Lesson Notes By Weeks and Term v3 - Senior Secondary 2
Differentiation cont.
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Differentiation cont.
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Subject: Further Mathematics
Class: Senior Secondary 2
Term: 1st Term
Week: 1
Theme: Pure Mathematics
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This topic builds upon students' foundational knowledge of differentiation (power rule, sum/difference rule) by introducing more advanced techniques required to differentiate complex functions. Mastering these rules is crucial for solving problems involving rates of change, optimization, and tangent/normal calculations, which have widespread applications in various fields such as engineering, economics, physics, and even in understanding market trends or resource management in Nigeria.
This section covers the fundamental rules and concepts for differentiating more complex functions beyond the basic power rule. A. The Product Rule The product rule is used to differentiate a function that is the product of two differentiable functions. If $y = uv$, where $u$ and $v$ are functions of $x$, then the derivative $\frac{dy}{dx}$ is given by: $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ In simpler terms: (first function) × (derivative of the second) + (second function) × (derivative of the first).
Example 1: Differentiate $y = (3x^2 + 5)(2x - 1)$. Let $u = 3x^2 + 5$ and $v = 2x - 1$. Then $\frac{du}{dx} = 6x$ and $\frac{dv}{dx} = 2$.
Applying the product rule: $\frac{dy}{dx} = (3x^2 + 5)(2) + (2x - 1)(6x)$ $\frac{dy}{dx} = 6x^2 + 10 + 12x^2 - 6x$ $\frac{dy}{dx} = 18x^2 - 6x + 10$ Example 2 (Nigerian Context): The cost $C$ (in Naira) of producing a certain commodity at a factory in Aba depends on the number of units $x$ produced daily, given by $C(x) = (2x + 10)\sqrt{x}$. Find the rate of change of cost with respect to the number of units produced. Given $C(x) = (2x + 10)x^{1/2}$. Let $u = 2x + 10$ and $v = x^{1/2}$. Then $\frac{du}{dx} = 2$ and $\frac{dv}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.
Applying the product rule: $\frac{dC}{dx} = (2x + 10)\left(\frac{1}{2\sqrt{x}}\right) + (x^{1/2})(2)$ $\frac{dC}{dx} = \frac{2x + 10}{2\sqrt{x}} + 2\sqrt{x}$ $\frac{dC}{dx} = \frac{x + 5}{\sqrt{x}} + 2\sqrt{x}$ $\frac{dC}{dx} = \frac{x + 5 + 2\sqrt{x}\sqrt{x}}{\sqrt{x}} = \frac{x + 5 + 2x}{\sqrt{x}} = \frac{3x + 5}{\sqrt{x}}$ This derivative represents the marginal cost, which is crucial for pricing and production decisions. B. The Quotient Rule The quotient rule is used to differentiate a function that is the quotient of two differentiable functions. If $y = \frac{u}{v}$, where $u$ and $v$ are functions of $x$ and $v \neq 0$, then the derivative $\frac{dy}{dx}$ is given by: $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$ In simpler terms: (denominator) × (derivative of numerator) - (numerator) × (derivative of denominator) all divided by (denominator) squared.
Example 3: Differentiate $y = \frac{4x - 3}{x^2 + 1}$. Let $u = 4x - 3$ and $v = x^2 + 1$. Then $\frac{du}{dx} = 4$ and $\frac{dv}{dx} = 2x$.
Applying the quotient rule: $\frac{dy}{dx} = \frac{(x^2 + 1)(4) - (4x - 3)(2x)}{(x^2 + 1)^2}$ $\frac{dy}{dx} = \frac{4x^2 + 4 - (8x^2 - 6x)}{(x^2 + 1)^2}$ $\frac{dy}{dx} = \frac{4x^2 + 4 - 8x^2 + 6x}{(x^2 + 1)^2}$ $\frac{dy}{dx} = \frac{-4x^2 + 6x + 4}{(x^2 + 1)^2}$ Example 4 (Nigerian Context): The concentration $C$ (in mg/L) of a pollutant in a river near Port Harcourt $t$ hours after spillage is given by $C(t) = \frac{3t}{t^2 + 2}$. Find the rate at which the concentration is changing after 1 hour. Given $C(t) = \frac{3t}{t^2 + 2}$. Let $u = 3t$ and $v = t^2 + 2$. Then $\frac{du}{dt} = 3$ and $\frac{dv}{dt} = 2t$.
Applying the quotient rule: $\frac{dC}{dt} = \frac{(t^2 + 2)(3) - (3t)(2t)}{(t^2 + 2)^2}$ $\frac{dC}{dt} = \frac{3t^2 + 6 - 6t^2}{(t^2 + 2)^2}$ $\frac{dC}{dt} = \frac{-3t^2 + 6}{(t^2 + 2)^2}$ At $t=1$: $\frac{dC}{dt}\Big|_{t=1} = \frac{-3(1)^2 + 6}{((1)^2 + 2)^2} = \frac{-3 + 6}{(1 + 2)^2} = \frac{3}{3^2} = \frac{3}{9} = \frac{1}{3}$ mg/L per hour. The concentration is increasing at a rate of 1/3 mg/L per hour after 1 hour. C. The Chain Rule The chain rule is used to differentiate composite functions, i.e., functions within functions. If $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$. Alternatively, if $y$ is a function of $u$, and $u$ is a function of $x$, i.e., $y = f(u)$ and $u = g(x)$, then: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ Example 5: Differentiate $y = (3x^2 - 4x + 1)^5$. Let $u = 3x^2 - 4x + 1$. Then $y = u^5$. $\frac{dy}{du} = 5u^4$. $\frac{du}{dx} = 6x - 4$.
Applying the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = (5u^4)(6x - 4)$ Substitute $u$ back: $\frac{dy}{dx} = 5(3x^2 - 4x + 1)^4 (6x - 4)$ $\frac{dy}{dx} = (30x - momentarily at rest? * Solution: a) Velocity is the first derivative of position: $v(t) = \frac{ds}{dt} = 3t^2 - 12t + 9$. b) Acceleration is the first derivative of velocity (second derivative of position): $a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} = 6t - 12$. c) The drone is momentarily at rest when its velocity is zero: $v(t) = 0 \implies 3t^2 - 12t + 9 = 0$ Divide by 3: $t^2 - 4t + 3 = 0$ Factor: $(t - 1)(t - 3) = 0$ So, $t = 1$ second or $t = 3$ seconds. The drone is momentarily at rest after 1 second and after 3 seconds. $y$ is a function of $u$, and $u$ is a function of $x$, i.e., $y = f(u)$ and $u = g(x)$, then: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ Example 5: Differentiate $y = (3x^2 - 4x + 1)^5$. Let $u = 3x^2 - 4x + 1$. Then $y = u^5$. $\frac{dy}{du} = 5u^4$. $\frac{du}{dx} = 6x - 4$.
Applying the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = (5u^4)(6x - 4)$ Substitute $u$ back: $\frac{dy}{dx} = 5(3x^2 - 4x + 1)^4 (6x - 4)$ $\frac{dy}{dx} = (30x - 20)(3x^2 - 4x + 1)^4$ Example 6 (Nigerian Context): The number of registered voters $N$ in a local government area in Kano state changes with time $t$ (in years) according to the formula $N(t) = 15000 + (t^2 + 5t)^{1/2}$. Find the rate of change of registered voters after 2 years. Given $N(t) = 15000 + (t^2 + 5t)^{1/2}$. The constant 15000 differentiates to
0. We need to differentiate $(t^2 + 5t)^{1/2}$. Let $u = t^2 + 5t$. Then $y = u^{1/2}$. $\frac{dy}{du} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}}$. $\frac{du}{dt} = 2t + 5$.
Applying the chain rule: $\frac{dN}{dt} = \frac{1}{2\sqrt{t^2 + 5t}}(2t + 5) = \frac{2t + 5}{2\sqrt{t^2 + 5t}}$. At $t=2$: $\frac{dN}{dt}\Big|_{t=2} = \frac{2(2) + 5}{2\sqrt{(2)^2 + 5(2)}} = \frac{4 + 5}{2\sqrt{4 + 10}} = \frac{9}{2\sqrt{14}}$. The rate of change is $\frac{9}{2\sqrt{14}}$ voters per year. D. Differentiation of Trigonometric Functions It is important for students to recall the standard derivatives of trigonometric functions. If $y = \sin x$, then $\frac{dy}{dx} = \cos x$. If $y = \cos x$, then $\frac{dy}{dx} = -\sin x$. If $y = \tan x$, then $\frac{dy}{dx} = \sec^2 x$. (This can be derived using the quotient rule for $\frac{\sin x}{\cos x}$). These can be combined with other rules (product, quotient, chain).
Example 7: Differentiate $y = \sin(4x + 3)$. This is a composite function. Let $u = 4x + 3$. Then $y = \sin u$. $\frac{dy}{du} = \cos u$. $\frac{du}{dx} = 4$.
Using the chain rule: $\frac{dy}{dx} = \cos(4x + 3) \cdot 4 = 4\cos(4x + 3)$.
Example 8: Differentiate $y = x^2 \cos x$. This is a product. Let $u = x^2$ and $v = \cos x$. $\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = -\sin x$.
Using the product rule: $\frac{dy}{dx} = x^2(-\sin x) + (\cos x)(2x)$ $\frac{dy}{dx} = -x^2\sin x + 2x\cos x$ $\frac{dy}{dx} = x(2\cos x - x\sin x)$ E. Higher-Order Derivatives (Second Derivative) The second derivative is the derivative of the first derivative. It is denoted as $\frac{d^2y}{dx^2}$ or $f''(x)$. The first derivative $\frac{dy}{dx}$ represents the instantaneous rate of change or the gradient of the tangent to the curve. The second derivative $\frac{d^2y}{dx^2}$ represents the rate of change of the gradient. It tells us about the concavity of the curve (whether it's curving upwards or downwards) and is crucial for identifying local maxima and minima in optimization problems.
Example 9: Find the first and second derivatives of $y = 5x^3 - 2x^2 + 7x - 1$.
First derivative: $\frac{dy}{dx} = 15x^2 - 4x + 7$.
Second derivative: $\frac{d^2y}{dx^2} = \frac{d}{dx}(15x^2 - 4x + 7) = 30x - 4$.
Example 10 (Nigerian Context): The position $s$ (in meters) of a drone used for surveying farmlands in Benue state, from its starting point, at time $t$ (in seconds) is given by $s(t) = t^3 - 6t^2 + 9t$. a) Find the velocity function, $v(t)$. b) Find the acceleration function, $a(t)$. c) When is the drone momentarily at rest?
Solution: a) Velocity is the first derivative of position: $v(t) = \frac{ds}{dt} = 3t^2 - 12t + 9$. b) Acceleration is the first derivative of velocity (second derivative of position): $a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} = 6t - 12$. c) The drone is momentarily at rest when its velocity is zero: $v(t) = 0 \implies 3t^2 - 12t + 9 = 0$ Divide by 3: $t^2 - 4t + 3 = 0$ Factor: $(t - 1)(t - 3) = 0$ So, $t = 1$ second or $t = Teacher Activities: Introduction & Recap (10 minutes): Begin by reviewing the basic rules of differentiation (power rule, sum/difference rule) and their geometric interpretation (gradient of tangent). Pose a problem that cannot be solved with basic rules (e.g., differentiating $(3x-1)(x^2+2)$ or $\frac{x}{x+1}$) to introduce the need for new rules. State the learning objectives clearly for the lesson. Product Rule Explanation and Demonstration (15 minutes): Introduce the product rule formula: $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$. Clearly identify $u$ and $v$ in an example, then demonstrate step-by-step calculation of $\frac{du}{dx}$, $\frac{dv}{dx}$, and the application of the rule. Work through Example 1 and Example 2 (Nigerian context) on the board, emphasizing each step and simplifying the final expression. Quotient Rule Explanation and Demonstration (15 minutes): Introduce the quotient rule formula: $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$. Stress the order of terms in the numerator. Demonstrate step-by-step application with an example, clearly identifying $u$ and $v$, calculating their derivatives, and substituting into the formula. Work through Example 3 and Example 4 (Nigerian context) on the board. Chain Rule Explanation and Demonstration (15 minutes): Introduce the concept of a "function of a function" (composite function).
Introduce the chain rule formula: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. Demonstrate how to identify the inner function ($u$) and the outer function ($y=f(u)$). Work through Example 5 and Example 6 (Nigerian context), carefully showing the substitution and differentiation steps. Differentiation of Trigonometric Functions (10 minutes): Remind students of the basic trigonometric derivatives (sin to cos, cos to -sin, tan to sec^2). Demonstrate how these combine with the chain rule for functions like $\sin(ax+b)$ or $\cos(x^2)$. Work through Example 7 and Example
8. Higher-Order Derivatives (10 minutes): Explain the concept of finding the derivative of a derivative (second derivative, third derivative, etc.). Explain the notation ($\frac{d^2y}{dx^2}$, $f''(x)$). Relate the second derivative to rates of change of gradient, concavity, and acceleration. Work through Example 9 and Example 10 (Nigerian context).
Consolidation and Group Work (15 minutes): Divide students into small groups. Assign a few mixed practice problems (combining different rules) for group discussion and solution. Circulate around the classroom, providing support, clarification, and correcting misconceptions.
Wrap-up and Homework (5 minutes): Briefly summarize the key differentiation rules learned. Assign independent practice questions as homework.
Student Activities: Active Listening and Note-Taking: Students will listen attentively to explanations, take detailed notes, and copy worked examples from the board.
Questioning and Clarification: Students will ask questions to clarify doubts during explanations.
Guided Practice Participation: Students will participate actively by suggesting steps, solving parts of examples on their rough notebooks, and verifying solutions during teacher demonstrations.
Group Problem-Solving: In groups, students will work collaboratively to solve assigned problems, discussing strategies and arriving at consensus solutions.
Independent Practice: Students will complete homework assignments to reinforce their understanding and practice the techniques learned.
Economic Analysis (Marginal Cost/Revenue/Profit): Nigerian businesses and economists use differentiation to determine marginal cost, marginal revenue, and marginal profit. For instance, a small-scale textile manufacturer in Kano can use the product rule to model how total production cost changes when both raw material costs and labor efficiency (which are functions of production volume) fluctuate. The quotient rule can be applied to find the rate of change of average cost per unit, helping to optimize production levels to maximize profit. Environmental Monitoring (Pollutant Concentration): Environmental scientists and government agencies (like NESREA) in Nigeria use differentiation to model and predict the spread and concentration of pollutants in rivers, air, or soil. For example, the rate of change of pollutant concentration in the Niger Delta waters after an oil spill can be described using the quotient rule, allowing authorities to understand how quickly a cleanup effort is impacting the environment. Higher-order derivatives can reveal if the rate of pollution decrease is slowing down or speeding up.
Agriculture and Resource Management: Agricultural researchers and farmers in Nigeria can apply differentiation to optimize crop yield or resource allocation. For instance, the chain rule can model how the growth rate of a specific crop (e.g., yam in Benue) depends on sunlight intensity, which in turn varies throughout the day. By differentiating the yield function, farmers can determine optimal planting times or watering schedules for maximum productivity.