MATH-181 (Calculus I) Curriculum

Session 08 - Chain Rule and Derivatives of Exponential, Logarithmic, and Inverse Trig Functions

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

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

Deconstruct complex composite functions into nested layers ("peeling the onion"), differentiating using both Leibniz notation ( $\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$ ) and Prime notation ( $f'(g(x))g'(x)$ ).
Differentiate natural and general exponential functions ( $e^u, b^u$ ), logarithmic functions ( $\ln u, \log_a u$ ), and inverse trigonometric functions ( $\sin^{-1}u, \tan^{-1}u, \sec^{-1}u$ ), applying pre-processing logarithmic properties to eliminate unnecessary quotient and product calculations.
Execute operational layering on deeply nested expressions combining Product, Quotient, and Chain rules, construct tangent lines to composite functions from point and derivative values, and verify the Quotient Rule via the negative power chain rule.

Micro-Lecture

1. The Composite Challenge: Why Basic Rules Fail on Nested Systems

Mathematical Foundation: In Session 07, we mastered differentiating standalone functions like $x^5$ , $\cos(x)$ , and $e^x$ . However, modern engineering systems are rarely standalone; variables interact through chained dependencies. In an expression like $y = \cos(x^3)$ or $y = \frac{1}{\sqrt{1+x^5}}$ , the interior term is not simply $x$ , but an entire inner function. Naively differentiating the outside without accounting for the inside (e.g., claiming the derivative of $\cos(x^3)$ is $-\sin(x^3)$ ) produces an incomplete rate. The Chain Rule states that the overall derivative is the rate of change of the outer shell evaluated at the unchanged inner state, multiplied by the rate of change of the inner system: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \qquad \text{or} \qquad \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ (see Chain Rule).
The "Peeling the Onion" Metaphor: Always work from the outermost operational shell inward. Differentiate the outer function first while preserving the entire inner argument untouched, and only then multiply by the derivative of the inner layer.

2. The Transcendental Catalog & Algebraic Pre-Processing

Mathematical Foundation: Transcendental functions—exponentials, logarithms, and inverse trigonometric functions—form the core analytical vocabulary of dynamical engineering:
- Natural & General Exponentials: $\frac{d}{dx}[e^u] = e^u u'$ and $\frac{d}{dx}[b^u] = \ln(b) b^u u'$ (see Derivatives of Exponential Functions).
- Natural & General Logarithms: $\frac{d}{dx}[\ln(u)] = \frac{u'}{u}$ and $\frac{d}{dx}[\log_a(u)] = \frac{u'}{u\ln(a)}$ (see Derivatives of Logarithmic Functions).
- Inverse Trigonometric Functions: $\frac{d}{dx}[\sin^{-1}u] = \frac{u'}{\sqrt{1-u^2}}$ and $\frac{d}{dx}[\tan^{-1}u] = \frac{u'}{1+u^2}$ (see Derivatives of Inverse Trigonometric Functions).
The Golden Rule of Logarithms: Before differentiating an expression containing logarithms, always expand the argument completely using logarithmic properties ( $\ln(AB) = \ln A + \ln B$ , $\ln(A/B) = \ln A - \ln B$ , $\ln(A^k) = k\ln A$ ) (see Logarithmic and Exponential Properties). Expanding turns painful Quotient Rule computations into elementary two-second single-term derivatives.

3. Operational Layering & Global Rule Prioritization

Mathematical Foundation: Engineering expressions frequently intertwine multiple rules simultaneously—such as a Product Rule whose second term is an inverse trigonometric function with an inner polynomial chain. The Mental Priority Algorithm requires engineers to inspect global structure first: identify the outermost operation (Product, Quotient, or Power), set up its empty governing blueprint, and execute the Chain Rule locally only as sub-layers require (see Product Rule and Quotient Rule).
Engineering Context: Composite derivatives allow aerospace and mechanical engineers to analyze dynamic aerodynamic drag decay ( $F_d(\rho(z(t)))$ ) during vertical rocket ascent, calculate temperature-induced clock drift in discharging capacitor circuits ( $V(t(\theta))$ ), optimize velocity gains in multi-stage launch vehicles via the Tsiolkovsky rocket equation, and command real-time angular slew velocities in automated radar tracking actuators.

Skill Block 1

Worked Example A: Two-Layer Deconstruction with Leibniz Notation

Find the derivative of $y = \cos(x^3)$ using Leibniz notation.

Worked Solution:

Governing Formula: The Chain Rule in Leibniz Form: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ (This rule is defined in Chain Rule.)
Assign intermediate variable: Let $u = x^3$ , so $y = \cos(u)$ .
Differentiate components: $\frac{dy}{du} = -\sin(u), \qquad \frac{du}{dx} = 3x^2$
Multiply rates and back-substitute: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = (-\sin(u))(3x^2) = -3x^2 \sin(x^3)$

Worked Example B: Radical Power Deconstruction with Leibniz Notation

Find the derivative of $h(x) = (1 + x^5)^{-\frac{1}{2}}$ using Leibniz notation.

Worked Solution:

Governing Formula: The Generalized Power Rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ (This rule is defined in Chain Rule and Power Rule.)
Assign intermediate variable: Let $u = 1 + x^5$ , so $y = u^{-\frac{1}{2}}$ .
Differentiate components: $\frac{dy}{du} = -\frac{1}{2}u^{-\frac{3}{2}}, \qquad \frac{du}{dx} = 5x^4$
Multiply and convert to radical form: $\frac{dy}{dx} = \left(-\frac{1}{2}u^{-\frac{3}{2}}\right)(5x^4) = -\frac{5x^4}{2(1 + x^5)^{\frac{3}{2}}} = -\frac{5x^4}{2\sqrt{(1 + x^5)^3}}$

Worked Example C: Exponential Deconstruction with Prime Notation

Find the derivative of $f(x) = e^{\cot(x)}$ using Prime notation.

Worked Solution:

Governing Formula: The Exponential Chain Rule: $\frac{d}{dx}[e^{g(x)}] = e^{g(x)} \cdot g'(x)$ (This rule is defined in Derivatives of Exponential Functions.)
Identify layers: Outer shell is $e^u$ , inner layer is $g(x) = \cot(x)$ .
Differentiate outer and inner layers: $f'(x) = e^{\cot(x)} \cdot \frac{d}{dx}[\cot(x)]$ Recall from Derivatives of Trigonometric Functions that $\frac{d}{dx}[\cot(x)] = -\csc^2(x)$ .
Assemble derivative: $f'(x) = e^{\cot(x)} \cdot (-\csc^2(x)) = -\csc^2(x)e^{\cot(x)}$

Practice Problem 1: Concept & Reading Check

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

(a) In the Conceptual Framework, explain the "Peeling the Onion" metaphor. Why must the interior argument remain completely unchanged when taking the derivative of the outer shell?
(b) Deconstruct Worked Example B: Identify the intermediate variable $u$ . What algebraic error occurs if a student differentiates both the inside and outside simultaneously, writing $-\frac{1}{2}(5x^4)^{-\frac{3}{2}}$ ?
(c) Review the linked Engineering Applications (RC Circuit Transient Surge (Electrical Engineering)). Why must the electrical engineer link rates via $\frac{dV}{d\theta} = \frac{dV}{dt} \cdot \frac{dt}{d\theta}$ rather than directly differentiating $V$ with respect to $\theta$ ?

Solution Key:

(a) The Chain Rule dictates that the derivative of $f(g(x))$ is $f'(g(x)) \cdot g'(x)$ . The outer derivative $f'$ must be evaluated at the instantaneous input it receives, which is $g(x)$ . Altering or differentiating the inside function during the first step changes the point of evaluation, destroying the functional composition.
(b) In Worked Example B, $u = 1 + x^5$ . Replacing the inner function with its derivative $5x^4$ inside the power rule yields $-\frac{1}{2}(5x^4)^{-3/2}$ , which is completely incorrect. The outer derivative must keep the base $1 + x^5$ intact: $-\frac{1}{2}(1 + x^5)^{-3/2} \cdot (5x^4)$ .
(c) In RC Circuit Transient Surge (Electrical Engineering), physical telemetry relates voltage decay to elapsed time ( $V(t) = 12e^{-2t}$ ), while thermal expansion and controller clock speed relate time to temperature ( $t(\theta) = 0.5\theta^2 + \theta$ ). Because $V$ is not given directly as an explicit formula of $\theta$ , the engineer must use the Chain Rule in Leibniz form: $\frac{dV}{d\theta} = \frac{dV}{dt} \cdot \frac{dt}{d\theta}$ to calculate thermal sensitivity without algebraically composing complicated fractional exponents.

Show your steps and calculations here.

Practice Problem 2: Leibniz Notation and Negative Fractional Exponents

Find the derivative of the following function with respect to $x$ using Leibniz notation. State the governing equation, define the intermediate variable $u$ , and express the final answer in radical form:

y = \frac{1}{\sqrt{1 + x^5}}

Solution Key:

Formally state the governing formula: By the Chain Rule in Leibniz Form:
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ (This formula and its operational steps are defined in Chain Rule.)
Pre-process the function into power form: Using negative and rational exponent laws:
$y = (1 + x^5)^{-\frac{1}{2}}$
Define intermediate variable $u$ and differentiate components: Let $u = 1 + x^5$ , which gives $y = u^{-\frac{1}{2}}$ .
- Differentiating $y$ with respect to $u$ using the Power Rule: $\frac{dy}{du} = -\frac{1}{2}u^{-\frac{1}{2} - 1} = -\frac{1}{2}u^{-\frac{3}{2}}$
- Differentiating $u$ with respect to $x$ using the Power Rule: $\frac{du}{dx} = 5x^4$
Combine components via the Chain Rule:
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \left(-\frac{1}{2}u^{-\frac{3}{2}}\right) \cdot (5x^4) = -\frac{5x^4}{2}u^{-\frac{3}{2}}$
Back-substitute $u$ and convert to radical form: Substitute $u = 1 + x^5$ :
$\frac{dy}{dx} = -\frac{5x^4}{2(1 + x^5)^{\frac{3}{2}}} = -\frac{5x^4}{2\sqrt{(1 + x^5)^3}}$

Show your steps and calculations here.

Practice Problem 3: Multi-Layer Nested Chain Rule

Find the derivative of the three-layer composite function using Prime notation:

h(x) = \cos^4(3x^2 - 1)

Solution Key:

Formally state the governing formula: By the Multi-Layer Chain Rule:
$\frac{d}{dx}[f(g(k(x)))] = f'(g(k(x))) \cdot g'(k(x)) \cdot k'(x)$ (This formula is defined in Chain Rule.)
Isolate the structural layers:
- Outer layer: $f(u) = u^4$ where $u = \cos(v)$
- Middle layer: $g(v) = \cos(v)$ where $v = 3x^2 - 1$
- Inner layer: $k(x) = 3x^2 - 1$
Differentiate from the outside inward:
- Layer 1 (Power Rule on outer shell): $\frac{d}{dx}\left[[\cos(3x^2 - 1)]^4\right] = 4[\cos(3x^2 - 1)]^3 \cdot \frac{d}{dx}[\cos(3x^2 - 1)]$
- Layer 2 (Trigonometric Rule on middle shell): Recall $\frac{d}{dv}[\cos(v)] = -\sin(v)$ (see Derivatives of Trigonometric Functions): $\frac{d}{dx}[\cos(3x^2 - 1)] = -\sin(3x^2 - 1) \cdot \frac{d}{dx}[3x^2 - 1]$
- Layer 3 (Power Rule on inner polynomial): $\frac{d}{dx}[3x^2 - 1] = 6x$
Multiply all layers together and simplify:
$h'(x) = 4\cos^3(3x^2 - 1) \cdot \left(-\sin(3x^2 - 1)\right) \cdot (6x)$ $h'(x) = -24x \cos^3(3x^2 - 1)\sin(3x^2 - 1)$

Show your steps and calculations here.

Regroup 1

Layer Integrity: Emphasize that students must never alter the interior argument during the outer derivative step.
Leibniz Agility: Verify that intermediate variables ( $u$ ) are fully eliminated in final engineering deliverables.

Skill Block 2

Worked Example A: Base Changes and General Exponentials

Differentiate the function:

f(x) = 3^x - 5e^x

Worked Solution:

Governing Formulas: The General Exponential Rule $\frac{d}{dx}[b^x] = \ln(b)b^x$ and Natural Exponential Rule $\frac{d}{dx}[e^x] = e^x$ (see Derivatives of Exponential Functions).
Differentiate term-by-term: $f'(x) = \frac{d}{dx}[3^x] - 5\frac{d}{dx}[e^x] = \ln(3)3^x - 5e^x$

Worked Example B: Logarithmic Pre-Processing

Differentiate $h(x) = \ln\left(\frac{x^2-x}{x^2+2x}\right)$ for $x > 1$ using logarithmic pre-processing.

Worked Solution:

Governing Formula: The Natural Logarithm Rule $\frac{d}{dx}[\ln(u)] = \frac{u'}{u}$ preceded by the Logarithmic Quotient Property $\ln(A/B) = \ln A - \ln B$ (see Logarithmic and Exponential Properties).
Step 1 (Pre-process and factor argument): $\frac{x^2 - x}{x^2 + 2x} = \frac{x(x - 1)}{x(x + 2)} = \frac{x - 1}{x + 2}$ $h(x) = \ln\left(\frac{x - 1}{x + 2}\right) = \ln(x - 1) - \ln(x + 2)$
Step 2 (Differentiate term-by-term): $h'(x) = \frac{1}{x - 1} - \frac{1}{x + 2}$
Step 3 (Consolidate over common denominator): $h'(x) = \frac{(x + 2) - (x - 1)}{(x - 1)(x + 2)} = \frac{3}{(x - 1)(x + 2)}$

Worked Example C: Inverse Trigonometric Differentiation

Find the derivative of $y = \tan^{-1}\left(\frac{3}{z}\right)$ with respect to $z$ .

Worked Solution:

Governing Formula: The Inverse Tangent Chain Rule: $\frac{d}{dz}[\tan^{-1}(u)] = \frac{1}{1 + u^2} \cdot \frac{du}{dz}$ (This rule is defined in Derivatives of Inverse Trigonometric Functions.)
Step 1 (Identify inner function): Let $u = \frac{3}{z} = 3z^{-1}$ .
Step 2 (Differentiate inner function): $\frac{du}{dz} = -3z^{-2} = -\frac{3}{z^2}$
Step 3 (Substitute into arctangent rule): $\frac{dy}{dz} = \frac{1}{1 + \left(\frac{3}{z}\right)^2} \cdot \left(-\frac{3}{z^2}\right) = \frac{1}{1 + \frac{9}{z^2}} \cdot \left(-\frac{3}{z^2}\right)$
Step 4 (Clear compound fraction): Multiply $z^2$ across the denominator: $\frac{dy}{dz} = \frac{-3}{z^2\left(1 + \frac{9}{z^2}\right)} = -\frac{3}{z^2 + 9}$

Practice Problem 4: Concept & Reading Check

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

(a) In the Conceptual Framework, explain the structural difference between the Power Rule and the Exponential Rule. Why is $\frac{d}{dx}[2^x] \neq x 2^{x-1}$ ?
(b) Deconstruct Worked Example B. Why is expanding $\ln\left(\frac{x-1}{x+2}\right) = \ln(x-1) - \ln(x+2)$ algebraically faster than computing $\frac{u'}{u}$ using the Quotient Rule directly on the rational fraction?
(c) Review the linked Engineering Applications (Rocket Velocity Optimization via Rocket Equation (Aerospace Engineering)). Why does the rocket's acceleration $\frac{dv}{dt}$ increase at $t = 60\text{ s}$ even though the propellant mass flow rate $\frac{dm}{dt} = -250\text{ kg/s}$ is completely constant?

Solution Key:

(a) The Power Rule $\frac{d}{dx}[x^n] = n x^{n-1}$ strictly applies when the base is the independent variable $x$ and the exponent $n$ is a constant real number. In $2^x$ , the base is constant and the exponent is variable, making it an exponential function. The derivative is $\frac{d}{dx}[2^x] = \ln(2)2^x$ . Treating $2^x$ as a polynomial is a foundational calculus error.
(b) In Worked Example B, applying $\frac{u'}{u}$ directly on $u = \frac{x-1}{x+2}$ requires computing $u' = \frac{(1)(x+2) - (x-1)(1)}{(x+2)^2} = \frac{3}{(x+2)^2}$ , and then dividing by $\frac{x-1}{x+2}$ . Using the Logarithmic Quotient Property expands the function into $\ln(x-1) - \ln(x+2)$ in one line, yielding immediate term-by-term derivatives $\frac{1}{x-1} - \frac{1}{x+2}$ .
(c) In Rocket Velocity Optimization via Rocket Equation (Aerospace Engineering), the Tsiolkovsky rocket equation gives $v(m) = 2500\ln(50,000/m)$ . Differentiating with respect to time gives $\frac{dv}{dt} = \frac{625,000}{m}$ . Because mass $m(t) = 50,000 - 250t$ continuously decreases as fuel burns, the denominator shrinks, causing the acceleration $\frac{dv}{dt}$ to rise from $12.5\text{ m/s}^2$ at launch to $17.86\text{ m/s}^2$ at $t = 60\text{ s}$ .

Show your steps and calculations here.

Practice Problem 5: Inverse Trigonometric Chain Rule

Find the derivative of the following inverse trigonometric function with respect to $x$ using Prime notation. State the governing formula and simplify all radicals:

f(x) = \sin^{-1}(5x^3)

Solution Key:

Formally state the governing formula: By the Inverse Sine Chain Rule:
$\frac{d}{dx}[\sin^{-1}(u)] = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} = \frac{u'}{\sqrt{1 - u^2}}$ (This formula and its geometric derivation are defined in Derivatives of Inverse Trigonometric Functions.)
Identify the inner function $u$ and differentiate: Let $u = 5x^3$ . By the Power Rule:
$u' = \frac{d}{dx}[5x^3] = 15x^2$
Substitute into the governing formula:
$f'(x) = \frac{15x^2}{\sqrt{1 - (5x^3)^2}}$
Simplify the radicand:
$(5x^3)^2 = 5^2 \cdot (x^3)^2 = 25x^6$ $f'(x) = \frac{15x^2}{\sqrt{1 - 25x^6}}$

Show your steps and calculations here.

Practice Problem 6

Differentiate the function using logarithmic pre-processing:

f(x) = \log_3(x^2 e^x)

Solution Key:

Pre-process using logarithmic properties: Using Product and Power properties (see Logarithmic and Exponential Properties): $f(x) = \log_3(x^2) + \log_3(e^x) = 2\log_3(x) + x\log_3(e)$
Differentiate term-by-term: By the General Logarithm Rule $\frac{d}{dx}[\log_a(x)] = \frac{1}{x\ln(a)}$ : $f'(x) = 2\left(\frac{1}{x\ln(3)}\right) + (1)\log_3(e)$
Convert $\log_3(e)$ using Change of Base formula: $\log_3(e) = \frac{\ln(e)}{\ln(3)} = \frac{1}{\ln(3)}$ $f'(x) = \frac{2}{x\ln(3)} + \frac{1}{\ln(3)}$
Combine over a common denominator: $f'(x) = \frac{2 + x}{x\ln(3)} = \frac{x + 2}{x\ln(3)}$

Show your steps and calculations here.

Practice Problem 7

Differentiate the function using logarithmic pre-processing:

h(x) = \ln\left(\frac{x^2 - x}{x^2 + 2x}\right)

Solution Key:

Pre-process argument algebraically: $\frac{x^2 - x}{x^2 + 2x} = \frac{x(x - 1)}{x(x + 2)} = \frac{x - 1}{x + 2}$
Expand using the Quotient Property of logarithms: $h(x) = \ln(x - 1) - \ln(x + 2)$
Differentiate term-by-term using the Natural Logarithm Rule: $h'(x) = \frac{1}{x - 1} - \frac{1}{x + 2}$
Find common denominator and simplify: $h'(x) = \frac{(x + 2) - (x - 1)}{(x - 1)(x + 2)} = \frac{x + 2 - x + 1}{(x - 1)(x + 2)} = \frac{3}{(x - 1)(x + 2)}$

Show your steps and calculations here.

Regroup 2

Pre-Processing Value: Reiterate that spending 15 seconds pre-expanding a logarithm saves 3 minutes of tedious quotient differentiation.
Base Care: Ensure students do not drop $\ln(b)$ factors when differentiating non-base- $e$ functions.

Skill Block 3

Worked Example A: Composite Product Rule and Tangent Line Construction

Suppose $f(3) = 2$ and $f'(3) = 8$ . Let $g(x) = x^2 f(3x)$ . Find the equation of the line tangent to $g(x)$ at $x = 1$ .

Worked Solution:

Governing Formulas:
- The Product Rule: $\frac{d}{dx}[u \cdot v] = u'v + uv'$ (see Product Rule).
- The Chain Rule: $\frac{d}{dx}[f(kx)] = k f'(kx)$ (see Chain Rule).
- The Point-Slope Equation: $y - y_1 = m(x - x_1)$ (see Tangent Lines).
Step 1 (Find the point of tangency $(x_1, y_1)$ ): Given $x_1 = 1$ : $y_1 = g(1) = (1)^2 \cdot f(3 \cdot 1) = 1 \cdot f(3)$ Substitute $f(3) = 2$ : $y_1 = 2 \implies \text{Point: } (1, 2)$
Step 2 (Differentiate $g(x)$ using Product and Chain Rules): Let $u = x^2$ and $v = f(3x)$ .
- $u' = 2x$
- $v' = f'(3x) \cdot \frac{d}{dx}[3x] = 3f'(3x)$ (by the Chain Rule) Assembling via the Product Rule: $g'(x) = u'v + uv' = (2x)f(3x) + x^2(3f'(3x)) = 2xf(3x) + 3x^2 f'(3x)$
Step 3 (Calculate tangent slope $m = g'(1)$ ): $m = g'(1) = 2(1)f(3(1)) + 3(1)^2 f'(3(1)) = 2f(3) + 3f'(3)$ Substitute given values $f(3) = 2$ and $f'(3) = 8$ : $m = 2(2) + 3(8) = 4 + 24 = 28$
Step 4 (Construct tangent line equation): $y - y_1 = m(x - x_1) \implies y - 2 = 28(x - 1)$ $y - 2 = 28x - 28 \implies y = 28x - 26$

Worked Example B: Proof of Quotient Rule via Product and Chain Rules

Use the Product Rule and the Chain Rule to prove that if $y = \frac{f(x)}{g(x)}$ , then:

\frac{dy}{dx} = \frac{f'(x)g(x) - g'(x)f(x)}{[g(x)]^2}

Worked Solution:

Governing Formulas:
- The Product Rule: $\frac{d}{dx}[u \cdot v] = u'v + uv'$ (see Product Rule).
- The Generalized Power Rule (Chain Rule): $\frac{d}{dx}[[g(x)]^n] = n[g(x)]^{n-1}g'(x)$ (see Chain Rule).
Step 1 (Restructure fraction as a product with a negative exponent): $y = f(x) \cdot [g(x)]^{-1}$
Step 2 (Apply the Product Rule): $\frac{dy}{dx} = \frac{d}{dx}[f(x)] \cdot [g(x)]^{-1} + f(x) \cdot \frac{d}{dx}\left[[g(x)]^{-1}\right]$ $\frac{dy}{dx} = f'(x)[g(x)]^{-1} + f(x) \cdot \frac{d}{dx}\left[[g(x)]^{-1}\right]$
Step 3 (Differentiate $[g(x)]^{-1}$ via the Chain Rule): Treating power $-1$ as the outer function and $g(x)$ as the inner function: $\frac{d}{dx}\left[[g(x)]^{-1}\right] = -1[g(x)]^{-2} \cdot g'(x) = -\frac{g'(x)}{[g(x)]^2}$
Step 4 (Substitute back and express with positive exponents): $\frac{dy}{dx} = f'(x)[g(x)]^{-1} + f(x)\left(-\frac{g'(x)}{[g(x)]^2}\right) = \frac{f'(x)}{g(x)} - \frac{f(x)g'(x)}{[g(x)]^2}$
Step 5 (Combine over common denominator $[g(x)]^2$ ): Multiply numerator and denominator of the first term by $g(x)$ : $\frac{dy}{dx} = \frac{f'(x)g(x)}{[g(x)]^2} - \frac{f(x)g'(x)}{[g(x)]^2} = \frac{f'(x)g(x) - g'(x)f(x)}{[g(x)]^2}$ The identity is proven. $\blacksquare$

Practice Problem 8: Concept & Reading Check

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

(a) Outline the Mental Priority Algorithm. When differentiating $y = x^2 \sin^{-1}(3x)$ , why must the Product Rule be initialized before the Chain Rule?
(b) Deconstruct Worked Example A: Identify where the factor of $3$ in $3x^2 f'(3x)$ originates. Why would forgetting the Chain Rule on $f(3x)$ invalidate the slope calculation?
(c) Review the linked Engineering Applications (Reaction Rate Kinematics and pH Buffering (Chemical Engineering)). What is the operational significance of finding $\frac{d(\text{pH})}{dt} = 0$ at $t = 4\text{ minutes}$ for automated neutralizing valves?

Solution Key:

(a) Under the Mental Priority Algorithm, engineers inspect the global architecture first. In $y = x^2 \sin^{-1}(3x)$ , the primary mathematical structure is a product of $u = x^2$ and $v = \sin^{-1}(3x)$ . Therefore, the Product Rule template $u'v + uv'$ must be established first. The Chain Rule is only triggered locally when evaluating $v' = \frac{d}{dx}[\sin^{-1}(3x)] = \frac{3}{\sqrt{1-9x^2}}$ .
(b) In Worked Example A, the second factor $v = f(3x)$ has an inner argument $3x$ . By the Chain Rule, $v' = f'(3x) \cdot \frac{d}{dx}[3x] = 3f'(3x)$ . Omitting the Chain Rule would leave $v' = f'(3x)$ , resulting in $g'(1) = 2f(3) + f'(3) = 2(2) + 8 = 12$ , an erroneous tangent slope that causes major geometric and mechanical misalignment.
(c) In Reaction Rate Kinematics and pH Buffering (Chemical Engineering), calculating $\frac{d(\text{pH})}{dt} = 0$ at $t = 4\text{ minutes}$ identifies a critical point where the chemical effluent reaches its peak acidity level before decay dominates. Automated PID controllers use derivative zero-crossings to transition buffer delivery valves from active ramp-up to holding mode, preventing chemical over-injection.

Show your steps and calculations here.

Practice Problem 9

Find the derivative of the complex composite exponential function:

y = 2^{\sec(5x)}

Solution Key:

Formally state governing formulas: By the General Exponential Chain Rule: $\frac{d}{dx}[a^u] = \ln(a)a^u \cdot \frac{du}{dx}$ (This rule is defined in Derivatives of Exponential Functions.) By the Secant Derivative Rule: $\frac{d}{dx}[\sec(v)] = \sec(v)\tan(v) \cdot \frac{dv}{dx}$ (This rule is defined in Derivatives of Trigonometric Functions.)
Differentiate outer exponential layer: Let $u = \sec(5x)$ . $y' = \ln(2) \cdot 2^{\sec(5x)} \cdot \frac{d}{dx}[\sec(5x)]$
Differentiate middle trigonometric layer: Let $v = 5x$ . $\frac{d}{dx}[\sec(5x)] = \sec(5x)\tan(5x) \cdot \frac{d}{dx}[5x] = 5\sec(5x)\tan(5x)$
Assemble the complete derivative: $y' = \ln(2) \cdot 2^{\sec(5x)} \cdot 5\sec(5x)\tan(5x) = 5\ln(2) \cdot 2^{\sec(5x)}\sec(5x)\tan(5x)$

Show your steps and calculations here.

Practice Problem 10

Find the derivative of the following composite product function for $|x| > 1$ :

f(x) = \sqrt{x^2 + 1} \csc^{-1}(x)

Solution Key:

Formally state governing formulas: By the Product Rule: $\frac{d}{dx}[u \cdot v] = u'v + uv'$ (This rule is defined in Product Rule.) By the Inverse Cosecant Derivative Rule: $\frac{d}{dx}[\csc^{-1}x] = -\frac{1}{|x|\sqrt{x^2 - 1}}$ (This rule is defined in Derivatives of Inverse Trigonometric Functions.)
Differentiate individual components:
- For $u = (x^2 + 1)^{\frac{1}{2}}$ , use the Generalized Power Rule: $u' = \frac{1}{2}(x^2 + 1)^{-\frac{1}{2}}(2x) = \frac{x}{\sqrt{x^2 + 1}}$
- For $v = \csc^{-1}(x)$ : $v' = -\frac{1}{|x|\sqrt{x^2 - 1}}$
Assemble via the Product Rule: $f'(x) = u'v + uv' = \left(\frac{x}{\sqrt{x^2 + 1}}\right)\csc^{-1}(x) + \sqrt{x^2 + 1}\left(-\frac{1}{|x|\sqrt{x^2 - 1}}\right)$ $f'(x) = \frac{x \csc^{-1}(x)}{\sqrt{x^2 + 1}} - \frac{\sqrt{x^2 + 1}}{|x|\sqrt{x^2 - 1}}$

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Practice Problem 11: Tangent Line to Composite Exponential Curve

Find the equation of the line tangent to the curve $y = e^{2x - x^2}$ at the point where $x = 2$ . Express the equation in slope-intercept form ( $y = mx + b$ ).

Solution Key:

Formally state governing formulas: By the Natural Exponential Chain Rule:
$\frac{d}{dx}[e^u] = e^u \cdot \frac{du}{dx}$ (This rule is defined in Derivatives of Exponential Functions.) By the Point-Slope Tangent Equation: $y - y_1 = m(x - x_1)$ (This formula is defined in Tangent Lines.)
Find the point of tangency $(x_1, y_1)$ : Evaluate $y$ at $x = 2$ :
$y_1 = y(2) = e^{2(2) - (2)^2} = e^{4 - 4} = e^0 = 1$ The point of tangency is $(2, 1)$ .
Compute the derivative $y'$ via the Chain Rule: Let $u = 2x - x^2$ .
$u' = \frac{d}{dx}[2x - x^2] = 2 - 2x$ $y' = e^{2x - x^2} \cdot (2 - 2x)$
Calculate tangent slope $m = y'(2)$ :
$m = y'(2) = e^{2(2) - (2)^2} \cdot (2 - 2(2)) = e^0 \cdot (2 - 4) = 1 \cdot (-2) = -2$
Assemble the tangent line equation:
$y - y_1 = m(x - x_1) \implies y - 1 = -2(x - 2)$ $y - 1 = -2x + 4$ $y = -2x + 5$

Show your steps and calculations here.

Regroup 3

Mental Priority: Review student work on nested problems to verify that students set up the global product/quotient framework before evaluating interior chain derivatives.
Algebraic Factoring: Ensure students factor out lowest powers and exponential terms before attempting numerical evaluation.

Synthesis Wrap-up

Core Takeaway: The Chain Rule is the engine of dynamical calculus, translating interconnected physical rates across layered systems.
Looking Ahead (Session 09): In Session 09, we will apply the Chain Rule to relations where $y$ cannot be isolated explicitly, introducing Implicit Differentiation and Logarithmic Differentiation.
Homework Alignment (Knewton Alta):
- The Chain Rule
- Derivatives of Exponential Functions
- Derivatives of Logarithmic Functions
- Derivatives of Inverse Trigonometric Functions