Lesson Notes By Weeks and Term v3 - Senior Secondary 2
Differentiation
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Differentiation
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Subject: Further Mathematics
Class: Senior Secondary 2
Term: 1st Term
Week: 1
Theme: Pure Mathematics
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Find limits of a function at a given point Differentiation from first principle Differentiating polynomials Differentiation of special functions like sin x, cos ax, eax Use rules of differentiation pply differentiation to
The concept of a limit is foundational to calculus. It describes the value that a function "approaches" as the input (or independent variable) "approaches" some value.
Notation: $\lim_{x \to a} f(x) = L$ means that as $x$ gets arbitrarily close to $a$ (but not necessarily equal to $a$), the value of $f(x)$ gets arbitrarily close to $L$.
Methods for Evaluating Limits: Direct Substitution: If $f(x)$ is a polynomial or a rational function where the denominator is not zero at $x=a$, the limit can be found by directly substituting $a$ into $f(x)$. Example 2.1.1: Find $\lim_{x \to 2} (3x^2 - 5x + 1)$ Solution: Substitute $x=2$ directly: $3(2)^2 - 5(2) + 1 = 3(4) - 10 + 1 = 12 - 10 + 1 = 3$.
Therefore, $\lim_{x \to 2} (3x^2 - 5x + 1) = 3$.
Factorization: If direct substitution results in an indeterminate form (e.g., $\frac{0}{0}$), factorization of the numerator and/or denominator may simplify the expression, allowing for cancellation of the term causing the indeterminacy. Example 2.1.2: Find $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$ Solution: Direct substitution yields $\frac{1^2 - 1}{1 - 1} = \frac{0}{0}$, which is indeterminate.
Factorize the numerator: $x^2 - 1 = (x - 1)(x + 1)$. The expression becomes $\lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1}$. For $x \neq 1$, we can cancel $(x - 1)$: $\lim_{x \to 1} (x + 1)$. Now, substitute $x=1$: $1 + 1 = 2$.
Therefore, $\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2$.
Rationalization: If the function involves square roots and direct substitution gives $\frac{0}{0}$, rationalizing the numerator or denominator by multiplying by the conjugate can simplify the expression. Example 2.1.3: Find $\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x}$ Solution: Direct substitution yields $\frac{\sqrt{0 + 1} - 1}{0} = \frac{0}{0}$. Multiply numerator and denominator by the conjugate of the numerator, $(\sqrt{x+1} + 1)$: $\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x} \times \frac{\sqrt{x + 1} + 1}{\sqrt{x + 1} + 1}$ $= \lim_{x \to 0} \frac{(x + 1) - 1}{x(\sqrt{x + 1} + 1)}$ $= \lim_{x \to 0} \frac{x}{x(\sqrt{x + 1} + 1)}$ For $x \neq 0$, cancel $x$: $\lim_{x \to 0} \frac{1}{\sqrt{x + 1} + 1}$. Now, substitute $x=0$: $\frac{1}{\sqrt{0 + 1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$.
Therefore, $\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x} = \frac{1}{2}$. Differentiation from first principle (also known as the delta method or definition of the derivative) is the formal way to find the derivative of a function. It is based on the concept of the gradient of a curve at a point. The gradient of a straight line is constant. For a curve, the gradient changes from point to point. To find the gradient at a specific point, consider a secant line connecting two points on the curve: $(x, f(x))$ and $(x + \Delta x, f(x + \Delta x))$. The gradient of this secant line is $\frac{f(x + \Delta x) - f(x)}{\Delta x}$. As the second point approaches the first point (i.e., as $\Delta x \to 0$), the secant line approaches the tangent line at $(x, f(x))$. The limit of the gradient of the secant line is the gradient of the tangent line, which is the derivative. The derivative of a function $y = f(x)$ with respect to $x$, denoted as $\frac{dy}{dx}$ or $f'(x)$, is given by: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ Steps: Write down $f(x)$. Find $f(x + \Delta x)$. Calculate $f(x + \Delta x) - f(x)$. Divide by $\Delta x$: $\frac{f(x + \Delta x) - f(x)}{\Delta x}$. Take the limit as $\Delta x \to 0$. Example 2.2.1: Differentiate $y = 4x^3 + 6x^2 + 2x + 1$ from first principles.
Solution: $f(x) = 4x^3 + 6x^2 + 2x + 1$ $f(x + \Delta x) = 4(x + \Delta x)^3 + 6(x + \Delta x)^2 + 2(x + \Delta x) + 1$ $= 4(x^3 + 3x^2\Delta x + 3x(\Delta x)^2 + (\Delta x)^3) + 6(x^2 + 2x\Delta x + (\Delta x)^2) + 2x + 2\Delta x + 1$ $= 4x^3 + 12x^2\Delta x + 12x(\Delta x)^2 + 4(\Delta x)^3 + 6x^2 + 12x\Delta x + 6(\Delta x)^2 + 2x + 2\Delta x + 1$ $f(x + \Delta x) - f(x) = (4x^3 + 12x^2\Delta x + 12x(\Delta x)^2 + 4(\Delta x)^3 + 6x^2 + 12x\Delta x + 6(\Delta x)^2 + 2x + 2\Delta x + 1) - (4x^3 + 6x^2 + 2x + 1)$ $= 12x^2\Delta x + 12x(\Delta x)^2 + 4(\Delta x)^3 + 12x\Delta x + 6(\Delta x)^2 + 2\Delta x$ $\frac{f(x + \Delta x) - f(x)}{\Delta x} = \frac{\Delta x (12x^2 + 12x\Delta x + 4(\Delta x)^2 + 12x + 6\Delta x + 2)}{\Delta x}$ $= 12x^2 + 12x\Delta x + 4(\Delta x)^2 + 12x + 6\Delta x + 2$ $\lim_{\Delta x \to 0} (12x^2 + 12x\Delta x + 4(\Delta x)^2 + 12x + 6\Delta x + 2)$ $= 12x^2 + 12x(0) + 4(0)^2 + 12x + 6(0) + 2$ $= 12x^2 + 12x + 2$ Therefore, $\frac{dy}{dx} = 12x^2 + 12x + 2$. The process of differentiating polynomials relies on a set of standard rules: Power Rule: If $y = ax^n$, then $\frac{dy}{dx} = anx^{n-1}$. This applies to any real number $n$.
Constant Rule: If $y = c$ (where $c$ is a constant), then $\frac{dy}{dx} = 0$.
Sum/Difference Rule: If $y = f(x) \pm g(x)$, then $\frac{dy}{dx} = \frac{df}{dx} \pm \frac{dg}{dx}$.
Constant Multiple Rule: If $y = c \cdot f(x)$, then $\frac{dy}{dx} = c \cdot \frac{df}{dx}$. Example 2.3.1: Differentiate $y = 4x^8 + 7x^6 – 5x^3 + 4/√x$.
Solution: First, rewrite $4/√x$ using index notation: $4x^{-1/2}$. So, $y = 4x^8 + 7x^6 – 5x^3 + 4x^{-1/2}$.
Apply the power rule to each term: $\frac{d}{dx}(4x^8) = 4 \times 8x^{8-1} = 32x^7$ $\frac{d}{dx}(7x^6) = 7 \times 6x^{6-1} = 42x^5$ $\frac{d}{dx}(-5x^3) = -5 \times 3x^{3-1} = -15x^2$ $\frac{d}{dx}(4x^{-1/2}) = 4 \times (-\frac{1}{2})x^{-1/2 - 1} = -2x^{-3/2}$ Summing these terms: $\frac{dy}{dx} = 32x^7 + 42x^5 - 15x^2 - 2x^{-3/2}$ This can also be written as $\frac{dy}{dx} = 32x^7 + 42x^5 - 15x^2 - \frac{2}{x\sqrt{x}}$. The derivatives of certain standard functions are important to memorize: $\frac{d}{dx}(\sin x) = \cos x$ $\frac{d}{dx}(\cos x) = -\sin x$ $\frac{d}{dx}(e^x) = e^x$ For composite special functions, the Chain Rule is applied: If $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$. $\frac{d}{dx}(\sin (ax+b)) = a \cos (ax+b)$ $\frac{d}{dx}(\cos (ax+b)) = -a \sin (ax+b)$ $\frac{d}{dx}(e^{ax+b}) = a e^{ax+b}$ Example 2.4.1: Differentiate $y = 4\cos 5x + e^x$.
Solution: Apply the sum rule: $\frac{dy}{dx} = \frac{d}{dx}(4\cos 5x) + \frac{d}{dx}(e^x)$. For $4\cos 5x$: Let $u = 5x$, so $\frac{du}{dx} = 5$. Then $y = 4\cos u$, so $\frac{dy}{du} = -4\sin u$.
Using chain rule: $\frac{d}{dx}(4\cos 5x) = \frac{dy}{du} \times \frac{du}{dx} = (-4\sin u) \times 5 = -20\sin 5x$. For $e^x$: $\frac{d}{dx}(e^x) = e^x$.
Therefore, $\frac{dy}{dx} = -20\sin 5x + e^x$.
Economic Analysis (e.g., Profit Maximization for Nigerian SMEs): Differentiation is used to find the maximum or minimum values of functions. A small-scale Nigerian entrepreneur (e.g., a tailor, a farmer, or a local artisan) can use calculus to model their cost and revenue functions. By differentiating the profit function (Revenue - Cost) and setting the derivative to zero, they can determine the production quantity that maximizes their profit or minimizes their costs. For example, a farmer might model crop yield as a function of fertilizer input, and use differentiation to find the optimal amount of fertilizer to maximize yield and profit. Population Dynamics and Disease Spread in Nigeria: Demographers and public health officials use differentiation to model population growth rates in different Nigerian states or the rate of spread of infectious diseases (e.g., cholera outbreaks, malaria prevalence). The derivative of a population function with respect to time gives the instantaneous growth rate, providing insights into population trends or the urgency of health interventions. Understanding these rates is crucial for planning public services and resource allocation. Vehicle Dynamics and Road Safety in Nigeria: For vehicle manufacturers or traffic engineers, differentiation is essential. The derivative of displacement with respect to time gives velocity (speed), and the derivative of velocity gives acceleration. This is used in designing vehicles for optimal performance, calculating braking distances, or analyzing traffic flow on major Nigerian highways like the Lagos-Ibadan expressway to improve safety and reduce congestion.