MATH-181 (Calculus I) Curriculum

Session 10 - Midterm Exam 1 Review

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

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

Evaluate algebraic and indeterminate limits ( $0/0$ ) using factoring, conjugate rationalization, and common denominators; determine infinite limits, vertical asymptotes, and limits at infinity with radicals; and verify continuity or solve for piecewise parameters using the three-part definition of continuity and the Intermediate Value Theorem.
Compute instantaneous rates of change using the limit definition of the derivative (difference quotients), evaluate differentiability versus continuity across non-smooth geometric features, and apply basic differentiation rules (Power, Product, Quotient) to construct tangent and normal line equations.
Execute the Chain Rule across composite transcendental functions (exponential, logarithmic, and inverse trigonometric), perform implicit differentiation on multi-variable curve relations to isolate tangent slopes, and apply logarithmic differentiation to evaluate variable-base variable-exponent towers and multi-factor rational systems.

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.

Skill Block 1

Worked Example A: Indeterminate Radical Conjugate Limit

Evaluate the following limit analytically:

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

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.

Solution Key:

(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.

Show your steps and calculations here.

Practice Problem 2: Factoring & Rational Limit

Evaluate the following limit analytically:

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

Solution Key:

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 ( $x = -3$ ): $\frac{-3 - 3}{2(-3) - 1} = \frac{-6}{-6 - 1} = \frac{-6}{-7} = \frac{6}{7}$

Show your steps and calculations here.

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}

Solution Key:

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 $x \to -\infty$ ): $\frac{-\sqrt{4 + 0}}{3 - 0} = \frac{-\sqrt{4}}{3} = -\frac{2}{3}$

Show your steps and calculations here.

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}

Solution Key:

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 $k$ : $4k + k = 10 - 4$ $5k = 6 \implies k = \frac{6}{5} = 1.2$

Show your steps and calculations here.

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.

Skill Block 2

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}

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?

Solution Key:

(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.

Show your steps and calculations here.

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}

Solution Key:

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 $h \to 0$ : $f'(4) = \lim_{h \to 0} \frac{1}{\sqrt{9 + h} + 3} = \frac{1}{\sqrt{9 + 0} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$

Show your steps and calculations here.

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.

Solution Key:

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: $f''(x) = [(6x + 6) + (3x^2 + 6x - 5)]e^x = (3x^2 + 12x + 1)e^x$

Show your steps and calculations here.

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.

Solution Key:

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: $y - \frac{1}{2} = \frac{1}{2}x - \frac{1}{2} \implies y = \frac{1}{2}x$

Show your steps and calculations here.

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.

Skill Block 3

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 $(3, 3)$ .

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.

Solution Key:

(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.

Show your steps and calculations here.

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)

Solution Key:

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: $f'(x) = 3\sin^2\left(e^{2x} + 4x\right) \cdot \cos\left(e^{2x} + 4x\right) \cdot 2(e^{2x} + 2)$ $f'(x) = 6(e^{2x} + 2)\sin^2\left(e^{2x} + 4x\right)\cos\left(e^{2x} + 4x\right)$

Show your steps and calculations here.

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}

Solution Key:

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

Show your steps and calculations here.

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}

Solution Key:

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

Show your steps and calculations here.

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.

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.
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