Session 10 - Midterm Exam 1 Review | Tony's Intertwingled Memex

📅 Micro-Lecture

1. Midterm Exam 1 Architecture & Engineering Problem-Solving Protocol

Mathematical Foundation: Unit 1 of Calculus I establishes the analytical foundation for all modern engineering modeling: transitioning from static algebraic relations to dynamic rates of change. Success on Midterm Exam 1 requires mastery across three distinct operational tiers:
1. The Language of Limits & Continuity: Evaluating behavior near singularities, horizontal/vertical asymptotic envelopes, and verifying continuous physical domains (see Limits, Continuity, and Intermediate Value Theorem).
2. The Limit Definition & Structural Differentiation: Bridging average rates of change to instantaneous rates of change via the difference quotient, and deploying algebraic rules (Power, Product, Quotient) to calculate tangent trajectories (see Definition of the Derivative, Power Rule, Product Rule, and Quotient Rule).
3. Advanced Chain Rule & Implicit Mechanics: Decoupling composite motions, finding tangent slopes along closed geometric constraints, and leveraging logarithms to linearize complex fractional rates (see Chain Rule, Implicit Differentiation, and Logarithmic Differentiation).
Engineering Context: Professional engineers never jump straight to numerical scratchwork. On an exam, you are graded on technical communication and quality control: formally state the governing formula, write out the symbolic relationship, and execute the algebraic steps with dimensional and sign integrity.

2. Exam Day Checklist & Formula Management

No Mystery Formulas: Every rule tested on Midterm Exam 1 originates from first principles. When executing Quotient Rule, recite "low d-high minus high d-low over low squared". When differentiating composite functions, identify the outer shell and multiply by the derivative of the interior.
Managing Indeterminate Forms: Encountering $0/0$ is not an error—it is an invitation to do algebra. Factor polynomials, multiply by radical conjugates, or combine fractions over common denominators until the latent zero-factor cancels.
The 4-Hour Effort Rule & Review Center: Remember our safety nets from the syllabus. Complete your active exam review, audit your redlines, and utilize the Test Review Center to secure homework buy-backs and exam replacement eligibility.

Segment 1 of 8

Skill Block 1

Real-Life Engineering Applications

🏗️

Civil Engineering

Structural Vibration Resonance & Dynamic Load Thresholds (Civil & Mechanical Engineering)

A structural dynamics engineer at UNLV is evaluating the aeroelastic response of an overhead pedestrian skybridge connecting the Thomas & Mack Center to campus parking structures during high-wind gusts. The steady-state dynamic amplification factor $M(\beta)$ of the bridge deck displacement as a function of the frequency ratio $\beta = \frac{\omega}{\omega_n}$ (where $\omega$ is the wind vortex shedding frequency and $\omega_n$ is the bridge's fundamental natural frequency) is governed by the undamped dynamic magnification model:

M(\beta) = \frac{1}{|1 - \beta^2|}

When an active mass damper system is engaged, the transient response amplitude near resonance (

\beta \to 1

) is characterized by the modified limit ratio:

A(\beta) = \frac{\sqrt{3 + \beta} - 2}{\beta - 1}

where

A(\beta)

represents the normalized modal acceleration ratio.

Evaluate the direct substitution behavior of $A(\beta)$ as the frequency ratio $\beta$ approaches resonance ( $\beta \to 1$ ).
Apply algebraic conjugate multiplication to evaluate the exact limit $\lim_{\beta \to 1} A(\beta)$ analytically.
Investigate the undamped behavior $\lim_{\beta \to 1^-} M(\beta)$ and $\lim_{\beta \to 1^+} M(\beta)$ , and explain the structural engineering significance of infinite limits versus finite conjugate limits when sizing tuned mass dampers.

💡 Show Worked Derivation

Problem Overview

We analyze the dynamic amplification and resonant modal acceleration of an engineered pedestrian bridge deck under wind-induced harmonic loading:

M(\beta) = \frac{1}{|1 - \beta^2|}, \qquad A(\beta) = \frac{\sqrt{3 + \beta} - 2}{\beta - 1}

as the operating frequency ratio

\beta = \frac{\omega}{\omega_n}

approaches the critical resonant condition

\beta \to 1

.

Step-by-Step Mathematical Solution

Step 1: Direct Substitution Test on $A(\beta)$

Substitute $\beta = 1$ directly into the damped modal acceleration function $A(\beta)$ :

A(1) = \frac{\sqrt{3 + (1)} - 2}{(1) - 1} = \frac{\sqrt{4} - 2}{0} = \frac{2 - 2}{0} = \frac{0}{0}

The expression yields the indeterminate form

\frac{0}{0}

. This indicates that direct substitution fails and that an algebraic factor of

(\beta - 1)

is latent within the numerator.

Step 2: Evaluate the Limit via Radical Conjugate Multiplication

The governing mathematical technique is Radical Conjugate Rationalization:

\lim_{x \to c} \frac{\sqrt{f(x)} - k}{g(x)} = \lim_{x \to c} \frac{(\sqrt{f(x)} - k)(\sqrt{f(x)} + k)}{g(x)(\sqrt{f(x)} + k)}

(This algebraic strategy is detailed in Limits.)

Multiply the numerator and denominator of $A(\beta)$ by the conjugate $(\sqrt{3 + \beta} + 2)$ :

\lim_{\beta \to 1} A(\beta) = \lim_{\beta \to 1} \frac{(\sqrt{3 + \beta} - 2)(\sqrt{3 + \beta} + 2)}{(\beta - 1)(\sqrt{3 + \beta} + 2)}

Apply the difference of squares identity $(a - b)(a + b) = a^2 - b^2$ :

\lim_{\beta \to 1} A(\beta) = \lim_{\beta \to 1} \frac{(3 + \beta) - 4}{(\beta - 1)(\sqrt{3 + \beta} + 2)} = \lim_{\beta \to 1} \frac{\beta - 1}{(\beta - 1)(\sqrt{3 + \beta} + 2)}

Cancel the common zero-factor $(\beta - 1)$ from the numerator and denominator:

\lim_{\beta \to 1} A(\beta) = \lim_{\beta \to 1} \frac{1}{\sqrt{3 + \beta} + 2}

Evaluate the simplified limit by direct substitution ( $\beta = 1$ ):

\lim_{\beta \to 1} A(\beta) = \frac{1}{\sqrt{3 + (1)} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4} = 0.25

Step 3: Investigate Undamped Resonant Magnification $M(\beta)$

Evaluate the one-sided limits of the undamped magnification model $M(\beta) = \frac{1}{|1 - \beta^2|}$ as $\beta \to 1$ :

As $\beta \to 1^-$ , $\beta^2 \to 1^-$ and $1 - \beta^2 \to 0^+$ , so: $\lim_{\beta \to 1^-} M(\beta) = \lim_{\beta \to 1^-} \frac{1}{|1 - \beta^2|} = \frac{1}{0^+} = +\infty$
As $\beta \to 1^+$ , $\beta^2 \to 1^+$ and $1 - \beta^2 \to 0^-$ , but the absolute value ensures $|1 - \beta^2| \to 0^+$ , so: $\lim_{\beta \to 1^+} M(\beta) = \lim_{\beta \to 1^+} \frac{1}{|1 - \beta^2|} = \frac{1}{0^+} = +\infty$ Because both one-sided limits approach positive infinity, $\lim_{\beta \to 1} M(\beta) = +\infty$ .

Engineering Conclusion

In structural civil engineering, an infinite limit $\lim_{\beta \to 1} M(\beta) = +\infty$ represents catastrophic harmonic resonance—the mathematical phenomenon that destroyed the original Tacoma Narrows Bridge. Without damping, input energy accumulates without bound, exceeding the elastic yield strength of structural steel members.

By installing an active tuned mass damper, engineers introduce phase-lagged viscous reaction forces that eliminate the infinite singularity. Calculating $\lim_{\beta \to 1} A(\beta) = 0.25$ proves that the active damper caps the resonant acceleration to a safe, finite threshold of $25\%$ of baseline uncoupled excitation. This guarantees that wind vortex shedding during Las Vegas desert windstorms will not cause structural fatigue or pedestrian discomfort on the pedestrian skybridge.

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

Limit Foundations, Algebraic Evaluation, & Continuity

General Overview
Skill Block 1 synthesizes algebraic techniques for resolving indeterminate limits ( $0/0$ ), evaluates directional behavior near vertical asymptotes, determines asymptotic horizontal limits at infinity ( $\pm\infty$ ), and enforces the formal three-part definition of continuity and the Intermediate Value Theorem.

Conceptual Framework

Indeterminate Forms ( $0/0$ ): Direct substitution of $x = c$ into a rational or radical fraction often yields $0/0$ . This indicates that $(x - c)$ is a common factor in both numerator and denominator. Techniques to reveal and cancel this factor include:
- Factoring: Factoring polynomials over $\mathbb{R}$ .
- Radical Conjugate Rationalization: Multiplying numerator and denominator by $(\sqrt{A} + B)$ to exploit the difference of squares $(A - B^2)$ .
- Common Denominators: Merging compound rational fractions.
Infinite Limits & Vertical Asymptotes: If substitution yields $\frac{\text{nonzero}}{0}$ , the limit diverges to $+\infty$ or $-\infty$ . Track the sign of the denominator as $x \to c^-$ versus $x \to c^+$ using sign analysis.
Limits at Infinity ( $x \to \pm\infty$ ): For rational functions, divide every term by the highest power of $x$ in the denominator. For expressions with square roots, recall that $\sqrt{x^2} = |x|$ , which equals $+x$ as $x \to +\infty$ and $-x$ as $x \to -\infty$ .
The Three-Part Continuity Rubric: A function $f(x)$ is continuous at $x = c$ if and only if:
1. $f(c)$ is defined ( $c$ is in the domain of $f$ ).
2. $\lim_{x \to c} f(x)$ exists ( $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L \in \mathbb{R}$ ).
3. $\lim_{x \to c} f(x) = f(c)$ .
Intermediate Value Theorem (IVT): If $f(x)$ is continuous on the closed interval $[a, b]$ and $u$ is any target value strictly between $f(a)$ and $f(b)$ , there exists at least one number $c \in (a, b)$ such that $f(c) = u$ .

Key Equations & Structural Rules

Difference of Squares Factorization: $A - B = (\sqrt{A} - \sqrt{B})(\sqrt{A} + \sqrt{B})$ (see Limits)
Absolute Value Radical Identity: $\sqrt{x^2} = |x| = \begin{cases} x, & x > 0 \\ -x, & x < 0 \end{cases}$ (see Limits at Infinity)
Formal Continuity Criterion: $\lim_{x \to c} f(x) = f(c)$ (see Continuity)
Intermediate Value Theorem (IVT): $f \in C[a, b] \text{ and } f(a) < u < f(b) \implies \exists c \in (a, b) \text{ such that } f(c) = u$ (see Intermediate Value Theorem)

Worked Example A: Indeterminate Radical Conjugate Limit

Evaluate the following limit analytically:

\lim_{x \to 4} \frac{\sqrt{x + 5} - 3}{x - 4}

💡 Reveal Worked Solution

Governing Strategy: Radical Conjugate Rationalization for $\frac{0}{0}$ Indeterminate Forms: $\lim_{x \to c} \frac{\sqrt{f(x)} - k}{x - c} = \lim_{x \to c} \frac{(\sqrt{f(x)} - k)(\sqrt{f(x)} + k)}{(x - c)(\sqrt{f(x)} + k)}$ (This algebraic technique is defined in Limits.)
Test direct substitution: $\frac{\sqrt{4 + 5} - 3}{4 - 4} = \frac{\sqrt{9} - 3}{0} = \frac{3 - 3}{0} = \frac{0}{0}$ Direct substitution yields an indeterminate form $\frac{0}{0}$ .
Multiply numerator and denominator by the conjugate $(\sqrt{x + 5} + 3)$ : $\lim_{x \to 4} \frac{(\sqrt{x + 5} - 3)(\sqrt{x + 5} + 3)}{(x - 4)(\sqrt{x + 5} + 3)}$
Expand the difference of squares in the numerator: $\lim_{x \to 4} \frac{(\sqrt{x + 5})^2 - (3)^2}{(x - 4)(\sqrt{x + 5} + 3)} = \lim_{x \to 4} \frac{(x + 5) - 9}{(x - 4)(\sqrt{x + 5} + 3)} = \lim_{x \to 4} \frac{x - 4}{(x - 4)(\sqrt{x + 5} + 3)}$
Cancel the latent zero-factor $(x - 4)$ : $\lim_{x \to 4} \frac{1}{\sqrt{x + 5} + 3}$
Evaluate the simplified limit by direct substitution: $\frac{1}{\sqrt{4 + 5} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$

Practice Problem 1: Concept & Reading Check

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

(a) Under the Conceptual Framework section, state the three mathematical conditions required for a function $f(x)$ to be continuous at $x = c$ .
(b) In Worked Example A, explain why multiplying numerator and denominator by the conjugate is algebraically valid, and identify the exact zero-factor that cancels.
(c) In the linked engineering application Structural Vibration Resonance & Dynamic Load Thresholds (Civil & Mechanical Engineering), explain how evaluating an indeterminate limit reveals the finite dynamic magnification factor at near-resonant frequencies.

Practice Space Solve this problem on paper or tablet.

Review the Three-Part Continuity Rubric and Worked Example A solution steps above.

Answer: * (a) (1) $f(c)$ is defined; (2) $\lim_{x \to c} f(x)$ exists; (3) $\lim_{x \to c} f(x) = f(c)$ .

(b) Multiplying numerator and denominator by the same nonzero expression is equivalent to multiplying by 1. The canceled factor is $(x - 4)$ .
(c) Canceling the singular zero-factor eliminates the infinite resonance singularity and reveals the finite damping acceleration envelope ( $A(\beta) = 0.25$ ).

Step-by-Step Execution

(a) A function $f(x)$ is continuous at $x = c$ if and only if: (1) $f(c)$ is defined ( $c$ is in the domain of $f$ ); (2) $\lim_{x \to c} f(x)$ exists as a finite real number; and (3) $\lim_{x \to c} f(x) = f(c)$ .
(b) Multiplying by $\frac{\sqrt{x+5}+3}{\sqrt{x+5}+3}$ is algebraically equivalent to multiplying the expression by $1$ , which preserves the value of the function for all $x \neq 4$ . Expanding the numerator via the difference of squares produces $(x - 4)$ , which cancels the identical zero-factor $(x - 4)$ in the denominator that was producing the $\frac{0}{0}$ division by zero.
(c) In structural engineering, undamped harmonic excitation predicts an infinite displacement amplitude at resonance ( $\beta = 1$ ). By incorporating active mass damping, the mathematical model transforms into an indeterminate form $\frac{0}{0}$ whose limit resolves to $0.25$ , demonstrating that the damper safely caps mechanical acceleration.

Practice Problem 2: Factoring & Rational Limit

Evaluate the following limit analytically:

\lim_{x \to -3} \frac{x^2 - 9}{2x^2 + 5x - 3}

Practice Space Solve this problem on paper or tablet.

State the Factoring Technique for Rational Limits. Notice that direct substitution gives $\frac{0}{0}$ . Factor the difference of squares in the numerator and factor the quadratic trinomial in the denominator to expose the common factor $(x + 3)$ .

Answer:

\lim_{x \to -3} \frac{x^2 - 9}{2x^2 + 5x - 3} = \frac{6}{7}

Step-by-Step Execution

Governing Formula: The Factoring Technique for Indeterminate Forms: $\lim_{x \to c} \frac{P(x)}{Q(x)} = \lim_{x \to c} \frac{(x - c)P_1(x)}{(x - c)Q_1(x)} = \lim_{x \to c} \frac{P_1(x)}{Q_1(x)}$ (This technique is defined in Limits.)
Test direct substitution: $\frac{(-3)^2 - 9}{2(-3)^2 + 5(-3) - 3} = \frac{9 - 9}{2(9) - 15 - 3} = \frac{0}{18 - 18} = \frac{0}{0}$ Direct substitution confirms the indeterminate form $\frac{0}{0}$ , meaning $(x - (-3)) = (x + 3)$ is a factor of both polynomials.
Factor numerator and denominator:
- Numerator: $x^2 - 9 = (x - 3)(x + 3)$
- Denominator: $2x^2 + 5x - 3 = (2x - 1)(x + 3)$
Cancel the common factor $(x + 3)$ : $\lim_{x \to -3} \frac{(x - 3)(x + 3)}{(2x - 1)(x + 3)} = \lim_{x \to -3} \frac{x - 3}{2x - 1}$
Evaluate by direct substitution ( $x = -3$ ): $\frac{-3 - 3}{2(-3) - 1} = \frac{-6}{-6 - 1} = \frac{-6}{-7} = \frac{6}{7}$

Practice Problem 3: Limit at Infinity with Square Root Radical

Evaluate the limit at negative infinity:

\lim_{x \to -\infty} \frac{\sqrt{4x^2 + 7}}{3x - 1}

Practice Space Solve this problem on paper or tablet.

State the Absolute Value Radical Identity: $\sqrt{x^2} = |x| = -x$ for $x < 0$ . Factor $x^2$ out of the radical as $-x$ , then divide every term in the numerator and denominator by $x$ .

Answer:

\lim_{x \to -\infty} \frac{\sqrt{4x^2 + 7}}{3x - 1} = -\frac{2}{3}

Step-by-Step Execution

Governing Formula: The Radical Limit Identity at Negative Infinity: $\sqrt{x^2} = |x| = -x \quad \text{for } x < 0$ (This rule is defined in Limits at Infinity.)
Factor $x^2$ inside the square root in the numerator: $\sqrt{4x^2 + 7} = \sqrt{x^2\left(4 + \frac{7}{x^2}\right)} = \sqrt{x^2}\sqrt{4 + \frac{7}{x^2}}$
Apply the negative infinity absolute value identity: Because $x \to -\infty$ , $x$ is negative, so $\sqrt{x^2} = -x$ : $\sqrt{4x^2 + 7} = -x\sqrt{4 + \frac{7}{x^2}}$
Rewrite the quotient and divide numerator and denominator by $x$ : $\lim_{x \to -\infty} \frac{-x\sqrt{4 + \frac{7}{x^2}}}{3x - 1} = \lim_{x \to -\infty} \frac{-\sqrt{4 + \frac{7}{x^2}}}{3 - \frac{1}{x}}$
Evaluate the limits of the reciprocal terms ( $\frac{1}{x} \to 0$ and $\frac{7}{x^2} \to 0$ as $x \to -\infty$ ): $\frac{-\sqrt{4 + 0}}{3 - 0} = \frac{-\sqrt{4}}{3} = -\frac{2}{3}$

Practice Problem 4: Piecewise Continuity Parameter Determination

Determine the constant $k$ such that the function $g(x)$ is continuous for all real numbers:

g(x) = \begin{cases} k x^2 + 2x, & x < 2 \\ 5x - k, & x \ge 2 \end{cases}

Practice Space Solve this problem on paper or tablet.

State the Definition of Continuity at a Point: $\lim_{x \to 2^-} g(x) = \lim_{x \to 2^+} g(x) = g(2)$ . Evaluate the left-hand limit using the $x < 2$ rule, evaluate the right-hand value using the $x \ge 2$ rule, set them equal, and solve for $k$ .

Answer:

k = \frac{6}{5} = 1.2

Step-by-Step Execution

Governing Formula: The Three-Part Continuity Criterion at a Piecewise Boundary: $\lim_{x \to c^-} g(x) = \lim_{x \to c^+} g(x) = g(c)$ (This criterion is defined in Continuity.)
Evaluate the left-hand limit ( $x \to 2^-$ ): Use the branch for $x < 2$ : $\lim_{x \to 2^-} g(x) = \lim_{x \to 2^-} \left(k x^2 + 2x\right) = k(2)^2 + 2(2) = 4k + 4$
Evaluate the right-hand limit and function value ( $x \to 2^+$ ): Use the branch for $x \ge 2$ : $g(2) = \lim_{x \to 2^+} g(x) = \lim_{x \to 2^+} (5x - k) = 5(2) - k = 10 - k$
Equate the one-sided limits to enforce continuity: $4k + 4 = 10 - k$
Solve the linear equation for $k$ : $4k + k = 10 - 4$ $5k = 6 \implies k = \frac{6}{5} = 1.2$

Segment 2 of 8

📅 Regroup 1

Conjugate Multiplication Discipline: Never distribute or expand the non-conjugate factor in the denominator during conjugate rationalization. Leaving $(x - c)(\sqrt{\cdots} + k)$ intact ensures the zero-factor $(x - c)$ is immediately visible for cancellation.
Negative Infinity Sign Trap: Forgetting that $\sqrt{x^2} = -x$ when $x \to -\infty$ is one of the most common point deductions on Midterm Exam 1. Always write out $|x| = -x$ explicitly.

Segment 3 of 8

Skill Block 2

Real-Life Engineering Applications

🏗️

Civil Engineering

Cantilever Beam Elastic Deflection & Tangent Stiffness (Civil Engineering)

A civil structural engineer is designing a steel cantilever balcony projection for a multi-story casino resort on the Las Vegas Strip. Under a uniform dead load $w_0$ (in $\text{kN/m}$ ), the downward vertical elastic deflection curve $y(x)$ along the length of the beam ( $0 \le x \le L$ ) is governed by the fourth-degree polynomial:

y(x) = \frac{w_0}{24EI}\left(x^4 - 4Lx^3 + 6L^2x^2\right)

where

E

is the elastic modulus of structural steel (

200\text{ GPa}

),

I

is the cross-sectional moment of inertia (

4.5 \times 10^{-5}\text{ m}^4

),

L = 6.0\text{ m}

is the total cantilever span, and

x

is the horizontal distance measured from the rigid wall support (

x = 0

) to the free tip (

x = L

).

Compute the first derivative $\theta(x) = \frac{dy}{dx}$ of the deflection curve using the Power and Constant Multiple Rules to determine the rotational slope of the beam.
Verify that the fixed clamp condition at the wall support ( $x = 0$ ) satisfies $\theta(0) = 0$ , and compute the exact slope of the beam at the free cantilever tip ( $x = L$ ).
Formulate the equation of the tangent line to the deflection curve at the mid-span location $x = \frac{L}{2} = 3.0\text{ m}$ in terms of the load parameter $C = \frac{w_0}{24EI}$ , and explain how structural engineers use tangent slopes to ensure glass curtain-wall facade panels do not bind or shatter under building sway.

💡 Show Worked Derivation

Problem Overview

We calculate the slope $\theta(x) = \frac{dy}{dx}$ and mid-span tangent line equation for an engineered structural steel cantilever beam governed by the fourth-order elastic deflection equation:

y(x) = \frac{w_0}{24EI}\left(x^4 - 4Lx^3 + 6L^2x^2\right)

over the domain

0 \le x \le L

, where

C = \frac{w_0}{24EI}

is the constant flexural rigidity scaling coefficient and

L = 6.0\text{ m}

.

Step-by-Step Mathematical Solution

Step 1: Compute the First Derivative $\frac{dy}{dx}$

The governing mathematical rules are the Constant Multiple Rule and Power Rule:

\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x), \qquad \frac{d}{dx}[x^n] = n x^{n-1}

(These rules are defined in Power Rule.)

Treating $C = \frac{w_0}{24EI}$ and $L$ as constants, differentiate $y(x)$ term-by-term with respect to $x$ :

\theta(x) = \frac{dy}{dx} = C \cdot \frac{d}{dx}\left[x^4 - 4Lx^3 + 6L^2x^2\right]

\frac{dy}{dx} = C \cdot \left[4x^3 - 4L(3x^2) + 6L^2(2x)\right]

\frac{dy}{dx} = C \cdot \left[4x^3 - 12Lx^2 + 12L^2x\right] = 4Cx\left(x^2 - 3Lx + 3L^2\right)

Substituting back $C = \frac{w_0}{24EI}$ :

\theta(x) = \frac{w_0}{6EI}\left(x^3 - 3Lx^2 + 3L^2x\right)

Step 2: Evaluate Boundary Slopes at Support ( $x = 0$ ) and Tip ( $x = L$ )

At the Wall Support ( $x = 0$ ): $\theta(0) = \frac{w_0}{6EI}\left(0^3 - 3L(0)^2 + 3L^2(0)\right) = 0$ This rigorously confirms the structural boundary condition for a rigid clamp: zero slope at the fixed root.
At the Free Cantilever Tip ( $x = L$ ): $\theta(L) = \frac{w_0}{6EI}\left(L^3 - 3L(L^2) + 3L^2(L)\right) = \frac{w_0}{6EI}\left(L^3 - 3L^3 + 3L^3\right) = \frac{w_0 L^3}{6EI}$ This is the classical structural engineering formula for the maximum tip rotation of a cantilever beam under uniform load.

Step 3: Formulate the Tangent Line at Mid-Span ( $x = L/2 = 3.0\text{ m}$ )

The governing formula is the Point-Slope Equation of a Tangent Line:

y - y_1 = m(x - x_1), \qquad \text{where } m = \left.\frac{dy}{dx}\right|_{x_1}

(This formula is defined in Definition of the Derivative.)

Evaluate the deflection height $y_1$ at $x_1 = L/2$ : $y(L/2) = C \left[\left(\frac{L}{2}\right)^4 - 4L\left(\frac{L}{2}\right)^3 + 6L^2\left(\frac{L}{2}\right)^2\right]$ $y(L/2) = C \left[\frac{L^4}{16} - 4L\left(\frac{L^3}{8}\right) + 6L^2\left(\frac{L^2}{4}\right)\right] = C L^4 \left[\frac{1}{16} - \frac{1}{2} + \frac{3}{2}\right] = C L^4 \left[\frac{1}{16} + 1\right] = \frac{17}{16} C L^4$
Evaluate the tangent slope $m$ at $x_1 = L/2$ : $m = \theta(L/2) = C \left[4\left(\frac{L}{2}\right)^3 - 12L\left(\frac{L}{2}\right)^2 + 12L^2\left(\frac{L}{2}\right)\right]$ $m = C \left[4\left(\frac{L^3}{8}\right) - 12L\left(\frac{L^2}{4}\right) + 6L^3\right] = C L^3 \left[\frac{1}{2} - 3 + 6\right] = \frac{7}{2} C L^3$
Write the Tangent Line Equation:
$y - \frac{17}{16} C L^4 = \frac{7}{2} C L^3 \left(x - \frac{L}{2}\right)$
For $L = 6.0\text{ m}$ :
- $L^3 = 216$ , $L^4 = 1296$ .
- $y_1 = \frac{17}{16} C (1296) = 1377 C$ .
- $m = \frac{7}{2} C (216) = 756 C$ . $y - 1377 C = 756 C (x - 3) \implies y = 756 C x - 891 C$

Engineering Conclusion

In structural civil engineering, computing the instantaneous slope $\frac{dy}{dx}$ is just as critical as computing total vertical displacement. When building multi-story cantilever decks, architectural facade glazing (such as floor-to-ceiling glass curtain walls) attaches directly to cantilever floor slabs. While the slab may safely withstand vertical bending stress, excessive rotational slope $\theta(x)$ imparts destructive in-plane racking and shear strain onto the rigid glass framing. By formulating tangent line equations, structural engineers evaluate the angular clearance envelopes at mid-span and tip connections, ensuring rubber expansion gaskets and slotted slip-joints absorb beam rotation without shattering architectural glass.

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

Difference Quotients & Structural Differentiation Rules

General Overview
Skill Block 2 focuses on evaluating derivatives directly from the definition of the derivative via difference quotients, establishing differentiability conditions, applying Power, Product, and Quotient rules, and constructing tangent and normal line equations.

Conceptual Framework

The Limit Definition of the Derivative: The instantaneous rate of change of $f(x)$ is defined by the difference quotient limit: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ Evaluating this limit algebraically requires canceling $h$ from the denominator before substituting $h = 0$ .
Differentiability vs. Continuity:
- $\text{Differentiable at } c \implies \text{Continuous at } c$ .
- $\text{Continuous at } c \not\implies \text{Differentiable at } c$ (Continuity does not guarantee differentiability).
- Continuous functions fail to be differentiable at sharp corners ( $|x|$ ), cusps ( $x^{2/3}$ ), and vertical tangents ( $x^{1/3}$ ).
The Structural Differentiation Toolbox:
- Power Rule: $\frac{d}{dx}[x^n] = n x^{n-1}$ for any real exponent $n$ .
- Product Rule: Differentiating a product requires two terms: $\frac{d}{dx}[uv] = u'v + uv'$ . Never write $(uv)' = u'v'$ .
- Quotient Rule: $\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}$ . Always square the denominator and subtract the numerator term.
Point-Slope Tangent & Normal Lines:
- Tangent Line: $y - y_1 = m_{\text{tan}}(x - x_1)$ , where $m_{\text{tan}} = f'(x_1)$ .
- Normal Line: $y - y_1 = m_{\text{norm}}(x - x_1)$ , where $m_{\text{norm}} = -\frac{1}{f'(x_1)}$ (provided $f'(x_1) \neq 0$ ).

Key Equations & Structural Rules

Limit Definition of Derivative: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ (see Definition of the Derivative)
Product Rule: $\frac{d}{dx}[u \cdot v] = \frac{du}{dx}v + u\frac{dv}{dx}$ (see Product Rule)
Quotient Rule: $\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{\frac{du}{dx}v - u\frac{dv}{dx}}{v^2}$ (see Quotient Rule)
Point-Slope Equation: $y - y_1 = m(x - x_1)$ (see Derivatives)

Worked Example A: Derivative via Limit Definition (Rational Difference Quotient)

Use the formal limit definition of the derivative $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ to compute $f'(x)$ for:

f(x) = \frac{1}{2x + 1}

💡 Reveal Worked Solution

Governing Formula: The Limit Definition of the Derivative: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ (This definition is detailed in Definition of the Derivative.)
Set up the difference quotient with $f(x) = \frac{1}{2x + 1}$ : $f(x + h) = \frac{1}{2(x + h) + 1} = \frac{1}{2x + 2h + 1}$ $f'(x) = \lim_{h \to 0} \frac{\frac{1}{2x + 2h + 1} - \frac{1}{2x + 1}}{h}$
Find a common denominator for the numerator fractions: The common denominator is $(2x + 2h + 1)(2x + 1)$ : $\frac{1}{2x + 2h + 1} - \frac{1}{2x + 1} = \frac{(2x + 1) - (2x + 2h + 1)}{(2x + 2h + 1)(2x + 1)}$ $(2x + 1) - (2x + 2h + 1) = 2x + 1 - 2x - 2h - 1 = -2h$
Divide by $h$ and cancel the zero-factor: $f'(x) = \lim_{h \to 0} \frac{-2h}{h(2x + 2h + 1)(2x + 1)} = \lim_{h \to 0} \frac{-2}{(2x + 2h + 1)(2x + 1)}$
Evaluate the limit by direct substitution ( $h = 0$ ): $f'(x) = \frac{-2}{(2x + 2(0) + 1)(2x + 1)} = \frac{-2}{(2x + 1)(2x + 1)} = -\frac{2}{(2x + 1)^2}$

Practice Problem 5: Concept & Reading Check

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

(a) Under the Conceptual Framework section, distinguish between an average rate of change and an instantaneous rate of change.
(b) In Worked Example A, explain why the factor of $h$ in the numerator must cancel with the denominator before setting $h = 0$ .
(c) In the linked engineering application Cantilever Beam Elastic Deflection & Tangent Stiffness (Civil Engineering), how does evaluating the first derivative of the deflection equation $y(x)$ yield the physical slope and rotational deflection of the beam?

Practice Space Solve this problem on paper or tablet.

Review the definitions of secant vs. tangent lines and the difference quotient cancellation steps in Worked Example A.

Answer: * (a) Average rate of change is the secant slope over a non-zero interval $[x, x+h]$ ; instantaneous rate of change is the limiting tangent slope as $h \to 0$ .

(b) Setting $h = 0$ prior to cancellation produces an undefined $\frac{0}{0}$ indeterminate form; canceling $h$ exposes the continuous underlying rational function.
(c) The first derivative $\theta(x) = \frac{dy}{dx}$ represents the angular rotation of the beam cross-section, which structural engineers limit to prevent window cracking.

Step-by-Step Execution

(a) An average rate of change measures the net change in output divided by the elapsed input over a finite interval $[x, x+h]$ (geometrically representing the slope of a secant line). An instantaneous rate of change is the limit of that average rate as the interval length $h$ shrinks to zero (geometrically representing the slope of the tangent line at a single point).
(b) Prior to cancellation, the difference quotient at $h = 0$ evaluates to $\frac{f(x)-f(x)}{0} = \frac{0}{0}$ . The factor of $h$ in the denominator is the mathematical cause of this singularity. Finding a common denominator allows factoring $h$ out of the numerator, and canceling $h/h = 1$ removes the singularity so direct substitution can be executed.
(c) In the Euler-Bernoulli beam equation, vertical deflection $y(x)$ describes beam displacement. Its first derivative $\theta(x) = \frac{dy}{dx}$ gives the mathematical tangent slope, which corresponds physically to the angular rotation of the beam's neutral axis in radians.

Practice Problem 6: Limit Definition of the Derivative with a Radical at a Point

Use the limit definition of the derivative to compute the exact numerical value of $f'(4)$ for:

f(x) = \sqrt{x + 5}

Practice Space Solve this problem on paper or tablet.

State the Definition of the Derivative at a Point: $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ . Substitute $a = 4$ and $f(4) = \sqrt{9} = 3$ , then multiply by the conjugate $(\sqrt{9+h} + 3)$ .

Answer:

f'(4) = \frac{1}{6}

Step-by-Step Execution

Governing Formula: The Limit Definition of the Derivative at a Fixed Coordinate: $f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$ (This formula is defined in Definition of the Derivative.)
Substitute $a = 4$ and $f(x) = \sqrt{x + 5}$ : $f(4) = \sqrt{4 + 5} = \sqrt{9} = 3$ $f(4 + h) = \sqrt{(4 + h) + 5} = \sqrt{9 + h}$ $f'(4) = \lim_{h \to 0} \frac{\sqrt{9 + h} - 3}{h}$
Multiply numerator and denominator by the conjugate $(\sqrt{9 + h} + 3)$ : $f'(4) = \lim_{h \to 0} \frac{(\sqrt{9 + h} - 3)(\sqrt{9 + h} + 3)}{h(\sqrt{9 + h} + 3)}$
Expand the difference of squares in the numerator: $(\sqrt{9 + h})^2 - (3)^2 = (9 + h) - 9 = h$ $f'(4) = \lim_{h \to 0} \frac{h}{h(\sqrt{9 + h} + 3)}$
Cancel $h$ and evaluate the limit as $h \to 0$ : $f'(4) = \lim_{h \to 0} \frac{1}{\sqrt{9 + h} + 3} = \frac{1}{\sqrt{9 + 0} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$

Practice Problem 7: Product Rule with Exponential and Polynomial Terms

Given the function $f(x) = (3x^2 - 5)e^x$ , compute the first derivative $f'(x)$ and the second derivative $f''(x)$ , expressing both in completely factored form.

Practice Space Solve this problem on paper or tablet.

State the Product Rule: $\frac{d}{dx}[uv] = u'v + uv'$ . Differentiate $u = 3x^2 - 5$ and $v = e^x$ , factor out $e^x$ , and repeat the Product Rule on $f'(x)$ to obtain $f''(x)$ .

Answer: * $f'(x) = (3x^2 + 6x - 5)e^x$

$f''(x) = (3x^2 + 12x + 1)e^x$

Step-by-Step Execution

Governing Formula: The Product Rule for Differentiation: $\frac{d}{dx}[u \cdot v] = \frac{du}{dx}v + u\frac{dv}{dx}, \qquad \frac{d}{dx}[e^x] = e^x$ (These rules are defined in Product Rule and Derivatives of Exponential Functions.)
Compute the first derivative $f'(x)$ : Let $u(x) = 3x^2 - 5 \implies u'(x) = 6x$ .
Let $v(x) = e^x \implies v'(x) = e^x$ . $f'(x) = (6x)e^x + (3x^2 - 5)e^x$ Factor out the common factor $e^x$ : $f'(x) = (3x^2 + 6x - 5)e^x$
Compute the second derivative $f''(x)$ : Apply the Product Rule to $f'(x)$ with $u_1(x) = 3x^2 + 6x - 5$ and $v_1(x) = e^x$ : $u_1'(x) = 6x + 6$ $v_1'(x) = e^x$ $f''(x) = (6x + 6)e^x + (3x^2 + 6x - 5)e^x$
Factor out $e^x$ and combine like terms: $f''(x) = [(6x + 6) + (3x^2 + 6x - 5)]e^x = (3x^2 + 12x + 1)e^x$

Practice Problem 8: Quotient Rule and Tangent Line Equation

Find the equation of the line tangent to the curve $y = \frac{2x - 1}{x^2 + 1}$ at the point where $x = 1$ . Express your final answer in both point-slope form and slope-intercept form.

Practice Space Solve this problem on paper or tablet.

Evaluate the coordinate point $y(1)$ . State the Quotient Rule: $\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}$ . Find $y'(1)$ to establish the slope $m$ , then write $y - y_1 = m(x - x_1)$ .

Answer: * Point-Slope Form: $y - \frac{1}{2} = \frac{1}{2}(x - 1)$

Slope-Intercept Form: $y = \frac{1}{2}x$

Step-by-Step Execution

Governing Formula: The Quotient Rule and Point-Slope Tangent Equation: $\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}, \qquad y - y_1 = m(x - x_1) \quad (m = y'(x_1))$ (These rules are defined in Quotient Rule and Definition of the Derivative.)
Find the point coordinates $(x_1, y_1)$ at $x_1 = 1$ : $y_1 = \frac{2(1) - 1}{(1)^2 + 1} = \frac{1}{2}$ The point of tangency is $\left(1, \frac{1}{2}\right)$ .
Differentiate using the Quotient Rule: Let $u = 2x - 1 \implies u' = 2$ .
Let $v = x^2 + 1 \implies v' = 2x$ . $\frac{dy}{dx} = \frac{2(x^2 + 1) - (2x - 1)(2x)}{(x^2 + 1)^2}$ $\frac{dy}{dx} = \frac{2x^2 + 2 - (4x^2 - 2x)}{(x^2 + 1)^2} = \frac{2x^2 + 2 - 4x^2 + 2x}{(x^2 + 1)^2} = \frac{-2x^2 + 2x + 2}{(x^2 + 1)^2}$
Evaluate the tangent slope $m$ at $x = 1$ : $m = \left.\frac{dy}{dx}\right|_{x = 1} = \frac{-2(1)^2 + 2(1) + 2}{(1^2 + 1)^2} = \frac{-2 + 2 + 2}{(2)^2} = \frac{2}{4} = \frac{1}{2}$
Construct the tangent line equations:
- Point-Slope Form: $y - \frac{1}{2} = \frac{1}{2}(x - 1)$
- Slope-Intercept Form: $y - \frac{1}{2} = \frac{1}{2}x - \frac{1}{2} \implies y = \frac{1}{2}x$

Segment 4 of 8

📅 Regroup 2

Differentiability Audit: A continuous curve with a sharp corner has no tangent line because the left and right difference quotient limits disagree. On exams, never state that a corner has "two slopes"—a derivative at a single point is either a single unique finite number or it Does Not Exist.
Quotient Rule Minus Sign: The negative sign in $\frac{u'v - uv'}{v^2}$ is a prime source of algebraic sign errors. Always wrap the second term in brackets before distributing.

Segment 5 of 8

Skill Block 3

Real-Life Engineering Applications

⚡

Electrical Engineering

Speed Mapping (Electrical & Mechanical Engineering)

An electric vehicle (EV) powertrain engineer at UNLV is modeling the electromagnetic torque and efficiency characteristics of a permanent-magnet synchronous motor (PMSM) across its field-weakening operating regime. The motor mechanical output power $P(\omega)$ (in kilowatts) as a function of angular rotor velocity $\omega$ (in hundreds of radians per second, $\text{rad/s}$ ) is described by the multi-factor product function:

P(\omega) = 40 \omega \cdot e^{-0.05\omega} \cdot \left(\omega^2 + 16\right)^{-1/2}

over the operational range

1 \le \omega \le 20

.

Compute the first derivative $\frac{dP}{d\omega}$ using Logarithmic Differentiation to bypass nested Product and Quotient rules.
Formulate the fractional power sensitivity function $\frac{1}{P}\frac{dP}{d\omega}$ and evaluate its exact numerical value at a cruising angular velocity of $\omega = 4.0\text{ (400 rad/s)}$ .
Explain how powertrain engineers utilize the sign of $\frac{dP}{d\omega}$ and logarithmic sensitivity to optimize regenerative braking torque transition thresholds and prevent thermal battery inverter overload.

💡 Show Worked Derivation

Problem Overview

We analyze the mechanical power curve of an electric vehicle traction motor:

P(\omega) = 40 \omega \cdot e^{-0.05\omega} \cdot \left(\omega^2 + 16\right)^{-1/2}

using Logarithmic Differentiation to calculate the marginal power derivative

\frac{dP}{d\omega}

and the fractional power sensitivity

\frac{1}{P}\frac{dP}{d\omega}

at

\omega = 4.0\text{ (400 rad/s)}

.

Step-by-Step Mathematical Solution

Step 1: Apply Natural Logarithms to Decompose Products and Powers

The governing strategy is Logarithmic Differentiation:

\ln[u \cdot v \cdot w^k] = \ln(u) + \ln(v) + k\ln(w)

\frac{d}{dx}[\ln y] = \frac{1}{y}\frac{dy}{dx}

(This technique is defined in Logarithmic Differentiation.)

Take the natural logarithm of both sides of $P(\omega)$ :

\ln P(\omega) = \ln\left[40 \omega \cdot e^{-0.05\omega} \cdot \left(\omega^2 + 16\right)^{-1/2}\right]

Expand using logarithm properties:

\ln P(\omega) = \ln(40) + \ln(\omega) + \ln\left(e^{-0.05\omega}\right) + \ln\left[\left(\omega^2 + 16\right)^{-1/2}\right]

\ln P(\omega) = \ln(40) + \ln(\omega) - 0.05\omega - \frac{1}{2}\ln\left(\omega^2 + 16\right)

Step 2: Differentiate Implicitly with Respect to $\omega$

Differentiate both sides term-by-term with respect to $\omega$ :

\frac{d}{d\omega}[\ln P(\omega)] = \frac{d}{d\omega}\left[\ln(40) + \ln(\omega) - 0.05\omega - \frac{1}{2}\ln\left(\omega^2 + 16\right)\right]

\frac{1}{P}\frac{dP}{d\omega} = 0 + \frac{1}{\omega} - 0.05 - \frac{1}{2}\left(\frac{2\omega}{\omega^2 + 16}\right)

\frac{1}{P}\frac{dP}{d\omega} = \frac{1}{\omega} - 0.05 - \frac{\omega}{\omega^2 + 16}

Step 3: Evaluate Fractional Sensitivity at $\omega = 4.0$

Substitute $\omega = 4.0$ directly into the fractional sensitivity formula:

\left.\frac{1}{P}\frac{dP}{d\omega}\right|_{\omega = 4} = \frac{1}{4} - 0.05 - \frac{4}{4^2 + 16}

\left.\frac{1}{P}\frac{dP}{d\omega}\right|_{\omega = 4} = 0.25 - 0.05 - \frac{4}{16 + 16} = 0.20 - \frac{4}{32} = 0.20 - 0.125 = +0.075

Step 4: Compute the Absolute Marginal Power Rate $\frac{dP}{d\omega}$

Evaluate Baseline Power $P(4)$ : $P(4) = 40(4) \cdot e^{-0.05(4)} \cdot \left(4^2 + 16\right)^{-1/2}$ $P(4) = 160 \cdot e^{-0.2} \cdot (32)^{-1/2} = \frac{160 \cdot e^{-0.2}}{\sqrt{32}} = \frac{160 e^{-0.2}}{4\sqrt{2}} = \frac{40 e^{-0.2}}{\sqrt{2}} \approx \frac{40(0.81873)}{1.4142} \approx 23.16\text{ kW}$
Multiply by the Fractional Sensitivity: $\left.\frac{dP}{d\omega}\right|_{\omega = 4} = P(4) \cdot (+0.075) \approx 23.16 \cdot 0.075 \approx +1.737\text{ kW / (100 rad/s)}$

Engineering Conclusion

In automotive powertrain design, evaluating $\frac{1}{P}\frac{dP}{d\omega} = +0.075$ reveals that at $400\text{ rad/s}$ (approx. $3820\text{ RPM}$ ), the motor is operating on the rising side of its power envelope: every $1\%$ increase in motor angular velocity yields a $0.075\%$ increase in delivered wheel power.

Powertrain electronic control units (ECUs) continuously sample this logarithmic derivative. When $\frac{dP}{d\omega} > 0$ , the controller allows wide-open throttle acceleration. However, as rotor speed increases further, back-electromotive force (back-EMF) and rotor windage losses drive the sensitivity toward zero ( $\frac{dP}{d\omega} = 0$ , the peak power operating point) and into negative territory ( $\frac{dP}{d\omega} < 0$ ). Detecting where $\frac{dP}{d\omega}$ changes sign allows powertrain calibration engineers to schedule seamless automatic gear ratio shifts in multi-speed EV transmissions and initiate field-weakening current injection to safeguard the high-voltage battery inverter against thermal runaway.

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

Chain Rule, Transcendental Functions, & Implicit / Log Differentiation

General Overview
Skill Block 3 synthesizes differentiation of composite structures via the Chain Rule, exponential and inverse trigonometric derivatives, implicit differentiation of curves, and logarithmic differentiation for complex products and variable exponent towers.

Conceptual Framework

The Chain Rule Engine: When functions are composed as $y = f(g(x))$ , differentiation evaluates the outer derivative evaluated at the unchanged inner argument, multiplied by the derivative of the inner function: $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$
Transcendental Derivative Rules:
- Natural Exponential: $\frac{d}{dx}[e^{u}] = e^u \frac{du}{dx}$ .
- Natural Logarithm: $\frac{d}{dx}[\ln|u|] = \frac{1}{u}\frac{du}{dx}$ .
- Inverse Tangent: $\frac{d}{dx}[\arctan(u)] = \frac{1}{1 + u^2}\frac{du}{dx}$ .
- Inverse Sine: $\frac{d}{dx}[\arcsin(u)] = \frac{1}{\sqrt{1 - u^2}}\frac{du}{dx}$ .
Implicit Differentiation: When $x$ and $y$ are constrained implicitly within an equation $F(x, y) = C$ , treat $y$ as an implicit function $y(x)$ . Differentiating $y^n$ triggers the Chain Rule, producing $n y^{n-1} \frac{dy}{dx}$ . Collect all $\frac{dy}{dx}$ terms on one side, factor out $\frac{dy}{dx}$ , and solve algebraically.
Logarithmic Differentiation: For multi-factor quotients or variable-base variable-exponent functions ( $y = [f(x)]^{g(x)}$ ), the Power Rule and Exponential Rule both fail. Taking $\ln$ of both sides converts powers into products: $\ln y = g(x) \ln(f(x))$ Differentiating implicitly produces $\frac{1}{y}\frac{dy}{dx}$ , allowing straightforward multiplication by $y$ to yield $\frac{dy}{dx}$ .

Key Equations & Structural Rules

General Chain Rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$ (see Chain Rule)
Implicit Derivative Operator: $\frac{d}{dx}[g(y)] = g'(y) \cdot \frac{dy}{dx}$ (see Implicit Differentiation)
Logarithmic Derivative Identity: $\frac{d}{dx}[\ln(y)] = \frac{1}{y}\frac{dy}{dx} \implies \frac{dy}{dx} = y \cdot \frac{d}{dx}[\ln(y)]$ (see Logarithmic Differentiation)
Inverse Tangent Rule: $\frac{d}{dx}[\arctan(u)] = \frac{1}{1 + u^2}\frac{du}{dx}$ (see Derivatives of Inverse Trigonometric Functions)

Worked Example A: Implicit Differentiation and Tangent Line to an Algebraic Curve

Find the equation of the line tangent to the curve $x^3 + y^3 = 6xy$ (the Folium of Descartes) at the coordinate point $(3, 3)$ .

💡 Reveal Worked Solution

Governing Formula: Implicit Differentiation with the Product Rule: $\frac{d}{dx}[y^n] = n y^{n-1}\frac{dy}{dx}, \qquad \frac{d}{dx}[6xy] = 6y + 6x\frac{dy}{dx}$ (These rules are defined in Implicit Differentiation and Product Rule.)
Differentiate both sides with respect to $x$ : $\frac{d}{dx}[x^3] + \frac{d}{dx}[y^3] = \frac{d}{dx}[6xy]$ $3x^2 + 3y^2\frac{dy}{dx} = 6(1 \cdot y) + 6\left(x \cdot \frac{dy}{dx}\right)$ $3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$
Divide the entire equation by 3 to simplify coefficients: $x^2 + y^2\frac{dy}{dx} = 2y + 2x\frac{dy}{dx}$
Collect all terms with $\frac{dy}{dx}$ on the left side: $y^2\frac{dy}{dx} - 2x\frac{dy}{dx} = 2y - x^2$ $\frac{dy}{dx}(y^2 - 2x) = 2y - x^2$
Isolate $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}$
Evaluate the slope $m$ at $(3, 3)$ : $m = \left.\frac{dy}{dx}\right|_{(3, 3)} = \frac{2(3) - (3)^2}{(3)^2 - 2(3)} = \frac{6 - 9}{9 - 6} = \frac{-3}{3} = -1$
Write the tangent line equation using Point-Slope Form: $y - 3 = -1(x - 3) \implies y - 3 = -x + 3 \implies y = -x + 6$

Practice Problem 9: Concept & Reading Check

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

(a) Under the Conceptual Framework section, state the Chain Rule formula for composite functions and identify what the "inner derivative" represents.
(b) In Worked Example A, explain why differentiating $6xy$ produces both $6y$ and $6x\frac{dy}{dx}$ .
(c) In the linked engineering application Electric Powertrain Torque-Speed Mapping (Electrical & Mechanical Engineering), explain why logarithmic differentiation is preferred over nested quotient rules when calculating relative power loss gradients.

Practice Space Solve this problem on paper or tablet.

Review the composite function notation in the framework and the Product Rule application to $6xy$ in Worked Example A.

Answer: * (a) $\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)$ ; the inner derivative $g'(x)$ represents the instantaneous rate of change of the inside variable.

(b) Because $6xy$ is a product of two variables $x$ and $y(x)$ , requiring the Product Rule: $6(1)y + 6x\frac{dy}{dx}$ .
(c) Logarithmic differentiation expands multi-factor products into simple additions and subtractions, immediately yielding fractional sensitivity $\frac{1}{P}\frac{dP}{d\omega}$ .

Step-by-Step Execution

(a) The Chain Rule formula is $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$ . The outer derivative $f'$ is evaluated at the unchanged inner quantity $g(x)$ , while the inner derivative $g'(x)$ accounts for the rate at which the inner argument itself changes with respect to $x$ .
(b) The expression $6xy$ is the product of $u = 6x$ and $v = y(x)$ . Applying the Product Rule $\frac{d}{dx}[uv] = u'v + uv'$ yields $(6)y + 6x\left(\frac{dy}{dx}\right) = 6y + 6x\frac{dy}{dx}$ . Treating $y$ as a constant and writing only $6y$ is an invalid algebraic error.
(c) In powertrain modeling, motor output power involves exponential damping, linear rotor velocity, and radical impedance denominators. Taking logarithms converts products into sums, isolating the fractional sensitivity $\frac{1}{P}\frac{dP}{d\omega} = \frac{1}{\omega} - 0.05 - \frac{\omega}{\omega^2+16}$ in one line without complex polynomial quotient expansions.

Practice Problem 10: Nested Chain Rule with Trigonometric and Exponential Functions

Compute the derivative of the function:

f(x) = \sin^3\left(e^{2x} + 4x\right)

Practice Space Solve this problem on paper or tablet.

State the General Chain Rule: identify the three concentric layers: outer power $(\square)^3$ , middle trigonometric function $\sin(\square)$ , and inner argument $e^{2x} + 4x$ . Differentiate layer-by-layer from outside to inside.

Answer:

f'(x) = 6(e^{2x} + 2)\sin^2\left(e^{2x} + 4x\right)\cos\left(e^{2x} + 4x\right)

Step-by-Step Execution

Governing Formula: The Multi-Layer Chain Rule: $\frac{d}{dx}[u^n] = n u^{n-1} \frac{du}{dx}, \qquad \frac{d}{dx}[\sin(v)] = \cos(v)\frac{dv}{dx}, \qquad \frac{d}{dx}[e^{kw}] = k e^{kw}$ (These rules are defined in Chain Rule.)
Deconstruct the nested layers:
- Outer layer: $F(u) = u^3$ , where $u = \sin\left(e^{2x} + 4x\right)$ .
- Middle layer: $u = \sin(v)$ , where $v = e^{2x} + 4x$ .
- Inner layer: $v = e^{2x} + 4x$ .
Differentiate the outer power: $\frac{df}{du} = 3 u^2 = 3\sin^2\left(e^{2x} + 4x\right)$
Differentiate the middle trigonometric function: $\frac{du}{dv} = \cos(v) = \cos\left(e^{2x} + 4x\right)$
Differentiate the inner function: $\frac{dv}{dx} = 2e^{2x} + 4 = 2(e^{2x} + 2)$
Multiply all three derivatives together: $f'(x) = 3\sin^2\left(e^{2x} + 4x\right) \cdot \cos\left(e^{2x} + 4x\right) \cdot 2(e^{2x} + 2)$ $f'(x) = 6(e^{2x} + 2)\sin^2\left(e^{2x} + 4x\right)\cos\left(e^{2x} + 4x\right)$

Practice Problem 11: Logarithmic Differentiation for Variable Towers

Use logarithmic differentiation to calculate $\frac{dy}{dx}$ for the function:

y = (\cos x)^{\sqrt{x}}, \quad \text{for } 0 < x < \frac{\pi}{2}

Practice Space Solve this problem on paper or tablet.

State the Logarithmic Differentiation Algorithm. Take $\ln$ of both sides: $\ln y = \sqrt{x}\ln(\cos x)$ . Differentiate implicitly with the Product Rule and Chain Rule, then multiply the entire result by $y$ .

Answer:

\frac{dy}{dx} = (\cos x)^{\sqrt{x}}\left[ \frac{\ln(\cos x)}{2\sqrt{x}} - \sqrt{x}\tan x \right]

Step-by-Step Execution

Governing Formula: Logarithmic Differentiation for Variable-Base Variable-Exponent Functions: $y = [f(x)]^{g(x)} \implies \ln y = g(x)\ln[f(x)] \implies \frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}[g(x)\ln(f(x))]$ (This procedure is defined in Logarithmic Differentiation.)
Apply the natural logarithm to both sides: $\ln y = \ln\left[(\cos x)^{\sqrt{x}}\right] = \sqrt{x} \ln(\cos x) = x^{1/2} \ln(\cos x)$
Differentiate implicitly with respect to $x$ using the Product Rule: $\frac{d}{dx}[\ln y] = \frac{d}{dx}\left[x^{1/2}\right]\ln(\cos x) + x^{1/2}\frac{d}{dx}[\ln(\cos x)]$ $\frac{1}{y}\frac{dy}{dx} = \left(\frac{1}{2}x^{-1/2}\right)\ln(\cos x) + x^{1/2}\left(\frac{1}{\cos x} \cdot (-\sin x)\right)$
Simplify trigonometric and algebraic terms: $\frac{1}{y}\frac{dy}{dx} = \frac{\ln(\cos x)}{2\sqrt{x}} - \sqrt{x}\left(\frac{\sin x}{\cos x}\right) = \frac{\ln(\cos x)}{2\sqrt{x}} - \sqrt{x}\tan x$
Multiply both sides by the original function $y$ : $\frac{dy}{dx} = y \left[ \frac{\ln(\cos x)}{2\sqrt{x}} - \sqrt{x}\tan x \right]$ $\frac{dy}{dx} = (\cos x)^{\sqrt{x}}\left[ \frac{\ln(\cos x)}{2\sqrt{x}} - \sqrt{x}\tan x \right]$

Practice Problem 12: Inverse Trigonometric and Rational Derivative

Find the derivative $\frac{dy}{dx}$ and simplify completely:

y = \arctan\left(\frac{x}{3}\right) + \frac{3x}{x^2 + 9}

Practice Space Solve this problem on paper or tablet.

State the Inverse Tangent Derivative Rule: $\frac{d}{dx}[\arctan(u)] = \frac{u'}{1 + u^2}$ . Differentiate the second term using the Quotient Rule, then combine both fractions over the common denominator $(x^2 + 9)^2$ .

Answer:

\frac{dy}{dx} = \frac{54}{(x^2 + 9)^2}

Step-by-Step Execution

Governing Formula: The Inverse Tangent Chain Rule and Quotient Rule: $\frac{d}{dx}[\arctan(u)] = \frac{1}{1 + u^2}\frac{du}{dx}, \qquad \frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}$ (These rules are defined in Derivatives of Inverse Trigonometric Functions and Quotient Rule.)
Differentiate the first term $y_1 = \arctan\left(\frac{x}{3}\right)$ : Here $u = \frac{x}{3} \implies u' = \frac{1}{3}$ . $\frac{dy_1}{dx} = \frac{\frac{1}{3}}{1 + \left(\frac{x}{3}\right)^2} = \frac{\frac{1}{3}}{1 + \frac{x^2}{9}} = \frac{\frac{1}{3}}{\frac{9 + x^2}{9}} = \frac{1}{3} \cdot \frac{9}{x^2 + 9} = \frac{3}{x^2 + 9}$
Differentiate the second term $y_2 = \frac{3x}{x^2 + 9}$ via Quotient Rule: Let $u = 3x \implies u' = 3$ .
Let $v = x^2 + 9 \implies v' = 2x$ . $\frac{dy_2}{dx} = \frac{3(x^2 + 9) - (3x)(2x)}{(x^2 + 9)^2} = \frac{3x^2 + 27 - 6x^2}{(x^2 + 9)^2} = \frac{27 - 3x^2}{(x^2 + 9)^2}$
Combine both derivatives over a common denominator $(x^2 + 9)^2$ : $\frac{dy}{dx} = \frac{3}{x^2 + 9} + \frac{27 - 3x^2}{(x^2 + 9)^2}$ Multiply the first fraction by $\frac{x^2 + 9}{x^2 + 9}$ : $\frac{dy}{dx} = \frac{3(x^2 + 9) + (27 - 3x^2)}{(x^2 + 9)^2} = \frac{3x^2 + 27 + 27 - 3x^2}{(x^2 + 9)^2}$
Cancel opposite terms in the numerator: $\frac{dy}{dx} = \frac{54}{(x^2 + 9)^2}$

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📅 Regroup 3

Variable Towers Require Logarithms: The Power Rule $\frac{d}{dx}[x^n] = n x^{n-1}$ requires a constant exponent. The Exponential Rule $\frac{d}{dx}[a^x] = \ln(a) a^x$ requires a constant base. If an expression has a variable in both base and exponent ( $x^x$ , $(\cos x)^{\sqrt{x}}$ ), logarithmic differentiation is non-negotiable.
Back-Substitution Check: Always substitute the original $y$ back into your final answer when performing logarithmic differentiation. Leaving $\frac{1}{y}$ on the left side is an incomplete derivative.

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📅 Synthesis Wrap-up

Midterm 1 Strategy: Begin the exam by skimming all problems. Secure high-confidence points on algebraic limits and standard differentiation before tackling algebraic difference quotients and implicit tangent lines.
QC / Redline Habit: Verify your units and signs. Slopes of decreasing curves must be negative; horizontal tangents require setting the numerator of $\frac{dy}{dx}$ to zero; vertical tangents occur where the denominator of $\frac{dy}{dx}$ vanishes while the numerator remains nonzero.
Safety Net Reminder: If needed, complete exam corrections for retroactive replacement eligibility, and log hours in the Test Review Center to buy back homework points.

Segment 8 of 8

Midterm Exam 1 Review

📅 Micro-Lecture

Skill Block 1

Real-Life Engineering Applications

Structural Vibration Resonance & Dynamic Load Thresholds (Civil & Mechanical Engineering)

Problem Overview

Step-by-Step Mathematical Solution

Step 1: Direct Substitution Test on A(β)A(\beta)A(β)

Step 2: Evaluate the Limit via Radical Conjugate Multiplication

Step 3: Investigate Undamped Resonant Magnification M(β)M(\beta)M(β)

Engineering Conclusion

Worked Example A: Indeterminate Radical Conjugate Limit

Practice Problem 1: Concept & Reading Check

Step-by-Step Execution

Practice Problem 2: Factoring & Rational Limit

Step-by-Step Execution

Practice Problem 3: Limit at Infinity with Square Root Radical

Step-by-Step Execution

Practice Problem 4: Piecewise Continuity Parameter Determination

Step-by-Step Execution

📅 Regroup 1

Skill Block 2

Real-Life Engineering Applications

Cantilever Beam Elastic Deflection & Tangent Stiffness (Civil Engineering)

Problem Overview

Step-by-Step Mathematical Solution

Step 1: Compute the First Derivative dydx\frac{dy}{dx}dxdy​

Step 2: Evaluate Boundary Slopes at Support (x=0x = 0x=0) and Tip (x=Lx = Lx=L)

Step 3: Formulate the Tangent Line at Mid-Span (x=L/2=3.0 mx = L/2 = 3.0\text{ m}x=L/2=3.0 m)

Engineering Conclusion

Worked Example A: Derivative via Limit Definition (Rational Difference Quotient)

Practice Problem 5: Concept & Reading Check

Step-by-Step Execution

Practice Problem 6: Limit Definition of the Derivative with a Radical at a Point

Step-by-Step Execution

Practice Problem 7: Product Rule with Exponential and Polynomial Terms

Step-by-Step Execution

Practice Problem 8: Quotient Rule and Tangent Line Equation

Step-by-Step Execution

📅 Regroup 2

Skill Block 3

Real-Life Engineering Applications

Speed Mapping (Electrical & Mechanical Engineering)

Problem Overview

Step-by-Step Mathematical Solution

Step 1: Apply Natural Logarithms to Decompose Products and Powers

Step 2: Differentiate Implicitly with Respect to ω\omegaω

Step 3: Evaluate Fractional Sensitivity at ω=4.0\omega = 4.0ω=4.0

Step 4: Compute the Absolute Marginal Power Rate dPdω\frac{dP}{d\omega}dωdP​

Engineering Conclusion

Worked Example A: Implicit Differentiation and Tangent Line to an Algebraic Curve

Practice Problem 9: Concept & Reading Check

Step-by-Step Execution

Practice Problem 10: Nested Chain Rule with Trigonometric and Exponential Functions

Step-by-Step Execution

Practice Problem 11: Logarithmic Differentiation for Variable Towers

Step-by-Step Execution

Practice Problem 12: Inverse Trigonometric and Rational Derivative

Step-by-Step Execution

📅 Regroup 3

📅 Synthesis Wrap-up

Step 1: Direct Substitution Test on $A(\beta)$

Step 3: Investigate Undamped Resonant Magnification $M(\beta)$

Step 1: Compute the First Derivative $\frac{dy}{dx}$

Step 2: Evaluate Boundary Slopes at Support ( $x = 0$ ) and Tip ( $x = L$ )

Step 3: Formulate the Tangent Line at Mid-Span ( $x = L/2 = 3.0\text{ m}$ )

Step 2: Differentiate Implicitly with Respect to $\omega$

Step 3: Evaluate Fractional Sensitivity at $\omega = 4.0$

Step 4: Compute the Absolute Marginal Power Rate $\frac{dP}{d\omega}$