Product Rule, Power Rule, and Quotient Rule Differentiation Notes

Derivative Notations and Basic Derivatives

• Prime notation $f'(x)$ is mathematically equivalent to Leibnizian notation $\frac{d}{dx}[f(x)]$:   $f'(x) = \frac{d}{dx}[f(x)]$

• Basic derivative operations:

  • For $f(x) = x - 6$:     $\frac{d}{dx}(x - 6) = \frac{d}{dx}(x) - \frac{d}{dx}(6) = 1 - 0 = 1$

  • For $g(x) = 3x - 5$:     $\frac{d}{dx}(3x - 5) = \frac{d}{dx}(3x) - \frac{d}{dx}(5) = 3 - 0 = 3$

The Power Rule

• The Power Rule is used to differentiate terms raised to a power $n$:   $\frac{d}{dx}(x^n) = n x^{n-1}$

• Scope of exponents:

  • Positive integers ($n \in \mathbb{N}$)

  • Negative real numbers

  • Non-natural real numbers

• Basic application examples:

  • $\frac{d}{dt}(t^2) = 2t^{2-1} = 2t$

  • $\frac{d}{dt}(t) = 1t^{1-1} = 1$

  • $\frac{d}{dt}(c) = 0$, where $c$ is any constant.

The Product Rule

• Definition: The Product Rule determines the derivative of a function expressed as the product of two individual functions $f(x)$ and $g(x)$.

• General Formula:   $\frac{d}{dx}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x)$

• Leibnizian Notation Representation:   $\frac{d}{dx}[f(x) \cdot g(x)] = \left(\frac{d}{dx}[f(x)]\right) \cdot g(x) + f(x) \cdot \left(\frac{d}{dx}[g(x)]\right)$

• Example 1: Differentiating $f(x) = (x - 6)(3x - 5)$

  • Set up the product rule structure:     $f'(x) = (x - 6)'(3x - 5) + (x - 6)(3x - 5)'$

  • Evaluate individual component derivatives:     $\frac{d}{dx}(x - 6) = 1$     $\frac{d}{dx}(3x - 5) = 3$

  • Substitute component derivatives into the setup:     $f'(x) = 1 \cdot (3x - 5) + (x - 6) \cdot 3$

  • Expand and simplify algebraically:     $f'(x) = 3x - 5 + 3x - 18$     $f'(x) = 6x - 23$

• Example 2: Differentiating $g(t) = (t^2 + t - 2)(t - 1)$

  • The choice of independent variable notation (such as $t$, $x$, $y$, or $\alpha$) does not alter the differentiation procedure.

  • Set up the product rule structure:     $g'(t) = (t^2 + t - 2)'(t - 1) + (t^2 + t - 2)(t - 1)'$

  • Compute component derivatives using the Power Rule:     $\frac{d}{dt}(t - 1) = 1 - 0 = 1$     $\frac{d}{dt}(t^2 + t - 2) = 2t + 1 - 0 = 2t + 1$

  • Substitute component derivatives into the setup:     $g'(t) = (2t + 1)(t - 1) + (t^2 + t - 2)(1)$

  • Perform algebraic expansion on the first product:     $(2t + 1)(t - 1) = 2t^2 - 2t + t - 1 = 2t^2 - t - 1$

  • Add the expanded terms to the remaining polynomial:     $g'(t) = (2t^2 - t - 1) + (t^2 + t - 2)$

  • Combine like terms by powers of $t$:

  • $t^2$ terms: $2t^2 + t^2 = 3t^2$

  • $t$ terms: $-t + t = 0$

  • Constant terms: $-1 - 2 = -3$

  • Final simplified result:     $g'(t) = 3t^2 - 3$

• Example 3: Differentiating $y = (x^3 + 6x + 5)(3 - 4x)$

  • Set up using Leibnizian operator notation:     $\frac{dy}{dx} = \left(\frac{d}{dx}[x^3 + 6x + 5]\right)(3 - 4x) + (x^3 + 6x + 5)\left(\frac{d}{dx}[3 - 4x]\right)$

  • Calculate component derivatives:     $\frac{d}{dx}[x^3 + 6x + 5] = 3x^2 + 6$     $\frac{d}{dx}[3 - 4x] = -4$

  • Substitute computed derivatives into the setup:     $\frac{dy}{dx} = (3x^2 + 6)(3 - 4x) + (x^3 + 6x + 5)(-4)$

  • Expand the products algebraically:     $(3x^2 + 6)(3 - 4x) = 9x^2 - 12x^3 + 18 - 24x$     $(x^3 + 6x + 5)(-4) = -4x^3 - 24x - 20$

  • Sum the expanded terms:     $\frac{dy}{dx} = -12x^3 + 9x^2 - 24x + 18 - 4x^3 - 24x - 20$

  • Group and combine like terms:

  • $x^3$ terms: $-12x^3 - 4x^3 = -16x^3$

  • $x^2$ terms: $9x^2$

  • $x$ terms: $-24x - 24x = -48x$

  • Constant terms: $18 - 20 = -2$

  • Final simplified result:     $\frac{dy}{dx} = -16x^3 + 9x^2 - 48x - 2$

Practical Applications and Connections

• Kinematics:

  • If $g(t)$ represents the position of an object at time $t$, the velocity at time $t = 5\,\text{s}$ is determined by taking the derivative $g'(t)$ and evaluating it at $t = 5$ (written as $g'(5)$).

• Electromagnetism:

  • If $g(t)$ represents the total flow of electric charge over time $t$, its derivative $g'(t)$ represents the electric current at time t$.\n\n* Connection to Integral Calculus:\n * Every differentiation rule maps directly to a corresponding integration rule.\n * The direct inverse operation of the Product Rule is **Integration by Parts**.\n\n* Compound Expressions:\n * Complex non-polynomial expressions, such as x^3 \cos^2(x^2 + 1), utilize the Product Rule as their outer structure alongside advanced differentiation techniques.\n\n# The Quotient Rule\n\n* Definition: The Quotient Rule provides the derivative of a function written as the ratio or quotient of two functions \frac{f(x)}{g(x)}.\n\n* General Formula:\n  \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\n\n* Key Rules for Application:\n * The denominator is the square of the original denominator function: [g(x)]^2\n * The numerator consists of the derivative of the top times the bottom minus the derivative of the bottom times the top.\n * Order matters in the numerator due to the non-symmetric minus sign.\n\n* Example 1: Differentiating h(x) = \frac{4 - 7x}{2x - x^2}\n * Set up the quotient rule formula:\n    h'(x) = \frac{(4 - 7x)'(2x - x^2) - (4 - 7x)(2x - x^2)'}{(2x - x^2)^2}\n * Compute component derivatives:\n    \frac{d}{dx}(4 - 7x) = -7\n    \frac{d}{dx}(2x - x^2) = 2 - 2x\n * Substitute component derivatives into the formula:\n    h'(x) = \frac{-7(2x - x^2) - (4 - 7x)(2 - 2x)}{(2x - x^2)^2}\n * Expand terms in the numerator step-by-step:\n * First product: -7(2x - x^2) = -14x + 7x^2\n * Second product: (4 - 7x)(2 - 2x) = 8 - 8x - 14x + 14x^2 = 8 - 22x + 14x^2\n * Apply the subtraction sign across the expanded second product:\n    \text{Numerator} = (-14x + 7x^2) - (8 - 22x + 14x^2)\n    \text{Numerator} = -14x + 7x^2 - 8 + 22x - 14x^2\n * Combine like terms in the numerator:\n * x^2$terms:$7x^2 - 14x^2 = -7x^2\n * x$terms:$-14x + 22x = 8x\n * Constant terms: -8\n    \text{Numerator} = -7x^2 + 8x - 8\n * Final simplified result:\n    h'(x) = \frac{-7x^2 + 8x - 8}{(2x - x^2)^2}\n\n# Questions & Discussion\n\n* **Evaluating Derivatives at Specific Values in Word Problems**:\n * Question: How is a functional request such as finding g(5) processed when analyzing real-world word problems?\n * Answer: When a practical situation demands a rate of change at a specific moment (e.g., velocity at time t = 5\,\text{s}$given position$g(t)$), first calculate the general derivative function$g'(t)$using differentiation rules. Then, substitute$t = 5$directly into the simplified derivative equation to solve for$g'(5)$$.