Session 09 - Implicit Differentiation and Logarithmic Differentiation

📅 Micro-Lecture

1. Implicit Relations & Hidden Physical Constraints

Mathematical Foundation: Up to this point, curves have been written in explicit form, $y = f(x)$ , where $y$ is isolated entirely on one side. However, physical constraints often bind variables implicitly, such as circles ( $x^2 + y^2 = r^2$ ) or closed kinematic loops ( $x^2y - 2 = 3y^3$ ). These curves often fail the Vertical Line Test, yet every localized smooth segment behaves as a differentiable function $y(x)$ . The foundational principle of implicit differentiation is treating $y$ as an implicit function of $x$ and applying the Chain Rule whenever differentiating a $y$ term: $\frac{d}{dx}[g(y)] = g'(y) \cdot \frac{dy}{dx}$ (see Implicit Differentiation).
Engineering Context: Mechanical and robotics engineers model closed linkage chains, four-bar mechanisms, and automated cutting paths along elliptical rails. Thermodynamics engineers rely on implicit equations of state (such as the Van der Waals equation) to calculate the instantaneous rate of pressure change ( $\frac{dP}{dV}$ ) during gas expansion.

2. Logarithmic Differentiation: Taming Towers and Multi-Factor Systems

Mathematical Foundation: Calculus provides the Power Rule for constant powers ( $\frac{d}{dx}[x^n] = n x^{n-1}$ ) and the Exponential Rule for constant bases ( $\frac{d}{dx}[b^x] = \ln(b) b^x$ ). Neither rule applies to variable bases with variable exponents ( $y = [f(x)]^{g(x)}$ , like $y = x^x$ ). Applying $\ln$ to both sides transforms powers into products: $\ln y = g(x) \cdot \ln(f(x))$ Differentiating implicitly gives $\frac{1}{y}\frac{dy}{dx}$ , converting an impossible exponential derivative into an elementary Product Rule problem (see Logarithmic Differentiation).
Engineering Context: In electrical engineering, transmission line signal decay models intertwine exponential decay, sinusoidal oscillations, and radical damping factors. Logarithmic differentiation bypasses nested quotient rules to instantly calculate signal attenuation loss. In materials science, the logarithmic derivative directly yields the fractional rate of thermal expansion ( $\frac{1}{v}\frac{dv}{dT}$ ).

3. Higher-Order Implicit Rates & Curvature Analysis

Mathematical Foundation: While the first derivative $\frac{dy}{dx}$ provides the tangent slope, the second derivative $\frac{d^2y}{dx^2}$ quantifies concavity and physical curvature. Differentiating $\frac{dy}{dx}$ with respect to $x$ triggers the Quotient Rule and introduces new $\frac{dy}{dx}$ terms. A complete solution requires back-substituting the first derivative and simplifying using the curve's original constraint equation (see Higher-Order Implicit Differentiation).
Engineering Context: Aerospace engineers compute the second derivative along circular viewport bulkheads to locate thermal stress concentrations during atmospheric re-entry. Automotive chassis engineers analyze $\frac{d^2y}{dx^2}$ along suspension geometry curves to control dynamic wheel camber angles and maintain tire grip during cornering.

Segment 1 of 8

Skill Block 1

Real-Life Engineering Applications

⚙️

Mechanical Engineering

Mechanical Kinematics (Robotic Linkages)

A mechanical automation engineer is programming an industrial CNC plasma cutter on a manufacturing assembly line. The cutter head is constrained to move along an implicit elliptical track described by the equation:

4x^2 + 9y^2 = 36

where

x

and

y

represent coordinate positions in meters. During operation, optical telemetry confirms that the cutting head has reached the coordinate position:

(x, y) = \left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)

Differentiate the implicit tracking equation with respect to $x$ using the Chain Rule to find an expression for the instantaneous tangent slope $\frac{dy}{dx}$ .
Evaluate $\frac{dy}{dx}$ at the operational position $\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)$ .
Determine the required orientation angle and normal direction for the plasma cutting nozzle so that it cuts completely perpendicular to the elliptical track.

💡 Show Worked Derivation

Problem Overview

We calculate the instantaneous tangent direction $\frac{dy}{dx}$ and normal cutting orientation for a CNC plasma cutter moving along an implicit elliptical track:

4x^2 + 9y^2 = 36

evaluated at the coordinate point

(x, y) = \left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)

.

Step-by-Step Mathematical Solution

Step 1: Formally Apply Implicit Differentiation

The governing equation is the Implicit Differentiation Chain Rule:

\frac{d}{dx}[g(y)] = g'(y) \cdot \frac{dy}{dx}

(This rule is defined in Implicit Differentiation.)

Differentiate both sides of $4x^2 + 9y^2 = 36$ with respect to $x$ :

\frac{d}{dx}[4x^2] + \frac{d}{dx}[9y^2] = \frac{d}{dx}[36]

8x + 18y\frac{dy}{dx} = 0

Step 2: Isolate the First Derivative $\frac{dy}{dx}$

Subtract $8x$ from both sides and divide by $18y$ :

18y\frac{dy}{dx} = -8x \implies \frac{dy}{dx} = -\frac{8x}{18y} = -\frac{4x}{9y}

Step 3: Evaluate at the Operational Coordinates $\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)$

Substitute $x = \frac{3}{\sqrt{2}}$ and $y = \sqrt{2}$ :

\left.\frac{dy}{dx}\right|_{\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)} = -\frac{4\left(\frac{3}{\sqrt{2}}\right)}{9(\sqrt{2})} = -\frac{\frac{12}{\sqrt{2}}}{9\sqrt{2}} = -\frac{12}{9 \cdot 2} = -\frac{12}{18} = -\frac{2}{3}

Step 4: Determine the Perpendicular Normal Direction

The tangent slope of the track is $m_{\text{tangent}} = -\frac{2}{3}$ . The normal line perpendicular to the track has slope:

m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{-2/3} = \frac{3}{2} = 1.5

Engineering Conclusion

At the position $\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)$ , the cutter head has a tangent track slope of $-\frac{2}{3}$ (meaning for every $3\text{ meters}$ of horizontal travel in $+x$ , the carriage moves $2\text{ meters}$ downward in $-y$ ). Industrial CNC controllers calculate the perpendicular normal slope $m_{\text{normal}} = 1.5$ and continuously command the robot's wrist stepper motors to align the plasma torch perpendicular to the cut line. This orientation eliminates taper defects, guarantees clean edge bevels, and prevents expensive scrap in structural steel fabrication.

🚀

Aerospace Engineering

Linear Fluids)

An aerospace propulsion engineer is modeling the expansion kinetics of a dense supercritical hydrocarbon propellant in a high-pressure rocket combustion chamber. Because the fluid operates near its critical point, it deviates significantly from ideal gas laws and obeys a simplified Van der Waals equation of state:

(P + V^2)(V - 1) = 12

where

P

represents pressure (in atmospheres) and

V

represents the molar volume (in liters per mole,

\text{L/mol}

).

Differentiate the implicit equation of state with respect to volume $V$ using the Product Rule, treating pressure $P$ as an implicit function $P(V)$ .
Derive a symbolic expression for the rate of pressure change with respect to volume ( $\frac{dP}{dV}$ ).
Calculate the instantaneous rate of pressure change ( $\frac{dP}{dV}$ ) when the gas expands through a volume of $V = 3\text{ L/mol}$ at an internal pressure of $P = 5\text{ atm}$ .

💡 Show Worked Derivation

Problem Overview

We calculate the instantaneous rate of pressure change with respect to volume ( $\frac{dP}{dV}$ ) for a non-ideal gas governed by the Van der Waals equation of state:

(P + V^2)(V - 1) = 12

evaluated at the operational condition

V = 3\text{ L/mol}

and

P = 5\text{ atm}

.

Step-by-Step Mathematical Solution

Step 1: Formally Apply the Product Rule with Implicit Differentiation

The governing equation is the Product Rule:

\frac{d}{dV}[u \cdot v] = \frac{du}{dV} \cdot v + u \cdot \frac{dv}{dV}

(This rule is defined in Product Rule and Implicit Differentiation.)

Let $u = P + V^2$ and $v = V - 1$ . Treating $P$ as an implicit function $P(V)$ :

\frac{d}{dV}[P + V^2] \cdot (V - 1) + (P + V^2) \cdot \frac{d}{dV}[V - 1] = \frac{d}{dV}[12]

\left(\frac{dP}{dV} + 2V\right)(V - 1) + (P + V^2)(1) = 0

Step 2: Algebraically Isolate $\frac{dP}{dV}$

Distribute and group like terms:

(V - 1)\frac{dP}{dV} + 2V(V - 1) + P + V^2 = 0

(V - 1)\frac{dP}{dV} = -2V(V - 1) - P - V^2

\frac{dP}{dV} = \frac{-2V(V - 1) - P - V^2}{V - 1}

Step 3: Substitute Operational Metrics ( $V = 3\text{ L/mol}, P = 5\text{ atm}$ )

\left.\frac{dP}{dV}\right|_{(3,5)} = \frac{-2(3)(3 - 1) - 5 - (3)^2}{3 - 1} = \frac{-6(2) - 5 - 9}{2} = \frac{-12 - 14}{2} = \frac{-26}{2} = -13\text{ atm/(L/mol)}

Engineering Conclusion

At $V = 3\text{ L/mol}$ , the rate of pressure drop is $-13\text{ atm/(L/mol)}$ . In rocket combustion chambers and pressure vessels, $\frac{dP}{dV}$ is directly linked to the isothermal compressibility $\beta = -\frac{1}{V}\frac{\partial V}{\partial P}$ . A steep negative derivative indicates that small volume expansions cause rapid pressure drops, providing critical data to design fast-acting injector valves that prevent combustion blow-out or acoustic pressure spikes in liquid rocket engines.

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

The Mechanics of Implicit Differentiation

General Overview
Skill Block 1 develops operational agility in finding the instantaneous rate of change $\frac{dy}{dx}$ of implicit curves. Students learn to treat $y$ as an implicit function $y(x)$ , execute the Chain Rule on mixed-variable terms, and isolate $\frac{dy}{dx}$ algebraically.

Conceptual Framework

Explicit vs. Implicit Curves:
- Explicit: $y = f(x)$ (dependent variable $y$ is isolated).
- Implicit: $F(x, y) = C$ (variables $x$ and $y$ are intertwined; cannot be easily solved for $y$ ).
The Core Chain Rule Mechanism:
- Every time an expression containing the dependent variable $y$ is differentiated with respect to $x$ , the Chain Rule dictates multiplying by the internal rate $\frac{dy}{dx}$ : $\frac{d}{dx}[y^n] = n y^{n-1} \cdot \frac{dy}{dx}, \qquad \frac{d}{dx}[\cos(y)] = -\sin(y) \cdot \frac{dy}{dx}$
The Four-Step Isolation Algorithm:
1. Differentiate: Take $\frac{d}{dx}$ of both sides of the equation term-by-term.
2. Collect: Group all terms containing $\frac{dy}{dx}$ on one side (typically the left) and move all non- $\frac{dy}{dx}$ terms to the other side.
3. Factor: Factor out $\frac{dy}{dx}$ as a common factor.
4. Solve: Divide by the remaining coefficient expression to cleanly isolate $\frac{dy}{dx}$ .

Key Equations & Structural Rules

Implicit Differentiation Chain Rule: $\frac{d}{dx}[g(y)] = g'(y) \cdot \frac{dy}{dx}$ (see Implicit Differentiation)
Implicit Product Rule: $\frac{d}{dx}[u(x) \cdot v(y)] = \frac{du}{dx} \cdot v(y) + u(x) \cdot \frac{dv}{dy} \frac{dy}{dx}$ (see Product Rule)
Point-Slope Tangent Line Equation: $y - y_1 = m(x - x_1), \qquad \text{where } m = \left.\frac{dy}{dx}\right|_{(x_1, y_1)}$

Worked Example A: Differentiating Basic Polynomial Relations with Product Rule

Find $\frac{dy}{dx}$ for the implicit relation $x^2y - 2 = 3y^3$ .

💡 Reveal Worked Solution

Governing Formula: The Product Rule and Implicit Chain Rule: $\frac{d}{dx}[uv] = \frac{du}{dx}v + u\frac{dv}{dx}, \qquad \frac{d}{dx}[g(y)] = g'(y)\frac{dy}{dx}$ (These rules are defined in Product Rule and Implicit Differentiation.)
Differentiate both sides with respect to $x$ : $\frac{d}{dx}[x^2y] - \frac{d}{dx}[2] = \frac{d}{dx}[3y^3]$ $\left(2x \cdot y + x^2 \frac{dy}{dx}\right) - 0 = 9y^2 \frac{dy}{dx}$ $2xy + x^2\frac{dy}{dx} = 9y^2\frac{dy}{dx}$
Collect terms containing $\frac{dy}{dx}$ on one side: $2xy = 9y^2\frac{dy}{dx} - x^2\frac{dy}{dx}$
Factor out $\frac{dy}{dx}$ and solve: $2xy = \frac{dy}{dx}(9y^2 - x^2) \implies \frac{dy}{dx} = \frac{2xy}{9y^2 - x^2}$

Worked Example B: Managing Trigonometric Chains in Implicit Relations

Find $\frac{dy}{dx}$ for the equation $\cos(y) = x^2 - y^2$ .

💡 Reveal Worked Solution

Governing Formula: The Trigonometric Chain Rule: $\frac{d}{dx}[\cos(y)] = -\sin(y)\frac{dy}{dx}$ (This rule is defined in Implicit Differentiation.)
Differentiate both sides with respect to $x$ : $-\sin(y)\frac{dy}{dx} = 2x - 2y\frac{dy}{dx}$
Collect terms containing $\frac{dy}{dx}$ on the left side: $2y\frac{dy}{dx} - \sin(y)\frac{dy}{dx} = 2x$
Factor out $\frac{dy}{dx}$ and solve: $\frac{dy}{dx}(2y - \sin y) = 2x \implies \frac{dy}{dx} = \frac{2x}{2y - \sin y}$

Worked Example C: Solving Linear Systems in $\frac{dy}{dx}$

Find $\frac{dy}{dx}$ for the relation $y^2 - 2x = 1 - 2y$ .

💡 Reveal Worked Solution

Governing Formula: The Power Chain Rule: $\frac{d}{dx}[y^n] = n y^{n-1}\frac{dy}{dx}$ (This rule is defined in Implicit Differentiation.)
Differentiate both sides with respect to $x$ : $2y\frac{dy}{dx} - 2 = 0 - 2\frac{dy}{dx}$
Collect $\frac{dy}{dx}$ terms on the left side and constants on the right side: $2y\frac{dy}{dx} + 2\frac{dy}{dx} = 2$
Factor out $\frac{dy}{dx}$ and simplify: $\frac{dy}{dx}(2y + 2) = 2 \implies \frac{dy}{dx} = \frac{2}{2(y + 1)} = \frac{1}{y + 1}$

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 why differentiating an expression with respect to $x$ treats $x^3$ and $y^3$ differently. What role does the Chain Rule play?
(b) Deconstruct Worked Example B: Why does the left-side derivative contain $-\sin(y)\frac{dy}{dx}$ rather than simply $-\sin(y)$ ? What common error does this prevent?
(c) Review the linked Engineering Applications (Mechanical Kinematics (Robotic Linkages)). Why must the industrial cutting robot compute the normal slope $m_{\text{normal}} = -\frac{1}{dy/dx}$ from the implicit ellipse rather than relying on an explicit function?

Practice Space Solve this problem on paper or tablet.

Review the Core Chain Rule Mechanism in the Conceptual Framework, inspect Step 1 of Worked Example B, and review Step 4 of Mechanical Kinematics (Robotic Linkages).

Answer: * (a) $x$ is the independent variable ( $\frac{dx}{dx} = 1$ ), whereas $y$ is an implicit function of $x$ ( $y = y(x)$ ), requiring the Chain Rule $\frac{d}{dx}[g(y)] = g'(y)\frac{dy}{dx}$ .

(b) Differentiating the outer cosine wrapper gives $-\sin(y)$ , which must be multiplied by the inner derivative $\frac{dy}{dx}$ . Dropping $\frac{dy}{dx}$ treats $y$ as an independent constant, which is invalid.
(c) Ellipses fail the Vertical Line Test and cannot be expressed as a single explicit function; evaluating $\frac{dy}{dx}$ implicitly allows the robot to determine the perpendicular cutting normal ( $m_{\text{normal}} = 1.5$ ) at any coordinate on the track.

Step-by-Step Execution

(a) In single-variable calculus with respect to $x$ , the independent variable changes freely, so $\frac{d}{dx}[x^3] = 3x^2 \cdot \frac{dx}{dx} = 3x^2(1) = 3x^2$ . However, $y$ is dependent on $x$ through the relation ( $y = y(x)$ ). By the Chain Rule, differentiating the outer layer with respect to $y$ must be chained with the instantaneous rate of change of the inner function with respect to $x$ : $\frac{d}{dx}[y^3] = 3y^2 \cdot \frac{dy}{dx}$ .
(b) In Worked Example B, the term $\cos(y)$ represents a composite function where the outer shell is $\cos(u)$ and the inner function is $u = y(x)$ . Applying the Chain Rule yields $\frac{d}{dx}[\cos(y)] = -\sin(y) \cdot \frac{dy}{dx}$ . Forgetting the factor $\frac{dy}{dx}$ is the single most common student error in implicit differentiation, as it leaves no derivative term to isolate.
(c) In Mechanical Kinematics (Robotic Linkages), the plasma cutter follows the closed ellipse $4x^2 + 9y^2 = 36$ . Because an ellipse has two $y$ -values for every $x$ -value, solving for $y = \pm \frac{2}{3}\sqrt{9-x^2}$ introduces cumbersome square root branches. Implicit differentiation yields $\frac{dy}{dx} = -\frac{4x}{9y}$ directly. Evaluating this at $\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)$ gives tangent slope $-\frac{2}{3}$ , so the automated wrist motor aligns to the normal slope $m_{\text{normal}} = \frac{3}{2} = 1.5$ perpendicular to the cut surface.

Practice Problem 2: Mixed Polynomial and Product Rule Implicit Derivative

Find $\frac{dy}{dx}$ for the function $y = f(x)$ implicitly defined by the polynomial relation:

x^5 + y^5 = 1 + xy

Practice Space Solve this problem on paper or tablet.

State the Implicit Differentiation Chain Rule and the Product Rule: $\frac{d}{dx}[xy] = 1\cdot y + x\frac{dy}{dx}$ . Differentiate both sides, collect all $\frac{dy}{dx}$ terms on the left, and isolate.

Answer:

\frac{dy}{dx} = \frac{y - 5x^4}{5y^4 - x}

Step-by-Step Execution

Governing Formula: The Product Rule and Implicit Chain Rule: $\frac{d}{dx}[uv] = u'v + uv', \qquad \frac{d}{dx}[y^n] = n y^{n-1}\frac{dy}{dx}$ (These rules are defined in Product Rule and Implicit Differentiation.)
Differentiate both sides with respect to $x$ : $\frac{d}{dx}[x^5] + \frac{d}{dx}[y^5] = \frac{d}{dx}[1] + \frac{d}{dx}[xy]$ $5x^4 + 5y^4\frac{dy}{dx} = 0 + \left(1 \cdot y + x\frac{dy}{dx}\right)$ $5x^4 + 5y^4\frac{dy}{dx} = y + x\frac{dy}{dx}$
Collect terms containing $\frac{dy}{dx}$ on the left side: $5y^4\frac{dy}{dx} - x\frac{dy}{dx} = y - 5x^4$
Factor out $\frac{dy}{dx}$ and solve: $\frac{dy}{dx}(5y^4 - x) = y - 5x^4 \implies \frac{dy}{dx} = \frac{y - 5x^4}{5y^4 - x}$

Practice Problem 3: Exponential and Product Chain Implicit Derivative

Find $\frac{dy}{dx}$ for the implicit relation:

e^{x-y} = 2xy

Practice Space Solve this problem on paper or tablet.

State the Exponential Chain Rule: $\frac{d}{dx}[e^u] = e^u \frac{du}{dx}$ with $u = x - y$ , and the Product Rule on the right side.

Answer:

\frac{dy}{dx} = \frac{e^{x-y} - 2y}{2x + e^{x-y}}

Step-by-Step Execution

Governing Formula: The Exponential Chain Rule and Product Rule: $\frac{d}{dx}[e^{g(x,y)}] = e^{g(x,y)} \frac{d}{dx}[g(x,y)], \qquad \frac{d}{dx}[2xy] = 2y + 2x\frac{dy}{dx}$ (These rules are defined in Derivatives of Exponential Functions and Implicit Differentiation.)
Differentiate both sides with respect to $x$ : $e^{x-y} \cdot \frac{d}{dx}[x - y] = 2\left(1 \cdot y + x\frac{dy}{dx}\right)$ $e^{x-y}\left(1 - \frac{dy}{dx}\right) = 2y + 2x\frac{dy}{dx}$
Distribute the exponential term: $e^{x-y} - e^{x-y}\frac{dy}{dx} = 2y + 2x\frac{dy}{dx}$
Group all $\frac{dy}{dx}$ terms on the right side: $e^{x-y} - 2y = 2x\frac{dy}{dx} + e^{x-y}\frac{dy}{dx}$ $e^{x-y} - 2y = \frac{dy}{dx}(2x + e^{x-y})$
Isolate $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{e^{x-y} - 2y}{2x + e^{x-y}}$

Practice Problem 4: Trigonometric Quotient Implicit Derivative

Find $\frac{dy}{dx}$ for the implicit equation:

\sin\left(\frac{y}{x}\right) = 2x^3 - 3y^2

Practice Space Solve this problem on paper or tablet.

State the Trigonometric Chain Rule: $\frac{d}{dx}[\sin(u)] = \cos(u)\frac{du}{dx}$ , and use the Quotient Rule on the inner argument $u = \frac{y}{x}$ : $\frac{d}{dx}\left[\frac{y}{x}\right] = \frac{x\frac{dy}{dx} - y}{x^2}$ . Clear fractions by multiplying through by $x^2$ .

Answer:

\frac{dy}{dx} = \frac{6x^4 + y\cos(y/x)}{x\cos(y/x) + 6x^2y}

Step-by-Step Execution

Governing Formula: The Trigonometric Chain Rule and Quotient Rule: $\frac{d}{dx}[\sin(u)] = \cos(u)\frac{du}{dx}, \qquad \frac{d}{dx}\left[\frac{y}{x}\right] = \frac{\frac{dy}{dx}\cdot x - y\cdot 1}{x^2}$ (These rules are defined in Quotient Rule and Implicit Differentiation.)
Differentiate both sides with respect to $x$ : $\cos\left(\frac{y}{x}\right) \cdot \left[ \frac{x\frac{dy}{dx} - y}{x^2} \right] = 6x^2 - 6y\frac{dy}{dx}$
Multiply through by $x^2$ to eliminate the fraction: $\cos\left(\frac{y}{x}\right) \cdot \left(x\frac{dy}{dx} - y\right) = 6x^4 - 6x^2y\frac{dy}{dx}$
Distribute cosine and gather $\frac{dy}{dx}$ terms on the left: $x\cos\left(\frac{y}{x}\right)\frac{dy}{dx} - y\cos\left(\frac{y}{x}\right) = 6x^4 - 6x^2y\frac{dy}{dx}$ $x\cos\left(\frac{y}{x}\right)\frac{dy}{dx} + 6x^2y\frac{dy}{dx} = 6x^4 + y\cos\left(\frac{y}{x}\right)$
Factor out $\frac{dy}{dx}$ and solve: $\frac{dy}{dx}\left[ x\cos\left(\frac{y}{x}\right) + 6x^2y \right] = 6x^4 + y\cos\left(\frac{y}{x}\right)$ $\frac{dy}{dx} = \frac{6x^4 + y\cos\left(\frac{y}{x}\right)}{x\cos\left(\frac{y}{x}\right) + 6x^2y}$

Practice Problem 5: Tangent Line to an Implicit Transcendental Curve

Consider the curve defined by the equation $y + \sin y = x + \pi$ . Find the equation of the tangent line to the curve at the point $(\pi, 2\pi)$ in both point-slope and slope-intercept forms.

Practice Space Solve this problem on paper or tablet.

Differentiate both sides implicitly with respect to $x$ to find $\frac{dy}{dx} = \frac{1}{1 + \cos y}$ . Evaluate the slope at $(\pi, 2\pi)$ using $\cos(2\pi) = 1$ , then apply Point-Slope Form: $y - y_1 = m(x - x_1)$ .

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

Slope-Intercept Form: $y = \frac{1}{2}x + \frac{3\pi}{2}$

Step-by-Step Execution

Governing Formula: The Point-Slope Equation of a Tangent Line: $y - y_1 = m(x - x_1), \qquad m = \left.\frac{dy}{dx}\right|_{(x_1, y_1)}$ (This formula is defined in Implicit Differentiation.)
Differentiate the curve equation with respect to $x$ : $\frac{d}{dx}[y] + \frac{d}{dx}[\sin y] = \frac{d}{dx}[x] + \frac{d}{dx}[\pi]$ $\frac{dy}{dx} + \cos(y)\frac{dy}{dx} = 1 + 0$
Isolate $\frac{dy}{dx}$ : $\frac{dy}{dx}(1 + \cos y) = 1 \implies \frac{dy}{dx} = \frac{1}{1 + \cos y}$
Evaluate slope at $(\pi, 2\pi)$ : $m = \left.\frac{dy}{dx}\right|_{(\pi, 2\pi)} = \frac{1}{1 + \cos(2\pi)} = \frac{1}{1 + 1} = \frac{1}{2}$
Assemble Tangent Line Equations:
- Point-Slope Form: $y - 2\pi = \frac{1}{2}(x - \pi)$
- Slope-Intercept Form: $y = \frac{1}{2}x - \frac{\pi}{2} + 2\pi \implies y = \frac{1}{2}x + \frac{3\pi}{2}$

Segment 2 of 8

📅 Regroup 1

Review student solutions on collecting $\frac{dy}{dx}$ terms. Emphasize that every term containing $\frac{dy}{dx}$ must be moved to one side before factoring.
Clarify common sign errors when applying the Product Rule on expressions like $-xy$ (distributing the negative sign: $\frac{d}{dx}[-xy] = -y - x\frac{dy}{dx}$ ).

Segment 3 of 8

Skill Block 2

Real-Life Engineering Applications

⚡

Electrical Engineering

Electrical Engineering (Signal Degradation & Attenuation)

An electrical engineer is analyzing high-frequency AC signal loss across a subsea telecommunications cable subject to inductive noise and dielectric dissipation. The measured output voltage signal $V(t)$ (in Volts) over time $t$ (in milliseconds) follows a multi-factor decaying wave:

V(t) = \frac{e^{-2t} \sin(3t)}{\sqrt{t^2 + 5}}

Apply the natural logarithm to both sides of the voltage expression and expand completely using logarithmic product, quotient, and power properties.
Differentiate implicitly with respect to $t$ to find an expression for the instantaneous rate of voltage change $\frac{dV}{dt}$ .
Evaluate the exact rate of voltage degradation $\frac{dV}{dt}$ at $t = 2\text{ ms}$ , given that the carrier voltage amplitude at that instant is $V(2) \approx 0.0104\text{ V}$ .

💡 Show Worked Derivation

Problem Overview

We calculate the rate of voltage change ( $\frac{dV}{dt}$ ) for an attenuated signal:

V(t) = \frac{e^{-2t} \sin(3t)}{\sqrt{t^2 + 5}}

evaluated at

t = 2\text{ ms}

with

V(2) \approx 0.0104\text{ V}

.

Step-by-Step Mathematical Solution

Step 1: Formally Apply Logarithmic Differentiation

The governing process is Logarithmic Differentiation:

\frac{d}{dt}[\ln V] = \frac{1}{V}\frac{dV}{dt} \implies \frac{dV}{dt} = V \cdot \frac{d}{dt}[\ln(V(t))]

(This rule is defined in Logarithmic Differentiation.)

Take the natural logarithm of both sides:

\ln V = \ln\left[ \frac{e^{-2t} \sin(3t)}{(t^2 + 5)^{1/2}} \right]

Step 2: Expand Using Logarithmic Properties

Using the Product, Quotient, and Power properties:

\ln V = \ln(e^{-2t}) + \ln(\sin(3t)) - \ln(t^2 + 5)^{1/2}

\ln V = -2t + \ln(\sin(3t)) - \frac{1}{2}\ln(t^2 + 5)

Step 3: Differentiate Implicitly with Respect to Time $t$

\frac{1}{V}\frac{dV}{dt} = \frac{d}{dt}[-2t] + \frac{d}{dt}[\ln(\sin(3t))] - \frac{1}{2}\frac{d}{dt}[\ln(t^2 + 5)]

Apply the Chain Rule to each term:

\frac{1}{V}\frac{dV}{dt} = -2 + \frac{3\cos(3t)}{\sin(3t)} - \frac{1}{2}\left(\frac{2t}{t^2 + 5}\right)

\frac{1}{V}\frac{dV}{dt} = -2 + 3\cot(3t) - \frac{t}{t^2 + 5}

Step 4: Multiply by $V$ to Isolate $\frac{dV}{dt}$

\frac{dV}{dt} = V(t) \cdot \left[ -2 + 3\cot(3t) - \frac{t}{t^2 + 5} \right]

Step 5: Evaluate at $t = 2\text{ ms}$ with $V(2) \approx 0.0104\text{ V}$

\left.\frac{dV}{dt}\right|_{t=2} = (0.0104) \cdot \left[ -2 + 3\cot(6\text{ rad}) - \frac{2}{2^2 + 5} \right]

In radian mode,

\cot(6) \approx -3.4364

:

\left.\frac{dV}{dt}\right|_{t=2} = (0.0104) \cdot \left[ -2 + 3(-3.4364) - \frac{2}{9} \right]

\left.\frac{dV}{dt}\right|_{t=2} = (0.0104) \cdot [ -2 - 10.3092 - 0.2222 ] = (0.0104) \cdot [-12.5314] \approx -0.1303\text{ V/ms}

Engineering Conclusion

At $t = 2\text{ ms}$ , the line voltage is collapsing at $-0.1303\text{ V/ms}$ . Electrical and telecommunications engineers monitor this instantaneous attenuation rate to calibrate repeater and amplifier spacing along undersea cables. Calculating this rate via logarithmic differentiation eliminates the massive arithmetic overhead of nested quotient and product expansions, providing a fast, direct formula for real-time DSP equalizer tuning.

⚙️

General Engineering

Linear Fractional Material Strain)

A materials scientist is evaluating an advanced carbon-fiber reinforced polymer matrix designed for high-altitude aerodynamic radomes. Under intense aerodynamic friction and thermal load, the material's volumetric expansion $v$ responds to the normalized operating temperature $T$ according to an empirical power-tower equation:

v(T) = (T^2 + 1)^{\ln(T)} \qquad (T > 1)

Apply logarithmic differentiation to derive an expression for the volumetric sensitivity rate $\frac{dv}{dT}$ .
Determine the fractional rate of thermal expansion (defined as $\frac{1}{v}\frac{dv}{dT}$ , measuring percent expansion per degree temperature rise).
Evaluate the exact fractional rate of expansion $\frac{1}{v}\frac{dv}{dT}$ at the critical temperature baseline $T = e$ .

💡 Show Worked Derivation

Problem Overview

We calculate the absolute rate $\frac{dv}{dT}$ and fractional rate of thermal expansion $\frac{1}{v}\frac{dv}{dT}$ for an aerodynamic polymer composite governed by:

v(T) = (T^2 + 1)^{\ln(T)}

evaluated at the temperature baseline

T = e

.

Step-by-Step Mathematical Solution

Step 1: Formally Apply Logarithmic Differentiation

The governing process is Logarithmic Differentiation:

\ln v = \ln\left[ (T^2 + 1)^{\ln T} \right]

(This technique is defined in Logarithmic Differentiation.)

Bring the variable exponent down using the Power Property of logarithms:

\ln v = \ln(T) \cdot \ln(T^2 + 1)

Step 2: Differentiate Both Sides with Respect to Temperature $T$

Apply the Product Rule on the right side with $u = \ln(T)$ and $w = \ln(T^2 + 1)$ :

\frac{d}{dT}[\ln v] = \frac{d}{dT}[\ln T] \cdot \ln(T^2 + 1) + \ln(T) \cdot \frac{d}{dT}[\ln(T^2 + 1)]

\frac{1}{v}\frac{dv}{dT} = \left(\frac{1}{T}\right) \cdot \ln(T^2 + 1) + \ln(T) \cdot \left(\frac{2T}{T^2 + 1}\right)

\frac{1}{v}\frac{dv}{dT} = \frac{\ln(T^2 + 1)}{T} + \frac{2T \ln T}{T^2 + 1}

Step 3: Identify the Fractional Rate of Expansion

Because the problem specifically asks for the fractional rate $\frac{1}{v}\frac{dv}{dT}$ , we do not need to multiply by $v$ :

\text{Fractional Rate} = \frac{1}{v}\frac{dv}{dT} = \frac{\ln(T^2 + 1)}{T} + \frac{2T \ln T}{T^2 + 1}

Step 4: Evaluate at $T = e$

Recall that $\ln(e) = 1$ :

\left.\frac{1}{v}\frac{dv}{dT}\right|_{T=e} = \frac{\ln(e^2 + 1)}{e} + \frac{2e \ln(e)}{e^2 + 1} = \frac{\ln(e^2 + 1)}{e} + \frac{2e}{e^2 + 1}

Numerically, since

e \approx 2.71828

and

e^2 + 1 \approx 8.389

:

\ln(8.389) \approx 2.1269 \implies \frac{2.1269}{2.7183} + \frac{5.4366}{8.389} \approx 0.7824 + 0.6481 \approx 1.4305\text{ units}^{-1}

Engineering Conclusion

At $T = e$ , the fractional volumetric expansion rate of the composite material is $\approx 1.43\text{ K}^{-1}$ (meaning volume expands by approximately $143\%$ per normalized temperature unit). In aerospace and materials engineering, capturing the logarithmic derivative directly yields the linear/volumetric coefficient of thermal expansion (CTE) without requiring cumbersome multiplication by the base volume function $v(T)$ . This enables stress engineers to predict thermal delamination thresholds along aircraft radome seams during high-speed atmospheric cruise.

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

Logarithmic Differentiation

General Overview
Skill Block 2 introduces Logarithmic Differentiation—a powerful technique that transforms difficult derivatives into manageable linear operations. Students learn to handle variable-base variable-exponent functions ( $y = [f(x)]^{g(x)}$ ) where standard rules fail, and pre-process multi-factor algebraic fractions into simple sums and differences before differentiating.

Conceptual Framework

When Standard Rules Fail ( $f(x)^{g(x)}$ ):
- The Power Rule requires a constant exponent ( $x^n$ ).
- The Exponential Rule requires a constant base ( $b^x$ ).
- A variable base with a variable exponent (like $x^x$ or $(\cos x)^{\sin x}$ ) satisfies neither requirement. Taking the natural log pulls the exponent down into a multiplicative coefficient via the Power Property: $\ln([f(x)]^{g(x)}) = g(x) \cdot \ln(f(x))$ .
Logarithmic Algebraic Pre-Processing:
- Complex expressions involving multiple radical roots, quotients, and products can be expanded into sums and differences using: $\ln(AB) = \ln A + \ln B, \qquad \ln\left(\frac{A}{B}\right) = \ln A - \ln B, \qquad \ln(A^k) = k \ln A$
- Expanding first eliminates massive Quotient and Product rule trees.
The Implicit Left Side & The Fractional Rate:
- Differentiating $\ln y$ with respect to $x$ always produces $\frac{1}{y}\frac{dy}{dx}$ by the Chain Rule.
- The ratio $\frac{1}{y}\frac{dy}{dx} = \frac{y'}{y}$ represents the fractional (or relative) rate of change. Multiplying by $y$ restores the absolute derivative $\frac{dy}{dx}$ .

Key Equations & Structural Rules

Logarithmic Differentiation Procedure: $\ln y = \ln[f(x)] \implies \frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}[\ln(f(x))] \implies \frac{dy}{dx} = f(x) \cdot \frac{d}{dx}[\ln(f(x))]$ (see Logarithmic Differentiation)
The Power Property of Logarithms: $\ln(A^B) = B \cdot \ln(A)$ (see Logarithmic and Exponential Properties)

Worked Example A: Handling Variable Base and Variable Exponent Functions

Find $\frac{dy}{dx}$ for the function $y = x^x$ (for $x > 0$ ).

💡 Reveal Worked Solution

Governing Formula: The Logarithmic Differentiation Process: $\ln y = \ln[f(x)], \qquad \frac{d}{dx}[\ln y] = \frac{1}{y}\frac{dy}{dx}$ (This technique is defined in Logarithmic Differentiation.)
Apply the natural logarithm to both sides: $\ln y = \ln(x^x)$
Bring down the exponent using the Power Property: $\ln y = x \cdot \ln x$
Differentiate implicitly with respect to $x$ : Apply the Product Rule on the right side: $\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}[x]\cdot \ln x + x\cdot \frac{d}{dx}[\ln x] = 1 \cdot \ln x + x\cdot\left(\frac{1}{x}\right) = \ln x + 1$
Multiply by $y$ and back-substitute $y = x^x$ : $\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$

Worked Example B: Simplifying Multi-Factor Radical and Quotient Expressions

Find $\frac{dy}{dx}$ for the function $y = \frac{x \sqrt[3]{2x-7}}{(x^3 + 1)^4}$ .

💡 Reveal Worked Solution

Governing Formula: Logarithmic Expansion Properties: $\ln\left[\frac{A \cdot B^{1/3}}{C^4}\right] = \ln A + \frac{1}{3}\ln B - 4\ln C$ (These properties are defined in Logarithmic and Exponential Properties and Logarithmic Differentiation.)
Take the natural logarithm of both sides: $\ln y = \ln\left[ \frac{x (2x-7)^{1/3}}{(x^3+1)^4} \right]$
Expand using log properties: $\ln y = \ln x + \frac{1}{3}\ln(2x-7) - 4\ln(x^3+1)$
Differentiate implicitly with respect to $x$ : $\frac{1}{y}\frac{dy}{dx} = \frac{1}{x} + \frac{1}{3}\left(\frac{2}{2x-7}\right) - 4\left(\frac{3x^2}{x^3+1}\right) = \frac{1}{x} + \frac{2}{3(2x-7)} - \frac{12x^2}{x^3+1}$
Multiply by $y$ and back-substitute: $\frac{dy}{dx} = \frac{x \sqrt[3]{2x-7}}{(x^3+1)^4} \left[ \frac{1}{x} + \frac{2}{3(2x-7)} - \frac{12x^2}{x^3+1} \right]$

Practice Problem 6: Concept & Reading Check

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

(a) In the Conceptual Framework, why does treating $y = x^x$ with the Power Rule ( $\frac{dy}{dx} = x \cdot x^{x-1} = x^x$ ) or Exponential Rule ( $\frac{dy}{dx} = \ln(x)x^x$ ) produce incorrect results?
(b) Deconstruct Worked Example B: Contrast the logarithmic approach with computing the derivative directly via the Quotient and Product rules. How does logarithmic pre-processing simplify the algebra?
(c) Review the linked Engineering Applications (Materials Engineering (Non-Linear Fractional Material Strain)). Why does the materials engineer evaluate the fractional rate of change $\frac{1}{v}\frac{dv}{dT}$ directly without multiplying by $v(T)$ ?

Practice Space Solve this problem on paper or tablet.

Review "When Standard Rules Fail" in the Conceptual Framework, compare Worked Example B with traditional quotient rule steps, and check Step 3 of Materials Engineering (Non-Linear Fractional Material Strain).

Answer: * (a) The Power Rule requires a constant numerical exponent; the Exponential Rule requires a constant base. $x^x$ has both a variable base and a variable exponent, requiring logarithmic differentiation to pull the exponent down.

(b) A direct approach requires a nested Quotient Rule whose numerator contains a multi-factor Product and Chain Rule. Logarithmic pre-processing converts products into additions and quotients into subtractions, yielding three independent single-term derivatives.
(c) The material's thermal expansion coefficient is defined physically as the fractional volume change per degree temperature ( $\frac{1}{v}\frac{dv}{dT}$ ). Since $\frac{d}{dT}[\ln v] = \frac{1}{v}\frac{dv}{dT}$ , the fractional rate is already isolated on the left side.

Step-by-Step Execution

(a) In elementary calculus, the Power Rule $\frac{d}{dx}[x^n] = n x^{n-1}$ is proven strictly under the condition that the power $n$ is a constant number. The Exponential Rule $\frac{d}{dx}[b^x] = \ln(b)b^x$ requires that the base $b$ is a constant. In $y = x^x$ , both the base and the exponent change simultaneously as $x$ varies. The correct derivative $x^x(\ln x + 1) = x^x\ln x + x^x$ is actually the sum of treating the base as variable plus treating the exponent as variable—a result that can only be proven via logarithmic differentiation.
(b) In Worked Example B, applying the Quotient Rule directly would require setting up $\frac{u'v - uv'}{v^2}$ where $u = x(2x-7)^{1/3}$ requires the Product Rule and $v = (x^3+1)^4$ requires the Chain Rule, producing massive fractions with compound common denominators. By taking the logarithm first, the expression expands into $\ln x + \frac{1}{3}\ln(2x-7) - 4\ln(x^3+1)$ . Each term is then differentiated in a single line without any common denominator algebra.
(c) In Materials Engineering (Non-Linear Fractional Material Strain), the volume function is $v(T) = (T^2+1)^{\ln T}$ . The engineer is tasked with finding the fractional rate of strain $\frac{1}{v}\frac{dv}{dT}$ , which represents the localized percentage expansion per degree Kelvin. Because differentiating $\ln v$ naturally yields $\frac{1}{v}\frac{dv}{dT}$ on the left side, the fractional rate is obtained immediately without needing to multiply back by the volume function $v(T)$ .

Practice Problem 7: Logarithmic Differentiation of Variable Exponent Function

Find the derivative $\frac{dy}{dx}$ of the function:

y = x^{\ln x} \qquad (x > 0)

Practice Space Solve this problem on paper or tablet.

Apply $\ln$ to both sides to get $\ln y = (\ln x)(\ln x) = (\ln x)^2$ . Differentiate implicitly using the Chain Rule on $(\ln x)^2$ , then multiply by $y$ and back-substitute.

Answer:

\frac{dy}{dx} = x^{\ln x} \cdot \left(\frac{2\ln x}{x}\right)

Step-by-Step Execution

Governing Formula: The Power Property of Logarithms and Logarithmic Differentiation: $\ln(A^B) = B\ln A, \qquad \frac{d}{dx}[\ln y] = \frac{1}{y}\frac{dy}{dx}$ (These rules are defined in Logarithmic Differentiation.)
Apply the natural logarithm to both sides: $\ln y = \ln\left(x^{\ln x}\right)$
Bring down the exponent using the Power Property: $\ln y = (\ln x) \cdot (\ln x) = (\ln x)^2$
Differentiate implicitly with respect to $x$ using the Chain Rule: $\frac{1}{y}\frac{dy}{dx} = 2(\ln x) \cdot \frac{d}{dx}[\ln x] = 2(\ln x) \cdot \frac{1}{x} = \frac{2\ln x}{x}$
Multiply by $y$ and back-substitute $y = x^{\ln x}$ : $\frac{dy}{dx} = y \cdot \left(\frac{2\ln x}{x}\right) = x^{\ln x} \cdot \frac{2\ln x}{x}$

Practice Problem 8: Logarithmic Differentiation of Trigonometric Base and Exponent

Find the derivative $\frac{dy}{dx}$ of the function:

y = (\cos x)^{\sin x} \qquad \left(0 < x < \frac{\pi}{2}\right)

Practice Space Solve this problem on paper or tablet.

Take the natural log of both sides to get $\ln y = \sin(x)\ln(\cos x)$ . Differentiate using the Product Rule on the right side with the Chain Rule for $\ln(\cos x)$ , then multiply by $y$ .

Answer:

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

Step-by-Step Execution

Governing Formula: Logarithmic Differentiation with Product Rule: $\ln y = g(x)\ln[f(x)] \implies \frac{1}{y}\frac{dy}{dx} = g'(x)\ln[f(x)] + g(x)\frac{f'(x)}{f(x)}$ (This formula is defined in Logarithmic Differentiation.)
Apply the natural logarithm to both sides: $\ln y = \ln\left[(\cos x)^{\sin x}\right] = \sin(x) \cdot \ln(\cos x)$
Differentiate implicitly with respect to $x$ using the Product Rule: $\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}[\sin x]\cdot \ln(\cos x) + \sin(x)\cdot \frac{d}{dx}[\ln(\cos x)]$ Recall that $\frac{d}{dx}[\ln(\cos x)] = \frac{-\sin x}{\cos x}$ : $\frac{1}{y}\frac{dy}{dx} = \cos(x)\ln(\cos x) + \sin(x)\left(\frac{-\sin x}{\cos x}\right) = \cos(x)\ln(\cos x) - \frac{\sin^2 x}{\cos x}$
Multiply by $y$ and back-substitute $y = (\cos x)^{\sin x}$ : $\frac{dy}{dx} = (\cos x)^{\sin x} \left[ \cos(x)\ln(\cos x) - \frac{\sin^2 x}{\cos x} \right]$

Practice Problem 9: Logarithmic Differentiation of Rational Base with Power Exponent

Find the derivative $\frac{dy}{dx}$ of the function:

y = \left(\frac{x+1}{x^2+1}\right)^{x^3}

Practice Space Solve this problem on paper or tablet.

Take $\ln$ of both sides and expand using the Quotient Property: $\ln y = x^3[\ln(x+1) - \ln(x^2+1)]$ . Apply the Product Rule to differentiate, then multiply by $y$ .

Answer:

\frac{dy}{dx} = \left(\frac{x+1}{x^2+1}\right)^{x^3} \left[ 3x^2\ln\left(\frac{x+1}{x^2+1}\right) + x^3\left(\frac{1}{x+1} - \frac{2x}{x^2+1}\right) \right]

Step-by-Step Execution

Governing Formula: Logarithmic Differentiation with Power and Quotient Properties: $\ln\left[\left(\frac{A}{B}\right)^C\right] = C[\ln A - \ln B]$ (These properties are defined in Logarithmic Differentiation.)
Apply the natural logarithm and expand: $\ln y = \ln\left[ \left(\frac{x+1}{x^2+1}\right)^{x^3} \right] = x^3 \cdot \left[ \ln(x+1) - \ln(x^2+1) \right]$
Differentiate implicitly with respect to $x$ using the Product Rule: $\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}[x^3]\cdot \left[\ln(x+1) - \ln(x^2+1)\right] + x^3 \cdot \frac{d}{dx}\left[\ln(x+1) - \ln(x^2+1)\right]$ $\frac{1}{y}\frac{dy}{dx} = 3x^2 \left[\ln(x+1) - \ln(x^2+1)\right] + x^3 \left[ \frac{1}{x+1} - \frac{2x}{x^2+1} \right]$
Recombine the logarithm and multiply by $y$ : $\frac{dy}{dx} = \left(\frac{x+1}{x^2+1}\right)^{x^3} \left\{ 3x^2 \ln\left(\frac{x+1}{x^2+1}\right) + x^3 \left[ \frac{1}{x+1} - \frac{2x}{x^2+1} \right] \right\}$

Segment 4 of 8

📅 Regroup 2

Review logarithmic expansion rules before differentiation. Ensure students do not attempt to differentiate $[f(x)]^{g(x)}$ by pulling $g(x)$ in front and subtracting 1 from the exponent (a common misapplication of the Power Rule).
Emphasize the back-substitution step: the final answer for $\frac{dy}{dx}$ in logarithmic differentiation must have $y$ replaced by the original function expression $f(x)$ so it remains purely in terms of $x$ .

Segment 5 of 8

Skill Block 3

Real-Life Engineering Applications

🚀

Aerospace Engineering

Aerospace Engineering (Structural Integrity under Thermal Expansion)

A structural aerospace engineer is inspecting a circular optical sensor viewing port integrated into the titanium forward bulkhead of a crew exploration vehicle. Due to aerodynamic friction during atmospheric re-entry, uneven thermal expansion deforms the window mounting bezel along an implicit circle:

x^2 + y^2 = 25

where

x

and

y

are coordinate displacements in centimeters. A stress concentration sensor focuses on the perimeter point

(x, y) = (3, 4)

.

Compute the first derivative $\frac{dy}{dx}$ using implicit differentiation.
Differentiate $\frac{dy}{dx}$ a second time with respect to $x$ using the Quotient Rule to establish an expression for the second derivative $\frac{d^2y}{dx^2}$ .
Back-substitute $\frac{dy}{dx}$ and simplify the numerator using the original circle constraint equation ( $x^2 + y^2 = 25$ ).
Evaluate $\frac{d^2y}{dx^2}$ at the physical inspection point $(3, 4)$ , and determine whether the window bezel is concave up or concave down.

💡 Show Worked Derivation

Problem Overview

We calculate the second derivative $\frac{d^2y}{dx^2}$ along an implicit circular viewport:

x^2 + y^2 = 25

evaluated at the coordinate point

(x, y) = (3, 4)

to determine localized curvature and concavity under thermal expansion stress.

Step-by-Step Mathematical Solution

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

Differentiate both sides with respect to $x$ :

\frac{d}{dx}[x^2] + \frac{d}{dx}[y^2] = \frac{d}{dx}[25]

2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}

(This rule is defined in Implicit Differentiation.)

Step 2: Differentiate a Second Time Using the Quotient Rule

Apply the Quotient Rule to $\frac{dy}{dx} = -\frac{x}{y}$ :

\frac{d^2y}{dx^2} = \frac{d}{dx}\left[-\frac{x}{y}\right] = -\left[ \frac{\frac{d}{dx}[x] \cdot y - x \cdot \frac{d}{dx}[y]}{y^2} \right] = -\frac{y - x\frac{dy}{dx}}{y^2} = \frac{-y + x\frac{dy}{dx}}{y^2}

Step 3: Back-Substitute $\frac{dy}{dx} = -\frac{x}{y}$

\frac{d^2y}{dx^2} = \frac{-y + x\left(-\frac{x}{y}\right)}{y^2} = \frac{-y - \frac{x^2}{y}}{y^2}

Step 4: Clear Fractions and Substitute Original Equation Constraint

Multiply numerator and denominator by $y$ :

\frac{d^2y}{dx^2} = \frac{-y^2 - x^2}{y^3} = \frac{-(x^2 + y^2)}{y^3}

Because the curve is constrained by

x^2 + y^2 = 25

:

\frac{d^2y}{dx^2} = -\frac{25}{y^3}

(This technique is detailed in Higher-Order Implicit Differentiation.)

Step 5: Evaluate at $(3, 4)$

Substitute $y = 4$ :

\left.\frac{d^2y}{dx^2}\right|_{(3,4)} = -\frac{25}{4^3} = -\frac{25}{64} \approx -0.3906\text{ cm}^{-1}

Engineering Conclusion

Because $\frac{d^2y}{dx^2} = -\frac{25}{64} < 0$ , the bulkhead viewport edge is strictly concave downward at $(3, 4)$ . In aerospace structural engineering, geometric curvature ( $\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}$ ) determines the severity of stress concentrations. If localized heating causes a boundary to buckle beyond its material curvature allowance, high shear stresses can crack silica glass viewports. Knowing that $\frac{d^2y}{dx^2} = -\frac{25}{64}$ enables stress analysts to verify that the titanium retaining ring distributes re-entry pressure loads evenly around the seal.

⚙️

General Engineering

Automotive Engineering (Suspension Camber & Curvature Analysis)

An automotive chassis and suspension engineer is designing a performance double-wishbone suspension linkage for an electric sports vehicle. As the vehicle corners aggressively over bumps, the tire contact patch moves along an implicit non-linear curve relative to the subframe:

x^3 - y^2 = 4

where

x

is the horizontal track width change (in decimeters,

\text{dm}

) and

y

is the vertical suspension travel (in

\text{dm}

). The vehicle's static ride-height baseline is calibrated at

(x, y) = (2, 2)

.

Use implicit differentiation to calculate the instantaneous slope $\frac{dy}{dx}$ of the suspension travel path at the baseline configuration $(2, 2)$ .
Apply the Quotient Rule and back-substitution to calculate the second derivative $\frac{d^2y}{dx^2}$ at $(2, 2)$ .
Interpret what the sign and numerical magnitude of $\frac{d^2y}{dx^2}$ tell the engineer regarding tire camber recovery and vehicle cornering grip.

💡 Show Worked Derivation

Problem Overview

We calculate the first derivative $\frac{dy}{dx}$ and second derivative $\frac{d^2y}{dx^2}$ along the suspension travel curve:

x^3 - y^2 = 4

evaluated at the baseline configuration

(x, y) = (2, 2)

.

Step-by-Step Mathematical Solution

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

Differentiate both sides with respect to $x$ :

\frac{d}{dx}[x^3] - \frac{d}{dx}[y^2] = \frac{d}{dx}[4]

3x^2 - 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{3x^2}{2y}

(This rule is defined in Implicit Differentiation.)

Evaluate at $(2, 2)$ :

\left.\frac{dy}{dx}\right|_{(2,2)} = \frac{3(2)^2}{2(2)} = \frac{12}{4} = 3

Step 2: Differentiate a Second Time Using the Quotient Rule

Apply the Quotient Rule to $\frac{dy}{dx} = \frac{3x^2}{2y}$ :

\frac{d^2y}{dx^2} = \frac{\frac{d}{dx}[3x^2] \cdot (2y) - (3x^2) \cdot \frac{d}{dx}[2y]}{(2y)^2} = \frac{(6x)(2y) - (3x^2)\left(2\frac{dy}{dx}\right)}{4y^2} = \frac{12xy - 6x^2\frac{dy}{dx}}{4y^2}

Step 3: Back-Substitute $\frac{dy}{dx} = \frac{3x^2}{2y}$

\frac{d^2y}{dx^2} = \frac{12xy - 6x^2\left(\frac{3x^2}{2y}\right)}{4y^2} = \frac{12xy - \frac{18x^4}{2y}}{4y^2} = \frac{12xy - \frac{9x^4}{y}}{4y^2}

Step 4: Clear Fractions

Multiply numerator and denominator by $y$ :

\frac{d^2y}{dx^2} = \frac{12xy^2 - 9x^4}{4y^3}

(This technique is detailed in Higher-Order Implicit Differentiation.)

Step 5: Evaluate at $(2, 2)$

Substitute $x = 2$ and $y = 2$ :

\left.\frac{d^2y}{dx^2}\right|_{(2,2)} = \frac{12(2)(2^2) - 9(2^4)}{4(2^3)} = \frac{12(2)(4) - 9(16)}{4(8)} = \frac{96 - 144}{32} = \frac{-48}{32} = -1.5\text{ dm}^{-1}

Engineering Conclusion

At $(2, 2)$ , the first derivative is $3$ (indicating rapid upward vertical compression per unit track width shift) and the second derivative is $-1.5\text{ dm}^{-1}$ . The negative second derivative confirms that the suspension kinematics are concave downward, causing the tire to gain negative camber as the chassis rolls into a turn. In high-performance automotive engineering, negative camber counters lateral tire rollover, maintaining maximum tire contact patch area against the road and preventing breakaway skids during high-speed maneuvering.

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

Higher-Order Implicit Differentiation

General Overview
Skill Block 3 extends implicit differentiation into higher orders, specifically calculating the second derivative $\frac{d^2y}{dx^2}$ . Students master the two critical milestones of higher-order implicit calculus: executing the nested Quotient Rule while handling internal $\frac{dy}{dx}$ factors, and back-substituting the first derivative along with the original curve constraint to compress solutions.

Conceptual Framework

Slope vs. Curvature:
- First Derivative $\frac{dy}{dx}$ : Instantaneous tangent slope at coordinate $(x, y)$ .
- Second Derivative $\frac{d^2y}{dx^2}$ : Rate of change of the tangent slope, determining curve concavity and geometric curvature $\kappa$ .
The Nested Quotient Rule:
- The first derivative $\frac{dy}{dx}$ of an implicit curve is typically a rational expression $\frac{u(x,y)}{v(x,y)}$ .
- Differentiating with respect to $x$ requires the Quotient Rule. Any derivative of $y$ within this step yields an internal factor of $\frac{dy}{dx}$ .
The Two Simplification Checkpoints:
1. Back-Substitution: Replace all internal $\frac{dy}{dx}$ terms with the isolated first derivative formula.
2. Constraint Substitution: In symmetric curves (circles, hyperbolas, ellipses), look for combinations matching the original governing equation in the numerator (e.g., $x^2 + y^2 = r^2$ or $x^2 - y^2 = 1$ ) to replace variable expressions with constants.

Key Equations & Structural Rules

Second Derivative Operator: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]$ (see Higher-Order Implicit Differentiation)
Quotient Rule for Second Derivatives: $\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{\frac{du}{dx}v - u\frac{dv}{dx}}{v^2}$ (see Quotient Rule)

Worked Example A: Second Derivative of a Circle with Constraint Simplification

Find $\frac{d^2y}{dx^2}$ for the circle relation $x^2 + y^2 = 1$ .

💡 Reveal Worked Solution

Governing Formula: The Quotient Rule for Higher-Order Derivatives: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right] = \frac{u'v - uv'}{v^2}$ (This formula is defined in Higher-Order Implicit Differentiation.)
Find the first derivative $\frac{dy}{dx}$ : $2x + 2y\frac{dy}{dx} = 0 \implies 2y\frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y}$
Differentiate a second time using the Quotient Rule: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left[-\frac{x}{y}\right] = -\left[ \frac{(1)(y) - (x)\left(\frac{dy}{dx}\right)}{y^2} \right] = \frac{-y + x\frac{dy}{dx}}{y^2}$
Back-substitute $\frac{dy}{dx} = -\frac{x}{y}$ : $\frac{d^2y}{dx^2} = \frac{-y + x\left(-\frac{x}{y}\right)}{y^2} = \frac{-y - \frac{x^2}{y}}{y^2}$
Clear fractions and substitute the original constraint ( $x^2 + y^2 = 1$ ): Multiply numerator and denominator by $y$ : $\frac{d^2y}{dx^2} = \frac{-y^2 - x^2}{y^3} = \frac{-(x^2 + y^2)}{y^3}$ Substitute $x^2 + y^2 = 1$ : $\frac{d^2y}{dx^2} = \frac{-(1)}{y^3} = -\frac{1}{y^3}$

Worked Example B: Second Derivative of Higher-Degree Polynomial Relations

Find $\frac{d^2y}{dx^2}$ for the implicit relation $2x^3 - 3y^2 = 8$ .

💡 Reveal Worked Solution

Governing Formula: The Quotient Rule with Implicit Chain Substitution: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{x^2}{y}\right]$ (This rule is defined in Higher-Order Implicit Differentiation.)
Find the first derivative $\frac{dy}{dx}$ : $6x^2 - 6y\frac{dy}{dx} = 0 \implies -6y\frac{dy}{dx} = -6x^2 \implies \frac{dy}{dx} = \frac{x^2}{y}$
Differentiate a second time using the Quotient Rule: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dx}[x^2]\cdot y - x^2\cdot \frac{d}{dx}[y]}{y^2} = \frac{2xy - x^2\left(\frac{dy}{dx}\right)}{y^2}$
Back-substitute $\frac{dy}{dx} = \frac{x^2}{y}$ : $\frac{d^2y}{dx^2} = \frac{2xy - x^2\left(\frac{x^2}{y}\right)}{y^2} = \frac{2xy - \frac{x^4}{y}}{y^2}$
Clear the complex fraction by multiplying top and bottom by $y$ : $\frac{d^2y}{dx^2} = \frac{2xy^2 - x^4}{y^3}$

Practice Problem 10: Concept & Reading Check

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

(a) In the Conceptual Framework, why is leaving a $\frac{dy}{dx}$ term inside your final answer for $\frac{d^2y}{dx^2}$ considered incomplete?
(b) Deconstruct Worked Example A: In Step 5, how does recognizing the circle's original equation $x^2 + y^2 = 1$ simplify the complex algebraic fraction $\frac{-y^2 - x^2}{y^3}$ into $-\frac{1}{y^3}$ ?
(c) Review the linked Engineering Applications (Aerospace Engineering (Structural Integrity under Thermal Expansion)). What physical property of the titanium bulkhead viewport does the second derivative $\frac{d^2y}{dx^2} = -\frac{25}{64}$ quantify during atmospheric re-entry?

Practice Space Solve this problem on paper or tablet.

Review the Two Simplification Checkpoints in the Conceptual Framework, inspect Step 5 of Worked Example A, and read the Engineering Conclusion of Aerospace Engineering (Structural Integrity under Thermal Expansion).

Answer: * (a) A second derivative must be expressed strictly in terms of the independent and dependent position coordinates $(x, y)$ .

(b) Factoring a negative sign out of the numerator yields $-(x^2 + y^2)$ ; substituting the constant constraint value $1$ eliminates both polynomial variables from the numerator.
(c) The second derivative quantifies the geometric concavity and structural curvature ( $\kappa$ ) of the viewport boundary, allowing engineers to calculate thermal stress concentrations and prevent frame cracking.

Step-by-Step Execution

(a) The purpose of computing $\frac{d^2y}{dx^2}$ is to establish a closed-form formula that evaluates the second derivative directly from coordinate data $(x, y)$ . Because $\frac{dy}{dx}$ is an intermediate rate, leaving it un-substituted leaves the formula unresolved. Performing back-substitution ensures that the second derivative is a self-contained function of the curve coordinates.
(b) In Worked Example A, clearing the complex fraction produced $\frac{-y^2 - x^2}{y^3}$ . Factoring out $-1$ gives $\frac{-(x^2 + y^2)}{y^3}$ . Because the point $(x, y)$ is constrained to lie on the unit circle $x^2 + y^2 = 1$ , the quantity $(x^2 + y^2)$ in the numerator can be replaced with the constant $1$ , yielding the elegant compressed result $-\frac{1}{y^3}$ .
(c) In Aerospace Engineering (Structural Integrity under Thermal Expansion), aerodynamic heating during atmospheric re-entry causes the titanium bulkhead window frame to warp along $x^2 + y^2 = 25$ . The negative second derivative $\frac{d^2y}{dx^2} = -\frac{25}{64} \approx -0.3906\text{ cm}^{-1}$ indicates downward concavity and directly determines the physical curvature $\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}$ . Curvature spikes amplify localized shear stresses; calculating this value confirms the window seal will not rupture under re-entry pressure loads.

Practice Problem 11: Second Derivative of a Hyperbola with Geometric Simplification

For the hyperbola implicitly defined by the equation $x^2 - y^2 = 1$ , find the second derivative $\frac{d^2y}{dx^2}$ and simplify your answer using the original equation constraint.

Practice Space Solve this problem on paper or tablet.

Differentiate implicitly to find $\frac{dy}{dx} = \frac{x}{y}$ . Differentiate using the Quotient Rule to get $\frac{d^2y}{dx^2} = \frac{y - x(dy/dx)}{y^2}$ . Back-substitute $\frac{dy}{dx}$ , clear fractions to get $\frac{y^2 - x^2}{y^3}$ , and note that $y^2 - x^2 = -(x^2 - y^2) = -1$ .

Answer:

\frac{d^2y}{dx^2} = -\frac{1}{y^3}

Step-by-Step Execution

Governing Formula: The Quotient Rule for Higher-Order Derivatives: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right] = \frac{u'v - uv'}{v^2}$ (This formula is defined in Higher-Order Implicit Differentiation.)
Find the first derivative $\frac{dy}{dx}$ : $\frac{d}{dx}[x^2] - \frac{d}{dx}[y^2] = \frac{d}{dx}[1]$ $2x - 2y\frac{dy}{dx} = 0 \implies 2y\frac{dy}{dx} = 2x \implies \frac{dy}{dx} = \frac{x}{y}$
Differentiate $\frac{dy}{dx}$ using the Quotient Rule: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dx}[x]\cdot y - x\cdot \frac{d}{dx}[y]}{y^2} = \frac{1\cdot y - x\left(\frac{dy}{dx}\right)}{y^2}$
Back-substitute $\frac{dy}{dx} = \frac{x}{y}$ : $\frac{d^2y}{dx^2} = \frac{y - x\left(\frac{x}{y}\right)}{y^2} = \frac{y - \frac{x^2}{y}}{y^2}$
Clear fractions and substitute the hyperbola constraint ( $x^2 - y^2 = 1$ ): Multiply numerator and denominator by $y$ : $\frac{d^2y}{dx^2} = \frac{y^2 - x^2}{y^3}$ Since $x^2 - y^2 = 1$ , multiplying by $-1$ gives $y^2 - x^2 = -1$ : $\frac{d^2y}{dx^2} = \frac{-1}{y^3} = -\frac{1}{y^3}$

Practice Problem 12: Second Derivative of a Transcendental Relation via Chain Rule

For the curve implicitly defined by $y + \cos y = x - \frac{\pi}{3}$ , find the second derivative $\frac{d^2y}{dx^2}$ in terms of $y$ .

Practice Space Solve this problem on paper or tablet.

Differentiate both sides with respect to $x$ to get $\frac{dy}{dx}(1 - \sin y) = 1$ , so $\frac{dy}{dx} = (1 - \sin y)^{-1}$ . Differentiate with respect to $x$ using the Power Chain Rule, and back-substitute $\frac{dy}{dx}$ .

Answer:

\frac{d^2y}{dx^2} = \frac{\cos y}{(1 - \sin y)^3}

Step-by-Step Execution

Governing Formula: The Power Chain Rule for Higher-Order Derivatives: $\frac{d}{dx}[(g(y))^{-1}] = -1(g(y))^{-2} \cdot g'(y)\frac{dy}{dx}$ (This rule is defined in Higher-Order Implicit Differentiation.)
Find the first derivative $\frac{dy}{dx}$ : $\frac{d}{dx}[y] + \frac{d}{dx}[\cos y] = \frac{d}{dx}[x] - \frac{d}{dx}\left[\frac{\pi}{3}\right]$ $\frac{dy}{dx} - \sin(y)\frac{dy}{dx} = 1 - 0$ $\frac{dy}{dx}(1 - \sin y) = 1 \implies \frac{dy}{dx} = \frac{1}{1 - \sin y} = (1 - \sin y)^{-1}$
Differentiate a second time with respect to $x$ using the Chain Rule: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left[(1 - \sin y)^{-1}\right] = -1(1 - \sin y)^{-2} \cdot \frac{d}{dx}[1 - \sin y]$ $\frac{d^2y}{dx^2} = -\frac{1}{(1 - \sin y)^2} \cdot \left(-\cos(y)\frac{dy}{dx}\right) = \frac{\cos y}{(1 - \sin y)^2} \cdot \frac{dy}{dx}$
Back-substitute $\frac{dy}{dx} = \frac{1}{1 - \sin y}$ : $\frac{d^2y}{dx^2} = \frac{\cos y}{(1 - \sin y)^2} \cdot \left(\frac{1}{1 - \sin y}\right) = \frac{\cos y}{(1 - \sin y)^3}$

Segment 6 of 8

📅 Regroup 3

Emphasize the algebraic payoff of simplifying higher-order derivatives with original curve equations ( $x^2 + y^2 = r^2$ , $x^2 - y^2 = 1$ ). Show how this geometric substitution turns unwieldy fractions into compact terms.
Review student work on Problem 12: demonstrate why writing $\frac{dy}{dx} = (1 - \sin y)^{-1}$ and using the Power Chain Rule is often faster and less error-prone than applying the Quotient Rule.

Segment 7 of 8

📅 Synthesis Wrap-up

Core Takeaways from Session 09:
1. Implicit Differentiation: Treat $y$ as $y(x)$ ; append $\frac{dy}{dx}$ whenever differentiating $y$ . Collect, factor, and solve.
2. Logarithmic Differentiation: The indispensable tool for variable towers ( $f(x)^{g(x)}$ ) and multi-factor rational systems. Expand with log properties before differentiating.
3. Higher-Order Derivatives: Differentiate $\frac{dy}{dx}$ via quotient or power chain rules, back-substitute $\frac{dy}{dx}$ , and simplify with original curve constraints.
Homework Preparation: Complete Knewton Alta assignments on Implicit Differentiation and Logarithmic Differentiation. Review tangent line setups and practice checking that all final derivatives for logarithmic differentiation are expressed purely in terms of $x$ .

Segment 8 of 8

Implicit Differentiation and Logarithmic Differentiation

📅 Micro-Lecture

Skill Block 1

Real-Life Engineering Applications

Mechanical Kinematics (Robotic Linkages)

Problem Overview

Step-by-Step Mathematical Solution

Step 1: Formally Apply Implicit Differentiation

Step 2: Isolate the First Derivative dydx\frac{dy}{dx}dxdy​

Step 3: Evaluate at the Operational Coordinates (32,2)\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)(2​3​,2​)

Step 4: Determine the Perpendicular Normal Direction

Engineering Conclusion

Linear Fluids)

Problem Overview

Step-by-Step Mathematical Solution

Step 1: Formally Apply the Product Rule with Implicit Differentiation

Step 2: Algebraically Isolate dPdV\frac{dP}{dV}dVdP​

Step 3: Substitute Operational Metrics (V=3 L/mol,P=5 atmV = 3\text{ L/mol}, P = 5\text{ atm}V=3 L/mol,P=5 atm)

Engineering Conclusion

Worked Example A: Differentiating Basic Polynomial Relations with Product Rule

Worked Example B: Managing Trigonometric Chains in Implicit Relations

Worked Example C: Solving Linear Systems in dydx\frac{dy}{dx}dxdy​

Practice Problem 1: Concept & Reading Check

Step-by-Step Execution

Practice Problem 2: Mixed Polynomial and Product Rule Implicit Derivative

Step-by-Step Execution

Practice Problem 3: Exponential and Product Chain Implicit Derivative

Step-by-Step Execution

Practice Problem 4: Trigonometric Quotient Implicit Derivative

Step-by-Step Execution

Practice Problem 5: Tangent Line to an Implicit Transcendental Curve

Step-by-Step Execution

📅 Regroup 1

Skill Block 2

Real-Life Engineering Applications

Electrical Engineering (Signal Degradation & Attenuation)

Problem Overview

Step-by-Step Mathematical Solution

Step 1: Formally Apply Logarithmic Differentiation

Step 2: Expand Using Logarithmic Properties

Step 3: Differentiate Implicitly with Respect to Time ttt

Step 4: Multiply by VVV to Isolate dVdt\frac{dV}{dt}dtdV​

Step 5: Evaluate at t=2 mst = 2\text{ ms}t=2 ms with V(2)≈0.0104 VV(2) \approx 0.0104\text{ V}V(2)≈0.0104 V

Engineering Conclusion

Linear Fractional Material Strain)

Problem Overview

Step-by-Step Mathematical Solution

Step 1: Formally Apply Logarithmic Differentiation

Step 2: Differentiate Both Sides with Respect to Temperature TTT

Step 3: Identify the Fractional Rate of Expansion

Step 4: Evaluate at T=eT = eT=e

Engineering Conclusion

Worked Example A: Handling Variable Base and Variable Exponent Functions

Worked Example B: Simplifying Multi-Factor Radical and Quotient Expressions

Practice Problem 6: Concept & Reading Check

Step-by-Step Execution

Practice Problem 7: Logarithmic Differentiation of Variable Exponent Function

Step-by-Step Execution

Practice Problem 8: Logarithmic Differentiation of Trigonometric Base and Exponent

Step-by-Step Execution

Practice Problem 9: Logarithmic Differentiation of Rational Base with Power Exponent

Step-by-Step Execution

📅 Regroup 2

Skill Block 3

Real-Life Engineering Applications

Aerospace Engineering (Structural Integrity under Thermal Expansion)

Problem Overview

Step-by-Step Mathematical Solution

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

Step 2: Differentiate a Second Time Using the Quotient Rule

Step 3: Back-Substitute dydx=−xy\frac{dy}{dx} = -\frac{x}{y}dxdy​=−yx​

Step 4: Clear Fractions and Substitute Original Equation Constraint

Step 5: Evaluate at (3,4)(3, 4)(3,4)

Engineering Conclusion

Automotive Engineering (Suspension Camber & Curvature Analysis)

Problem Overview

Step-by-Step Mathematical Solution

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

Step 2: Differentiate a Second Time Using the Quotient Rule

Step 3: Back-Substitute dydx=3x22y\frac{dy}{dx} = \frac{3x^2}{2y}dxdy​=2y3x2​

Step 2: Isolate the First Derivative $\frac{dy}{dx}$

Step 3: Evaluate at the Operational Coordinates $\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)$

Step 2: Algebraically Isolate $\frac{dP}{dV}$

Step 3: Substitute Operational Metrics ( $V = 3\text{ L/mol}, P = 5\text{ atm}$ )

Worked Example C: Solving Linear Systems in $\frac{dy}{dx}$

Step 3: Differentiate Implicitly with Respect to Time $t$

Step 4: Multiply by $V$ to Isolate $\frac{dV}{dt}$

Step 5: Evaluate at $t = 2\text{ ms}$ with $V(2) \approx 0.0104\text{ V}$

Step 2: Differentiate Both Sides with Respect to Temperature $T$

Step 4: Evaluate at $T = e$

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

Step 3: Back-Substitute $\frac{dy}{dx} = -\frac{x}{y}$

Step 5: Evaluate at $(3, 4)$

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

Step 3: Back-Substitute $\frac{dy}{dx} = \frac{3x^2}{2y}$

Step 5: Evaluate at $(2, 2)$