 \left(e^{\beta V} - 1\right) = 0

where:

$I$ is output current in Amperes ( $\text{A}$ ),
$V$ is terminal voltage in Volts ( $\text{V}$ ),
$I_{sc} = 8.0\text{ A}$ is the short-circuit illumination current,
$I_0 = 1.0 \times 10^{-6}\text{ A}$ is the reverse saturation dark current,
$\beta = \frac{q}{n k T} = 25.0\text{ V}^{-1}$ is the thermal voltage scaling coefficient.

At the target operating point, the cell operates at voltage $V = 0.60\text{ V}$ , which yields a current of $I \approx 4.73\text{ A}$ .

State the multivariable implicit differentiation formula for the incremental voltage-current derivative $\frac{dV}{dI}$ .
Compute the partial derivatives $F_I = \frac{\partial F}{\partial I}$ and $F_V = \frac{\partial F}{\partial V}$ .
Calculate the small-signal dynamic output resistance $R_d = -\frac{dV}{dI}$ of the solar cell at $V = 0.60\text{ V}$ .
Explain how MPPT control loops use the ratio $\frac{dV}{dI}$ to verify that the array is operating at peak power ( $P_{\max}$ ).

Step-by-Step Solution:

State Governing Multivariable Implicit Differentiation Formula: For a level curve defined implicitly by $F(I, V) = 0$ :
$\frac{dV}{dI} = -\frac{F_I}{F_V} = -\frac{\frac{\partial F}{\partial I}}{\frac{\partial F}{\partial V}}$ (Defined in Multivariable Chain Rule.)
Compute Partial Derivatives of $F(I, V)$ : Treating $V$ as a constant:
$F_I = \frac{\partial}{\partial I}\left[ I - I_{sc} + I_0(e^{\beta V} - 1) \right] = 1$ Treating $I$ as a constant: $F_V = \frac{\partial}{\partial V}\left[ I - I_{sc} + I_0(e^{\beta V} - 1) \right] = I_0 \cdot (\beta e^{\beta V}) = \beta I_0 e^{\beta V}$
Assemble the Symbolic Derivative:
$\frac{dV}{dI} = -\frac{1}{\beta I_0 e^{\beta V}}$
Evaluate at Operating Point $V = 0.60\text{ V}$ :
- Exponent argument: $\beta V = (25.0\text{ V}^{-1})(0.60\text{ V}) = 15.0$
- Exponential term: $e^{15.0} \approx 3.2690 \times 10^6$
- Denominator evaluation: $F_V = (25.0)(1.0 \times 10^{-6})(3.2690 \times 10^6) = 25.0 \times 3.2690 \approx 81.725\text{ A/V (or }\Omega^{-1}\text{)}$
- Incremental rate of change: $\frac{dV}{dI} = -\frac{1}{81.725} \approx -0.012236\text{ V/A}$
- Dynamic Resistance: $R_d = -\frac{dV}{dI} = \frac{1}{81.725} \approx 0.0122\text{ }\Omega = 12.2\text{ m}\Omega$
MPPT Power Optimization Meaning: Electric power is $P = V \cdot I$ . Differentiating with respect to $V$ :
$\frac{dP}{dV} = I + V \frac{dI}{dV} = 0 \implies -\frac{dV}{dI} = \frac{V}{I}$ At maximum power transfer, the dynamic incremental resistance $R_d = -\frac{dV}{dI}$ must match the static load resistance $R_{\text{load}} = \frac{V}{I}$ . Multivariable implicit differentiation allows microcontroller firmware to compute this dynamic impedance in microseconds without solving transcendental equations.

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Chain Rule for Several Variables & Implicit Differentiation

General Overview
Skill Block 2 expands the multivariable chain rule to systems with two or more independent parameters (such as surface parameterizations and coordinate changes $x(u, v), y(u, v)$ ) and establishes the multivariable technique for implicit differentiation. Students learn how to draw multi-tiered dependency trees, convert spatial derivatives between coordinate systems (e.g., Cartesian to Polar), and compute tangent slopes of implicit curves $F(x, y) = 0$ and surface gradients of implicit level sets $F(x, y, z) = 0$ .

Conceptual Framework

Chain Rule for Two Independent Parameters (Case 2): Given $z = f(x, y)$ where $x = g(u, v)$ and $y = h(u, v)$ : $\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$ $\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial v}$ (Defined in Multivariable Chain Rule.)
Implicit Differentiation for Curves ( $F(x, y) = 0$ ): $\frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}, \quad \text{where } F_y \neq 0$
Implicit Differentiation for Surfaces ( $F(x, y, z) = 0$ ): $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \qquad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z}, \quad \text{where } F_z \neq 0$

Key Equations & Structural Rules

Tree Diagram Construction: The number of independent variables at the bottom level dictates the number of separate partial derivative formulas required.
The Implicit Differentiation Minus Sign: The negative sign in $-\frac{F_x}{F_y}$ is an algebraic consequence of moving $F_x$ across the equals sign in $F_x dx + F_y dy = 0$ .

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 $(u, v) = (2, 1)$ .

💡 Reveal 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?

Practice Space Solve this problem on paper or tablet.

Review the Conceptual Framework, tree diagram rules, and the solutions to Worked Example A and the linked environmental engineering application. (Defined in Multivariable Chain Rule.)

Answer: * (a) Three product terms: $\frac{\partial w}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial u} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial u}$ .

(b) From isolating $\frac{dy}{dx}$ in the total differential equation $dF = F_x dx + F_y dy = 0 \implies F_y dy = -F_x dx$ .
(c) Because the contaminant plume has perfect circular (isotropic) symmetry, meaning concentration does not vary along circular concentric contours.

Step-by-Step Execution

(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 $\frac{\partial C}{\partial \theta} = (-4x e^{-0.02r^2})(-r\sin\theta) + (-4y e^{-0.02r^2})(r\cos\theta) = 4r^2 e^{-0.02r^2}(\cos\theta\sin\theta - \sin\theta\cos\theta) = 0$ .

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 $(r, \theta) = (3, \frac{\pi}{6})$ .

Practice Space Solve this problem on paper or tablet.

State the Multivariable Chain Rule for Several Variables: $\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$ . Compute $z_x = 8x, z_y = 2y$ , $\frac{\partial x}{\partial r} = \cos\theta$ , $\frac{\partial y}{\partial r} = \sin\theta$ , and evaluate at $(r, \theta) = (3, \pi/6)$ . (Defined in Multivariable Chain Rule.)

Answer: 1. $\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\cos(\theta) + \frac{\partial z}{\partial y}\sin(\theta) = 8r\cos^2(\theta) + 2r\sin^2(\theta)$ 2. $\frac{\partial z}{\partial r}\Big|_{(3, \pi/6)} = \frac{39}{2} = 19.5$

Step-by-Step Execution

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 $r = 3$ : $\frac{\partial z}{\partial r} = 8(3)\left(\frac{3}{4}\right) + 2(3)\left(\frac{1}{4}\right) = 24\left(\frac{3}{4}\right) + 6\left(\frac{1}{4}\right) = 18 + 1.5 = 19.5 = \frac{39}{2}$

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)

.

Practice Space Solve this problem on paper or tablet.

State the Multivariable Implicit Differentiation Formula for Curves: $\frac{dy}{dx} = -\frac{F_x}{F_y}$ . Define $F(x, y) = x^3 + y^3 - 6xy$ , compute $F_x$ and $F_y$ , and evaluate at $(3, 3)$ . (Defined in Multivariable Chain Rule.)

Answer:

\frac{dy}{dx}\Big|_{(3, 3)} = -1

Step-by-Step Execution

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 $y = x$ .

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)

.

Practice Space Solve this problem on paper or tablet.

State the Multivariable Implicit Differentiation Formulas for Surfaces: $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$ and $\frac{\partial z}{\partial y} = -\frac{F_y}{F_z}$ . Set $F(x, y, z) = x^2 y + y^2 z + z^2 x - 7 = 0$ , compute $F_x, F_y, F_z$ , and evaluate at $(1, 1, 2)$ . (Defined in Multivariable Chain Rule.)

Answer:

\frac{\partial z}{\partial x}\Big|_{(1, 1, 2)} = -\frac{6}{5}, \qquad \frac{\partial z}{\partial y}\Big|_{(1, 1, 2)} = -1

Step-by-Step Execution

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:
$\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{6}{5}$ $\frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = -\frac{5}{5} = -1$

Segment 4 of 8

📅 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!).

Segment 5 of 8

Skill Block 3

Real-Life Engineering Applications

🏗️

Civil Engineering

Structural Column Strain Energy Stability (Civil & Mechanical Engineering)

A load-bearing structural steel support column in a commercial building framework experiences internal strain energy $U$ (in Joules, $\text{J}$ ) due to combined axial mechanical compression loads and solar thermal expansion. The total strain energy is modeled as a function of the axial force load $x$ (in kilonewtons, $\text{kN}$ ) and the surface temperature shift $y$ (in degrees Celsius, $^\circ\text{C}$ ):

U(x, y) = 0.5x^2 + 0.04x^2 y + 3y^2

Compute the first-order partial derivatives $U_x$ and $U_y$ .
Compute all four second-order partial derivatives ( $U_{xx}, U_{yy}, U_{xy}, U_{yx}$ ).
Verify Clairaut's Theorem for the mixed partial derivatives.
Construct the $2 \times 2$ Hessian matrix $\mathbf{H}_U(x, y)$ and evaluate its determinant $D = \det(\mathbf{H}_U)$ at nominal service load $(x, y) = (10\text{ kN}, 25^\circ\text{C})$ .
Interpret the physical structural meaning of the mixed partial derivative $U_{xy}$ and explain why $U_{xx} > 0$ confirms structural stability against buckling.

💡 Show Worked Derivation

A load-bearing structural steel support column in a commercial building framework experiences internal strain energy $U$ (in Joules, $\text{J}$ ) due to combined axial mechanical compression loads and solar thermal expansion. The total strain energy is modeled as a function of the axial force load $x$ (in kilonewtons, $\text{kN}$ ) and the surface temperature shift $y$ (in degrees Celsius, $^\circ\text{C}$ ):

U(x, y) = 0.5x^2 + 0.04x^2 y + 3y^2

Compute the first-order partial derivatives $U_x$ and $U_y$ .
Compute all four second-order partial derivatives ( $U_{xx}, U_{yy}, U_{xy}, U_{yx}$ ).
Verify Clairaut's Theorem for the mixed partial derivatives.
Construct the $2 \times 2$ Hessian matrix $\mathbf{H}_U(x, y)$ and evaluate its determinant $D = \det(\mathbf{H}_U)$ at nominal service load $(x, y) = (10\text{ kN}, 25^\circ\text{C})$ .
Interpret the physical structural meaning of the mixed partial derivative $U_{xy}$ and explain why $U_{xx} > 0$ confirms structural stability against buckling.

Step-by-Step Solution:

Compute First-Order Partial Derivatives: State the First-Order Partial Derivative Definitions (see Partial Derivatives):
$U_x = \frac{\partial}{\partial x}[0.5x^2 + 0.04x^2 y + 3y^2] = 1.0x + 0.08xy$ $U_y = \frac{\partial}{\partial y}[0.5x^2 + 0.04x^2 y + 3y^2] = 0.04x^2 + 6y$
Compute Pure and Mixed Second-Order Partials: State the Second-Order Partial Derivative Formulas (see Second-Order Partial Derivatives):
- Pure Partials: $U_{xx} = \frac{\partial}{\partial x}[x + 0.08xy] = 1.0 + 0.08y$ $U_{yy} = \frac{\partial}{\partial y}[0.04x^2 + 6y] = 6$
- Mixed Partials: $U_{xy} = \frac{\partial}{\partial y}[x + 0.08xy] = 0.08x$ $U_{yx} = \frac{\partial}{\partial x}[0.04x^2 + 6y] = 0.08x$
Verify Clairaut's Theorem:
$U_{xy} = U_{yx} = 0.08x$ Since $U(x, y)$ is a smooth polynomial, its second partials are continuous everywhere on $\mathbb{R}^2$ , verifying Clairaut's Theorem.
Construct Hessian Matrix and Evaluate Determinant: State the Hessian Matrix definition:
$\mathbf{H}_U(x, y) = \begin{bmatrix} U_{xx} & U_{xy} \\ U_{yx} & U_{yy} \end{bmatrix} = \begin{bmatrix} 1.0 + 0.08y & 0.08x \\ 0.08x & 6 \end{bmatrix}$
- Evaluate at $(x, y) = (10, 25)$ :
  - $U_{xx}(10, 25) = 1.0 + 0.08(25) = 1.0 + 2.0 = 3.0\text{ J/kN}^2$
  - $U_{yy}(10, 25) = 6.0\text{ J/}(^\circ\text{C})^2$
  - $U_{xy}(10, 25) = 0.08(10) = 0.8\text{ J/(kN}\cdot^\circ\text{C)}$ $\mathbf{H}_U(10, 25) = \begin{bmatrix} 3.0 & 0.8 \\ 0.8 & 6.0 \end{bmatrix}$
- Determinant: $D = \det(\mathbf{H}_U) = (3.0)(6.0) - (0.8)^2 = 18.0 - 0.64 = 17.36 > 0$
Physical & Structural Interpretation:
- Mixed Partial $U_{xy}$ : Physically represents the thermal-mechanical coupling stiffness. It measures how thermal expansion increases the column's mechanical stiffness response to compressive loading.
- Stability & Convexity: Because $D = 17.36 > 0$ and $U_{xx} = 3.0 > 0$ , the strain energy function is strictly convex (concave upward) in all directions. In structural mechanics, positive curvature of the strain energy potential proves that the column is in a stable equilibrium state, guaranteeing resistance against sudden elastic buckling under these combined thermal-load service conditions.

⚡

Electrical Engineering

Automated Audio Amplifier Loop Gain (Electrical & Computer Engineering)

In professional audio electronics design, an automated gain control circuit stabilizes output volume using an active feedback amplifier loop. The loop amplification gain factor $G$ (dimensionless ratio) depends on an input control bias voltage $v$ (in Volts, $\text{V}$ ) and a digital potentiometer feedback resistance $r$ (in kilo-ohms, $\text{k}\Omega$ , for $r > 0$ ):

G(v, r) = v^3 \ln(r)

Compute the first-order partial derivatives $G_v$ and $G_r$ .
Compute the second-order partial derivatives ( $G_{vv}, G_{rr}, G_{vr}, G_{rv}$ ).
Verify Clairaut's Theorem for the mixed partials $G_{vr}$ and $G_{rv}$ .
Construct the $2 \times 2$ Hessian matrix $\mathbf{H}_G(v, r)$ and evaluate it at the reference bias operating point $(v, r) = (2\text{ V}, e\text{ k}\Omega)$ .
Calculate the Hessian determinant $D = \det(\mathbf{H}_G)$ at $(2, e)$ and determine whether the gain surface exhibits a local extremum or a saddle behavior at this operating point.

💡 Show Worked Derivation

In professional audio electronics design, an automated gain control circuit stabilizes output volume using an active feedback amplifier loop. The loop amplification gain factor $G$ (dimensionless ratio) depends on an input control bias voltage $v$ (in Volts, $\text{V}$ ) and a digital potentiometer feedback resistance $r$ (in kilo-ohms, $\text{k}\Omega$ , for $r > 0$ ):

G(v, r) = v^3 \ln(r)

Compute the first-order partial derivatives $G_v$ and $G_r$ .
Compute the second-order partial derivatives ( $G_{vv}, G_{rr}, G_{vr}, G_{rv}$ ).
Verify Clairaut's Theorem for the mixed partials $G_{vr}$ and $G_{rv}$ .
Construct the $2 \times 2$ Hessian matrix $\mathbf{H}_G(v, r)$ and evaluate it at the reference bias operating point $(v, r) = (2\text{ V}, e\text{ k}\Omega)$ .
Calculate the Hessian determinant $D = \det(\mathbf{H}_G)$ at $(2, e)$ and determine whether the gain surface exhibits a local extremum or a saddle behavior at this operating point.

Step-by-Step Solution:

Compute First-Order Partial Derivatives: State the First-Order Partial Derivative Formulas (see Partial Derivatives):
$G_v = \frac{\partial}{\partial v}[v^3 \ln(r)] = 3v^2 \ln(r)$ $G_r = \frac{\partial}{\partial r}[v^3 \ln(r)] = v^3 \left(\frac{1}{r}\right) = \frac{v^3}{r}$
Compute Second-Order Partial Derivatives: State the Second-Order Partial Derivative Formulas (see Second-Order Partial Derivatives):
- Pure Partials: $G_{vv} = \frac{\partial}{\partial v}[3v^2 \ln(r)] = 6v \ln(r)$ $G_{rr} = \frac{\partial}{\partial r}\left[v^3 r^{-1}\right] = v^3 (-r^{-2}) = -\frac{v^3}{r^2}$
- Mixed Partials: $G_{vr} = \frac{\partial}{\partial r}[3v^2 \ln(r)] = 3v^2 \left(\frac{1}{r}\right) = \frac{3v^2}{r}$ $G_{rv} = \frac{\partial}{\partial v}\left[\frac{v^3}{r}\right] = \frac{3v^2}{r}$
Verify Clairaut's Theorem:
$G_{vr} = G_{rv} = \frac{3v^2}{r}$ For all $r > 0$ , the function $G(v, r)$ is infinitely differentiable ( $C^\infty$ ), confirming Clairaut's Theorem.
Construct and Evaluate Hessian Matrix at $(v, r) = (2, e)$ :
$\mathbf{H}_G(v, r) = \begin{bmatrix} G_{vv} & G_{vr} \\ G_{rv} & G_{rr} \end{bmatrix} = \begin{bmatrix} 6v \ln(r) & \frac{3v^2}{r} \\ \frac{3v^2}{r} & -\frac{v^3}{r^2} \end{bmatrix}$
- Since $\ln(e) = 1$ :
  - $G_{vv}(2, e) = 6(2)(1) = 12$
  - $G_{rr}(2, e) = -\frac{2^3}{e^2} = -\frac{8}{e^2}$
  - $G_{vr}(2, e) = \frac{3(2^2)}{e} = \frac{12}{e}$ $\mathbf{H}_G(2, e) = \begin{bmatrix} 12 & \frac{12}{e} \\ \frac{12}{e} & -\frac{8}{e^2} \end{bmatrix}$
Compute Hessian Determinant and Classify Geometry:
$D = \det(\mathbf{H}_G) = G_{vv} G_{rr} - (G_{vr})^2 = (12)\left(-\frac{8}{e^2}\right) - \left(\frac{12}{e}\right)^2$ $D = -\frac{96}{e^2} - \frac{144}{e^2} = -\frac{240}{e^2} \approx -\frac{240}{7.389} \approx -32.48$ Because $D < 0$ , the gain surface possesses negative Gaussian curvature, confirming that the operating point $(2, e)$ is on a hyperbolic saddle geometry. In feedback control loop design, negative Hessian determinants indicate parameter trade-offs: increasing voltage boosts gain while increasing resistance suppresses gain, requiring active closed-loop stabilization.

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Higher-Order Partial Derivatives, Clairaut's Theorem & The Hessian Matrix

General Overview
Skill Block 3 covers the computation and geometric interpretation of higher-order partial derivatives. Students learn how pure second partials ( $f_{xx}, f_{yy}$ ) quantify surface concavity along coordinate axes, how mixed partials ( $f_{xy}, f_{yx}$ ) quantify cross-variable twisting, how to verify Clairaut's Theorem on the commutativity of mixed partials, and how to construct the Hessian matrix to evaluate structural stability and local curvature.

Conceptual Framework

Pure Second-Order Partial Derivatives: $f_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(f_x), \qquad f_{yy} = \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}(f_y)$ (Defined in Second-Order Partial Derivatives.)
Mixed Second-Order Partial Derivatives: $f_{xy} = \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y}(f_x), \qquad f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x}(f_y)$
Clairaut's Theorem (Schwarz's Theorem): If $f_{xy}$ and $f_{yx}$ are continuous on an open region, then: $f_{xy} = f_{yx}$ (Defined in Clairaut's Theorem.)
The Hessian Matrix & Curvature Determinant: $\mathbf{H}_f(x, y) = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix}, \qquad D = \det(\mathbf{H}_f) = f_{xx} f_{yy} - (f_{xy})^2$

Key Equations & Structural Rules

Lagrange Subscript Order: $f_{xy}$ differentiates with respect to $x$ first, then $y$ .
Leibniz Operator Order: $\frac{\partial^2 f}{\partial y \partial x}$ differentiates with respect to $x$ first (inner operator), then $y$ (outer operator).
Higher-Order Commutativity ( $n \ge 3$ ): For smooth functions ( $C^n$ ), mixed partials commute across all permutations of order (e.g., $f_{xxy} = f_{xyx} = f_{yxx}$ ).

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.

💡 Reveal 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.

Practice Space Solve this problem on paper or tablet.

Review the Conceptual Framework and the solutions to Worked Example A and the linked structural column application. (Defined in Clairaut's Theorem and Second-Order Partial Derivatives.)

Answer: * (a) Both mixed partial derivatives $f_{xy}$ and $f_{yx}$ must be continuous on an open disk containing the point.

(b) Because both variables $x$ and $y$ appear simultaneously in the leading coefficient $2xy$ and inside the exponential argument $x^2 y$ .
(c) $U_{xy}$ represents thermal-mechanical cross-coupling stiffness, and $D > 0$ with $U_{xx} > 0$ confirms that the strain energy is strictly convex (a stable potential energy minimum).

Step-by-Step Execution

(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.

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.

Practice Space Solve this problem on paper or tablet.

State the Second-Order Partial Derivative Formulas (see Second-Order Partial Derivatives). Compute $f_x$ and $f_y$ , taking care to use the single-variable chain rule on $\sin(xy)$ . When computing mixed partials, use the product rule on the trigonometric terms $y\cos(xy)$ and $x\cos(xy)$ . (Defined in Clairaut's Theorem.)

Answer: 1. $f_x = 2xy^3 + 4y + y\cos(xy), \quad f_y = 3x^2 y^2 + 4x + x\cos(xy)$ 2. $f_{xx} = 2y^3 - y^2\sin(xy), \quad f_{yy} = 6x^2 y - x^2\sin(xy)$ 3. $f_{xy} = f_{yx} = 6xy^2 + 4 + \cos(xy) - xy\sin(xy)$

Step-by-Step Execution

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.

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

Practice Space Solve this problem on paper or tablet.

State the Second-Order Partial Derivative Formulas (see Second-Order Partial Derivatives). Compute $f_x$ and $f_y$ using the power rule, then differentiate each with respect to $x$ and $y$ . (Defined in Clairaut's Theorem.)

Answer: - $f_{xx} = 12xy^2$

$f_{yy} = 4x^3$
$f_{xy} = f_{yx} = 12x^2 y + 5$

Step-by-Step Execution

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, $f_{xy} = f_{yx} = 12x^2 y + 5$ .

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.

Practice Space Solve this problem on paper or tablet.

State the Hessian Matrix Formula: $\mathbf{H}_f(x, y) = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix}$ . Compute first and second partials. To find $f_{xxy}$ , differentiate $f_{xx}$ with respect to $y$ . To find $f_{xyx}$ , differentiate $f_{xy}$ with respect to $x$ . (Defined in Second-Order Partial Derivatives and Clairaut's Theorem.)

Answer: 1.

\mathbf{H}_f(x, y) = \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}

2.

f_{xxy} = f_{xyx} = 24x^2 y - 12y^2

Step-by-Step Execution

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.

Segment 6 of 8

📅 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.

Segment 7 of 8

📅 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

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