Lesson Notes By Weeks and Term v3 - Senior Secondary 3
Differentiation of Algebraic fractions
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Differentiation of Algebraic fractions
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Subject: General Mathematics
Class: Senior Secondary 3
Term: 1st Term
Week: 5
Theme: Introductory Calculus
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Explain the meaning of differentiation Differentiate from the first principle Recognize the standard derivatives of some basic functions Apply the rules of differentiation of functions Apply differentiation of capital market and real life situations.
High = u, Dee = derivative)
Example 3: Differentiate $y = \frac{3x-2}{x+5}$.
Step 1: Identify $u$ and $v$. Let $u = 3x-2$ (numerator) and $v = x+5$ (denominator).
Step 2: Find $\frac{du}{dx}$ and $\frac{dv}{dx}$. $\frac{du}{dx} = \frac{d}{dx}(3x-2) = 3$ $\frac{dv}{dx} = \frac{d}{dx}(x+5) = 1$ Step 3: Apply the Quotient Rule formula. $\frac{dy}{dx} = \frac{(x+5)(3) - (3x-2)(1)}{(x+5)^2}$ Step 4: Simplify the expression. $\frac{dy}{dx} = \frac{3x + 15 - 3x + 2}{(x+5)^2} = \frac{17}{(x+5)^2}$ Example 4: Differentiate $y = \frac{x^2}{2x-1}$.
Step 1: Identify $u$ and $v$. Let $u = x^2$ and $v = 2x-1$.
Step 2: Find $\frac{du}{dx}$ and $\frac{dv}{dx}$. $\frac{du}{dx} = 2x$ $\frac{dv}{dx} = 2$ Step 3: Apply the Quotient Rule formula. $\frac{dy}{dx} = \frac{(2x-1)(2x) - (x^2)(2)}{(2x-1)^2}$ Step 4: Simplify the expression. $\frac{dy}{dx} = \frac{4x^2 - 2x - 2x^2}{(2x-1)^2} = \frac{2x^2 - 2x}{(2x-1)^2} = \frac{2x(x-1)}{(2x-1)^2}$ Example 5: Differentiate $y = \frac{x^3 - 4}{x^2+1}$.
Step 1: Identify $u = x^3 - 4$ and $v = x^2+1$.
Step 2: Find $\frac{du}{dx} = 3x^2$ and $\frac{dv}{dx} = 2x$.
Step 3: Apply the Quotient Rule formula. $\frac{dy}{dx} = \frac{(x^2+1)(3x^2) - (x^3-4)(2x)}{(x^2+1)^2}$ * Step 4: Simplify the expression. $\frac{dy}{dx} = \frac{3x^4 + 3x^2 - (2x^4 - 8x)}{(x^2+1)^2}$ $\frac{dy}{dx} = \frac{3x^4 + 3x^2 - 2x^4 + 8x}{(x^2+1)^2}$ $\frac{dy}{dx} = \frac{x^4 + 3x^2 + 8x}{(x^2+1)^2}$ This section provides a detailed explanation of the core concepts, definitions, and rules necessary for differentiating algebraic fractions. 2.
1. Introduction to Differentiation Differentiation is a process of finding the derivative of a function. The derivative represents the instantaneous rate of change of a function with respect to one of its variables. Geometrically, the derivative at a point represents the gradient (slope) of the tangent to the curve at that specific point. If a function is given as $y = f(x)$, its derivative is commonly denoted as $\frac{dy}{dx}$, $f'(x)$, or $y'$. 2.
2. Differentiation from First Principle (Also known as differentiation by definition) The derivative of a function $y = f(x)$ with respect to $x$ is defined as: $\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ Example 1: Differentiate $f(x) = x^2$ from the first principle.
Step 1: Find $f(x+h)$. $f(x+h) = (x+h)^2 = x^2 + 2xh + h^2$ Step 2: Find $f(x+h) - f(x)$. $f(x+h) - f(x) = (x^2 + 2xh + h^2) - x^2 = 2xh + h^2$ Step 3: Divide by $h$. $\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2}{h} = \frac{h(2x+h)}{h} = 2x+h$ Step 4: Take the limit as $h \to 0$. $\frac{dy}{dx} = \lim_{h \to 0} (2x+h) = 2x+0 = 2x$ Thus, the derivative of $x^2$ is $2x$. 2.
3. Standard Derivatives and Rules of Differentiation These rules simplify the process of differentiation, making it unnecessary to use the first principle for every function.
Rule 1: Derivative of a Constant: If $y = c$ (where $c$ is a constant), then $\frac{dy}{dx} = 0$.
Example: If $y=5$, $\frac{dy}{dx}=0$.
Rule 2: Power Rule: If $y = x^n$ (where $n$ is any real number), then $\frac{dy}{dx} = nx^{n-1}$.
Example: If $y = x^3$, $\frac{dy}{dx} = 3x^{3-1} = 3x^2$.
Example: If $y = x$, $\frac{dy}{dx} = 1x^{1-1} = 1x^0 = 1$.
Example: If $y = \frac{1}{x} = x^{-1}$, $\frac{dy}{dx} = -1x^{-1-1} = -x^{-2} = -\frac{1}{x^2}$.
Example: If $y = \sqrt{x} = x^{1/2}$, $\frac{dy}{dx} = \frac{1}{2}x^{\frac{1}{2}-1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.
Rule 3: Constant Multiple Rule: If $y = cf(x)$, then $\frac{dy}{dx} = c \frac{df}{dx}$.
Example: If $y = 3x^4$, $\frac{dy}{dx} = 3(4x^3) = 12x^3$.
Rule 4: Sum and Difference Rule: If $y = f(x) \pm g(x)$, then $\frac{dy}{dx} = \frac{df}{dx} \pm \frac{dg}{dx}$.
Example: If $y = 2x^3 + 5x - 7$, $\frac{dy}{dx} = 2(3x^2) + 5(1) - 0 = 6x^2 + 5$. 2.
4. Differentiation of Algebraic Fractions: Product Rule and Quotient Rule These rules are crucial for differentiating functions that are products or quotients of other functions. A. The Product Rule Used when a function $y$ is a product of two differentiable functions, say $u(x)$ and $v(x)$, i.e., $y = u \cdot v$.
The derivative is given by: $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$ Example 2: Differentiate $y = (2x+1)(x^2-3)$.
Step 1: Identify $u$ and $v$. Let $u = 2x+1$ and $v = x^2-3$.
Step 2: Find $\frac{du}{dx}$ and $\frac{dv}{dx}$. $\frac{du}{dx} = \frac{d}{dx}(2x+1) = 2$ $\frac{dv}{dx} = \frac{d}{dx}(x^2-3) = 2x$ Step 3: Apply the Product Rule formula. $\frac{dy}{dx} = (2x+1)(2x) + (x^2-3)(2)$ Step 4: Simplify the expression. $\frac{dy}{dx} = 4x^2 + 2x + 2x^2 - 6 = 6x^2 + 2x - 6$ B. The Quotient Rule This rule is specifically used for differentiating algebraic fractions, where a function $y$ is a quotient of two differentiable functions, say $u(x)$ and $v(x)$, i.e., $y = \frac{u(x)}{v(x)}$.
The derivative is given by: $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$ (A common mnemonic is "Low Dee High minus High Dee Low, over Low squared" where Low = v, High = u, Dee = derivative)
Example 3: Differentiate $y = \frac{3x-2}{x+5}$.
Step 1: Identify $u$ and $v$. Let $u = 3x-2$ (numerator) and $v = x+5$ (denominator).
Step 2: Find $\frac{du}{dx}$ and $\frac{dv}{dx}$. $\frac{du}{dx} = \frac{d}{dx}(3x-2) = 3$ $\frac{dv}{dx} = \frac{d}{dx}(x+5) = 1$ Step 3: Apply the Quotient Rule formula. $\frac{dy}{dx} = \frac{(x+5)(3) - (3x-2)(1)}{(x+5)^2}$ Step 4: Simplify the expression. $\frac{dy}{dx} = \frac{3x + 15 - 3x + 2}{(x+5)^2} = \frac{17}{(x+5)^2}$ Example 4: Differentiate $y = \frac{x^2}{2x-1}$.
Step 1: Identify $u$ and $v$. Let $u = 3.
1. Teacher Activities Introduction (10 minutes): Begin by reviewing the concept of gradient from coordinate geometry (change in y / change in x). Introduce the idea of an instantaneous gradient for curves, where the "change in x" becomes infinitesimally small. This leads to the concept of differentiation. Briefly state the importance of differentiation in various fields, linking it to real-life applications (e.g., speed, acceleration, profit maximization in businesses, like a textile factory in Kaduna optimizing cloth production). State the lesson objectives clearly. First Principle and Basic Rules (15 minutes): Present the definition of differentiation from first principles. Work through a simple example (e.g., $f(x)=x^2$) to demonstrate the process. Introduce the power rule and other basic rules (constant, sum/difference, constant multiple). Guide students through 2-3 simple examples using these rules.
Product Rule (15 minutes): Introduce the Product Rule formula. Work through Example 2 (e.g., $y = (2x+1)(x^2-3)$) step-by-step on the board, explaining each part. Pose a similar problem for students to attempt in pairs. Quotient Rule for Algebraic Fractions (25 minutes): Introduce the Quotient Rule formula, emphasizing its application for algebraic fractions. Explain the mnemonic "Low Dee High minus High Dee Low, over Low squared" to aid memorization. Work through Example 3 and 4 (e.g., $y = \frac{3x-2}{x+5}$ and $y = \frac{x^2}{2x-1}$) in detail, highlighting common pitfalls like incorrect subtraction order or squaring the denominator. Stress the importance of simplifying the resulting expression. Work through Example 5 which is slightly more complex. Guided Practice and Consolidation (15 minutes): Distribute pre-prepared questions for guided practice (as in Section 4). Circulate among students, providing support and correcting misconceptions. Review solutions to guided practice questions on the board, allowing students to check their work.
Application Discussion (10 minutes): Facilitate a brief discussion on how differentiation of algebraic fractions might apply in practical Nigerian scenarios (e.g., finding the rate of change of average cost per bag of cement produced by Dangote Cement as production volume changes; analyzing the rate of change of population density in Lagos). 3.
2. Student Activities Active Listening and Note-Taking: Students listen attentively to explanations and take comprehensive notes.
Problem Solving: Participate in solving example problems presented by the teacher.
Pair/Group Work: Work in pairs or small groups to attempt practice problems, discussing strategies and solutions.
Questioning: Ask clarifying questions when concepts are unclear.
Guided Practice: Solve the guided practice questions independently or with minimal teacher intervention, then verify answers during class review.
Independent Practice: Attempt the independent practice questions as homework or in-class exercises to solidify understanding.
Discussion: Engage in discussions about real-life applications of differentiation.
Question 1: Differentiate $f(x) = 3x^2 + 2x - 1$ from the first principle.
Solution: $f(x) = 3x^2 + 2x - 1$ $f(x+h) = 3(x+h)^2 + 2(x+h) - 1$ $f(x+h) = 3(x^2+2xh+h^2) + 2x+2h - 1$ $f(x+h) = 3x^2 + 6xh + 3h^2 + 2x + 2h - 1$ $f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 + 2x + 2h - 1) - (3x^2 + 2x - 1)$ $f(x+h) - f(x) = 6xh + 3h^2 + 2h$ $\frac{f(x+h) - f(x)}{h} = \frac{h(6x + 3h + 2)}{h} = 6x + 3h + 2$ $\frac{dy}{dx} = \lim_{h \to 0} (6x + 3h + 2) = 6x + 0 + 2 = 6x+2$
Commentary: This question assesses the understanding of differentiation from first principles, ensuring students grasp the fundamental definition before applying rules.
Question 2: Find the derivative of $y = (x^3+2)(4x-5)$.
Solution: Let $u = x^3+2 \implies \frac{du}{dx} = 3x^2$ Let $v = 4x-5 \implies \frac{dv}{dx} = 4$ Using the Product Rule: $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ $\frac{dy}{dx} = (x^3+2)(4) + (4x-5)(3x^2)$ $\frac{dy}{dx} = 4x^3 + 8 + 12x^3 - 15x^2$ $\frac{dy}{dx} = 16x^3 - 15x^2 + 8$
Commentary: This question tests the application of the Product Rule, a prerequisite for understanding the Quotient Rule, as components of fractions can be products.
Question 3: Differentiate $y = \frac{x}{x^2+1}$.
Solution: Let $u = x \implies \frac{du}{dx} = 1$ Let $v = x^2+1 \implies \frac{dv}{dx} = 2x$ Using the Quotient Rule: $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$ $\frac{dy}{dx} = \frac{(x^2+1)(1) - (x)(2x)}{(x^2+1)^2}$ $\frac{dy}{dx} = \frac{x^2+1 - 2x^2}{(x^2+1)^2}$ $\frac{dy}{dx} = \frac{1 - x^2}{(x^2+1)^2}$
Commentary: This is a straightforward application of the Quotient Rule with polynomial terms, requiring careful substitution and algebraic simplification.
Question 4: A Nigerian poultry farmer observes that the average cost (in Naira) of producing an egg, $C(x)$, is given by $C(x) = \frac{50x+2000}{x}$, where $x$ is the number of eggs produced (in thousands). Find the rate of change of the average cost with respect to the number of eggs produced, i.e., find $\frac{dC}{dx}$.
Solution: The function is $C(x) = \frac{50x+2000}{x}$. Let $u = 50x+2000 \implies \frac{du}{dx} = 50$ Let $v = x \implies \frac{dv}{dx} = 1$ Using the Quotient Rule: $\frac{dC}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$ $\frac{dC}{dx} = \frac{(x)(50) - (50x+2000)(1)}{x^2}$ $\frac{dC}{dx} = \frac{50x - 50x - 2000}{x^2}$ $\frac{dC}{dx} = \frac{-2000}{x^2}$
Commentary: This question demonstrates a practical application of differentiating an algebraic fraction in an economic context, familiar to Nigerian students. It also highlights that the answer can be negative, indicating a decreasing average cost as production increases.
Economic Analysis (e.g., in a Nigerian agricultural business or manufacturing sector): Differentiation of algebraic fractions is crucial for calculating marginal functions. For instance, if a company's total cost $C(x)$ and total revenue $R(x)$ are known, the average cost $AC(x) = \frac{C(x)}{x}$ or average revenue $AR(x) = \frac{R(x)}{x}$ are often algebraic fractions. Their derivatives, $\frac{d(AC)}{dx}$ and $\frac{d(AR)}{dx}$, represent the rate of change of average cost/revenue, helping businesses optimize production to minimize average costs or maximize average profits (e.g., for a cement factory in Obajana trying to find the optimal number of cement bags to produce daily). Population Dynamics and Environmental Studies: The rate of change of population density is often modeled using functions that result in algebraic fractions. For example, if the number of people $N(t)$ and the land area $A(t)$ in a growing city (like Port Harcourt) are both functions of time, the population density $D(t) = \frac{N(t)}{A(t)}$ is an algebraic fraction. Differentiating $D(t)$ helps understand how fast the density is changing, which has implications for urban planning, resource management, and addressing environmental concerns like waste disposal and traffic congestion.
Capital Market and Investment: While complex, introductory ideas can be discussed. For example, the "efficiency ratio" of a bank (like Zenith Bank) might be defined as $\frac{\text{Operating Expenses}}{\text{Operating Income}}$. If these components change over time, differentiating this ratio can show how quickly the bank's operational efficiency is improving or declining. This helps analysts understand the bank's performance and make investment decisions.