Exhaustive Calculus Notes: Product Rule, Quotient Rule, Chain Rule, Higher-Order Derivatives, and Implicit Differentiation

Product Rule and Algebraic Rewriting

• Product Rule Definition

  • Formula: $\frac{d}{dx}[f(x) \times g(x)] = f(x) \times g'(x) + g(x) \times f'(x)$
  • Verbal statement: The derivative of a product of two functions is the first function times the derivative of the second function, plus the second function times the derivative of the first function.

• Differentiating $f(x) = (3x + 2)(5x^2 - 7)$ via Product Rule

  • Identification of terms:$\text{First} = 3x + 2$, $\text{Second} = 5x^2 - 7$.
  • Individual derivatives: $\frac{d}{dx}(3x + 2) = 3$, $\frac{d}{dx}(5x^2 - 7) = 10x$.
  • Unsimplified derivative using the product rule: $f'(x) = (3x + 2)(10x) + (5x^2 - 7)(3)$
  • Algebraic distribution and simplification:
  • $f'(x) = 30x^2 + 20x + 15x^2 - 21$
  • Combining like terms yields: $f'(x) = 45x^2 + 20x - 21$

• Differentiating $f(x) = (3x + 2)(5x^2 - 7)$ via Pre-Multiplication

  • Multiply the terms prior to differentiation: $f(x) = 15x^3 - 21x + 10x^2 - 14 = 15x^3 + 10x^2 - 21x - 14$
  • Apply the power rule directly: $f'(x) = 45x^2 + 20x - 21$
  • Both methods produce identical results.

• Differentiating Functions with Rational Components

  • Example problem: $f(x) = \left(\frac{3}{x} + 2\right)(5x^2 - 7)$
  • Rewrite step prior to differentiation: $f(x) = (3x^{-1} + 2)(5x^2 - 7)$
  • Differentiation applying the product rule:
  • $\text{First} = 3x^{-1} + 2$
  • $\text{Derivative of Second} = 10x$
  • $\text{Second} = 5x^2 - 7$
  • $\text{Derivative of First} = -3x^{-2}$
  • Formulated derivative: $f'(x) = (3x^{-1} + 2)(10x) + (5x^2 - 7)(-3x^{-2})$

Quotient Rule Applications

• Quotient Rule Definition

  • Formula: $\frac{d}{dx}\left[\frac{T(x)}{B(x)}\right] = \frac{B(x) \times T'(x) - T(x) \times B'(x)}{[B(x)]^2}$
  • Verbal statement: The derivative of a quotient is the bottom function times the derivative of the top function minus the top function times the derivative of the bottom function, all divided by the bottom function squared.

• Differentiating $f(x) = \frac{2x - 3}{x + 1}$

  • Components: $\text{Top} = 2x - 3$, $\text{Bottom} = x + 1$.
  • Derivatives of components: $\text{Top}' = 2$, $\text{Bottom}' = 1$.
  • Substitution into quotient rule formula: $f'(x) = \frac{(x + 1)(2) - (2x - 3)(1)}{(x + 1)^2}$
  • Simplification of the numerator:
  • $(x + 1)(2) = 2x + 2$
  • $-(2x - 3)(1) = -2x + 3$
  • $2x + 2 - 2x + 3 = 5$
  • Final derivative: $f'(x) = \frac{5}{(x + 1)^2}$

• Advanced Algebraic Expansion of Numerators

  • Setup for expressions involving standard polynomial quotients:
  • $\text{Numerator} = \text{Bottom} \times (2x) - (x^2 + 5)(2)$
  • Distribution steps:
  • $4x^2 - 6x - (2x^2 + 10)$
  • $4x^2 - 6x - 2x^2 - 10 = 2x^2 - 6x - 10$
  • Fully expanding and simplifying the numerator is required when preparing expressions to find higher-order derivatives such as second or third derivatives.

Chain Rule and Combined Differentiation Techniques

• Chain Rule Definition

  • Used for composite functions of the form $y = [u(x)]^n$.
  • Formula: $\frac{d}{dx}[u(x)]^n = n \times [u(x)]^{n-1} \times u'(x)$
  • Verbal statement: Bring the exponent to the front, keep the inside function unchanged, decrease the power by 1, and multiply by the derivative of the inside function.

• Basic Chain Rule Application

  • Example: Find derivative of $f(x) = (3x^2 + 5)^4$
  • Outside operation: Power of $4$
  • Inside function: $3x^2 + 5$
  • Derivative of inside: $6x$
  • Resulting derivative: $f'(x) = 4(3x^2 + 5)^3 \times (6x)$

• Handling Fractional Exponents with Chain Rule

  • Example: Find derivative of $f(x) = 3(x^2 + 4)^{1/5}$
  • Power multiplication step: $3 \times \frac{1}{5} = \frac{3}{5}$
  • Exponent subtraction arithmetic: $\frac{1}{5} - 1 = \frac{1}{5} - \frac{5}{5} = -\frac{4}{5}$
  • Derivative of inside expression: $\frac{d}{dx}(x^2 + 4) = 2x$
  • Product combination: $\frac{3}{5}(x^2 + 4)^{-4/5} \times (2x)$
  • Final derivative form: $f'(x) = \frac{6x}{5}(x^2 + 4)^{-4/5}$

• Combining Product Rule and Chain Rule

  • Example: Find derivative of $f(x) = 3x \times \sqrt[5]{x^2 + 4}$
  • Rewrite standard radical into rational exponent: $f(x) = 3x(x^2 + 4)^{1/5}$
  • Identify terms: $\text{First} = 3x$, $\text{Second} = (x^2 + 4)^{1/5}$.
  • Apply product rule combined with chain rule for the second factor:
  • $f'(x) = 3x \times \left[\frac{1}{5}(x^2 + 4)^{-4/5} \times (2x)\right] + (x^2 + 4)^{1/5} \times (3)$
  • Simplify terms:
  • $3x \times \frac{1}{5} \times 2x = \frac{6x^2}{5}$
  • Result: $f'(x) = \frac{6x^2}{5}(x^2 + 4)^{-4/5} + 3(x^2 + 4)^{1/5}$

• Combining Quotient Rule and Chain Rule

  • Example: Find derivative of $f(x) = \frac{2x}{(3x + 5)^2}$
  • Set up quotient rule structure:
  • $\text{Bottom} = (3x + 5)^2$
  • $\text{Top} = 2x$
  • $\text{Derivative of Top} = 2$
  • $\text{Derivative of Bottom (Chain Rule)} = 2(3x + 5)^1 \times (3)$
  • Formula assembly: $f'(x) = \frac{(3x + 5)^2(2) - 2x \times [2(3x + 5)(3)]}{((3x + 5)^2)^2}$
  • Simplifying powers in denominator: $((3x + 5)^2)^2 = (3x + 5)^4$
  • Simplified expression setup: $f'(x) = \frac{2(3x + 5)^2 - 12x(3x + 5)}{(3x + 5)^4}$

Higher-Order Derivatives

• Notation Conventions

  • First derivative: $f'(x)$
  • Second derivative: $f''(x)$
  • Third derivative: $f'''(x)$
  • Fourth derivative: $f^{(4)}(x)$
  • Fifth derivative: $f^{(5)}(x)$
  • Prime tick marks are standard through the third derivative; numerical superscripts enclosed in parentheses are used for fourth order and higher.

• Polynomial Higher-Order Derivatives

  • Given function: $f(x) = 5x^3 + 4x^2 - 7x + 4$
  • First derivative: $f'(x) = 15x^2 + 8x - 7$
  • Second derivative: $f''(x) = 30x + 8$

• Negative Exponents in Higher-Order Derivatives

  • Given function: $f(x) = \frac{3}{5x^2}$
  • Correct algebraic rewriting: Move denominator variable to numerator: $f(x) = \frac{3}{5}x^{-2}$
  • Note: The coefficient $\frac{3}{5}$ remains unaffected by the negative exponent.
  • First derivative: $f'(x) = \frac{3}{5}(-2)x^{-3} = -\frac{6}{5}x^{-3}$
  • Second derivative: $f''(x) = -\frac{6}{5}(-3)x^{-4} = \frac{18}{5}x^{-4}$
  • Third derivative: $f'''(x) = \frac{18}{5}(-4)x^{-5} = -\frac{72}{5}x^{-5}$

• Successive Chain Rule Derivations from a Given Derivative

  • Given second derivative: $f''(x) = 3(x - 3)^{1/2}$
  • Third derivative: $f'''(x) = 3 \times \frac{1}{2}(x - 3)^{-1/2}(1) = \frac{3}{2}(x - 3)^{-1/2}$
  • Fourth derivative: $f^{(4)}(x) = \frac{3}{2}\left(-\frac{1}{2}\right)(x - 3)^{-3/2}(1) = -\frac{3}{4}(x - 3)^{-3/2}$
  • Fifth derivative: $f^{(5)}(x) = -\frac{3}{4}\left(-\frac{3}{2}\right)(x - 3)^{-5/2}(1) = \frac{9}{8}(x - 3)^{-5/2}$

• Evaluation of Higher-Order Derivatives at a Given Value

  • Find $g'''(0)$ for $g(x) = 3x^3(x^2 - 3x + 2)$
  • Pre-differentiation distribution step: $g(x) = 3x^5 - 9x^4 + 6x^3$
  • First derivative: $g'(x) = 15x^4 - 36x^3 + 18x^2$
  • Second derivative: $g''(x) = 60x^3 - 108x^2 + 36x$
  • Third derivative: $g'''(x) = 180x^2 - 216x + 36$
  • Evaluation at $x = 0$: $g'''(0) = 180(0)^2 - 216(0) + 36 = 36$

• Solving for $x$ Where Derivative Equals Zero vs. Evaluating at Zero

  • Distinction:
  • Evaluating $f''(0)$ means substituting $x = 0$ into the derivative equation.
  • Finding where $f''(x) = 0$ means setting the derivative equation equal to zero and solving for $x$
  • Example: Find where $f''(x) = 0$ for $f(x) = x^3 - 6x^2 + 18x - 18$
  • First derivative: $f'(x) = 3x^2 - 12x + 18$
  • Second derivative: $f''(x) = 6x - 12$
  • Setting equal to zero: $6x - 12 = 0 \implies 6x = 12 \implies x = 2$

Implicit Differentiation

• Implicit Differentiation Rules

  • Applied to implicit equations containing both $x$ and $y$ variables.
  • Take the derivative of both sides of the equation with respect to $x$
  • Differentiate $x$ terms normally.
  • Differentiate $y$ terms using the chain rule, multiplying by $\frac{dy}{dx}$ every time a $y$ term is differentiated.
  • Isolate all terms containing $\frac{dy}{dx}$ on one side of the equation and factor out $\frac{dy}{dx}$.

• Single Derivative Factor Example

  • Find $\frac{dy}{dx}$ for $3x^3 - 2y = 4$
  • Differentiate both sides: $9x^2 - 2\frac{dy}{dx} = 0$
  • Subtract $9x^2$ from both sides: $-2\frac{dy}{dx} = -9x^2$
  • Divide by $-2$: $\frac{dy}{dx} = \frac{9x^2}{2}$

• Multiple Derivative Factor Example

  • Find $\frac{dy}{dx}$ for $5x^2 - y^3 = 4x + 2y$
  • Differentiate both sides: $10x - 3y^2 \frac{dy}{dx} = 4 + 2\frac{dy}{dx}$
  • Rearrange terms to group $\frac{dy}{dx}$ terms on the left side:
  • Subtract $10x$ from both sides.
  • Subtract $2\frac{dy}{dx}$ from both sides.
  • $-3y^2 \frac{dy}{dx} - 2\frac{dy}{dx} = -10x + 4$
  • Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(-3y^2 - 2) = -10x + 4$
  • Divide to isolate $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-10x + 4}{-3y^2 - 2} = \frac{10x - 4}{3y^2 + 2}$

• Implicit Product Rule Example

  • Find $\frac{dy}{dx}$ for $x^2 y^3 + 6x + 2y = 7$
  • Product rule required on $x^2 y^3$ term:
  • $\text{First} = x^2$, $\text{Second} = y^3$
  • $\frac{d}{dx}(x^2 y^3) = x^2 \left(3y^2 \frac{dy}{dx}\right) + y^3(2x) = 3x^2 y^2 \frac{dy}{dx} + 2x y^3$
  • Differentiate full equation: $3x^2 y^2 \frac{dy}{dx} + 2x y^3 + 6 + 2\frac{dy}{dx} = 0$
  • Group $\frac{dy}{dx}$ terms: $3x^2 y^2 \frac{dy}{dx} + 2\frac{dy}{dx} = -2x y^3 - 6$
  • Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(3x^2 y^2 + 2) = -2x y^3 - 6$
  • Final expression: $\frac{dy}{dx} = \frac{-2x y^3 - 6}{3x^2 y^2 + 2}$

Tangent Line Applications and Slopes

• Determining Tangent Line Slope from Implicit Functions

  • The slope $m$ of a tangent line at a given point $(x, y)$ equals the value of $\frac{dy}{dx}$ evaluated at that coordinate.

• Example 1: Single Coordinate Evaluation

  • Find the slope of the tangent line for $x^3 - y^2 = -63$ at the point $(1, 8)$
  • Differentiate implicitly: $3x^2 - 2y \frac{dy}{dx} = 0$
  • Isolate $\frac{dy}{dx}$: $-2y \frac{dy}{dx} = -3x^2 \implies \frac{dy}{dx} = \frac{3x^2}{2y}$
  • Substitute point $(x, y) = (1, 8)$:
  • $\text{Numerator} = 3(1)^2 = 3$
  • $\text{Denominator} = 2(8) = 16$
  • Slope $m = \frac{3}{16}$

• Example 2: Product Rule Implicit Evaluation

  • Find the slope of the tangent line for $x^2 + y^3 = xy$ at the point $(2, 3)$
  • Differentiate both sides (apply product rule to $xy$ on the right-hand side):
  • $2x + 3y^2 \frac{dy}{dx} = x\frac{dy}{dx} + y(1)$
  • Group $\frac{dy}{dx}$ terms on the left:
  • $3y^2 \frac{dy}{dx} - x\frac{dy}{dx} = y - 2x$
  • Factor out $\frac{dy}{dx}$:
  • $\frac{dy}{dx}(3y^2 - x) = y - 2x$
  • Isolated derivative equation: $\frac{dy}{dx} = \frac{y - 2x}{3y^2 - x}$
  • Substitute point $(x, y) = (2, 3)$:
  • $\text{Numerator} = 3 - 2(2) = 3 - 4 = -1$
  • $\text{Denominator} = 3(3)^2 - 2 = 3(9) - 2 = 27 - 2 = 25$
  • Slope $m = -\frac{1}{25}$