Calculus Notes: Differentiability, Product Rule, and Quotient Rule

Differentiability and Absolute Value Functions

• The transformation of absolute value functions begins with the basic parent function $y = |x|$.
• The graph of $y = |x|$ forms a V-shape centered at the origin $(0,0)$, where a sharp corner occurs.
• At the sharp corner $x = 0$, the function is not differentiable because the direction of the tangent line changes abruptly.
• For the composite function $f(x) = |2x - 4|$, differentiability is analyzed by evaluating the limit definition of the derivative:
  • $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$
  • An issue arises at the point where the expression inside the absolute value equals zero, namely when $2x - 4 = 0$ or $x = 2$.
• To evaluate differentiability rigorously, $f(x) = |2x - 4|$ is rewritten as a piecewise function:
  • $f(x) = \begin{cases} 2x - 4, & \text{if } 2x - 4 \ge 0 \\ -(2x - 4), & \text{if } 2x - 4 < 0 \end{cases}$
• Simplifying the domain conditions yields:
  • $f(x) = \begin{cases} 2x - 4, & \text{if } x \ge 2 \\ -2x + 4, & \text{if } x < 2 \end{cases}$
• Calculating the one-sided derivatives at $x = 2$:
  • The derivative from the right (for $x \ge 2$) is equal to $2$
  • The derivative from the left (for $x < 2$) is equal to $-2$
• Because the derivative from the right ($2$) does not equal the derivative from the left ($-2$), the limit defining the derivative does not exist at $x = 2$, making $f(x)$ not differentiable at $x = 2$.

The Product Rule for Differentiation

• The Product Rule is used to compute the derivative of a product of two functions, $f(x)$ and $g(x)$.
• The formal equation for the Product Rule is:
  • $\frac{d}{dx} (f(x) \cdot g(x)) = f'(x) \cdot g(x) + f(x) \cdot g'(x)$
• A structured criss-cross method can be used to perform Product Rule calculations systematically:
  • Line up the first function $f(x)$ and the second function $g(x)$ horizontally.
  • Below each function, write its corresponding derivative, $f'(x)$ and $g'(x)$.
  • Cross-multiply the terms ($f'(x)$ with $g(x)$, and $f(x)$ with $g'(x)$) and add the resulting products.
• Worked Example 1: Finding the derivative of $x^2 e^x$
  • Identify the 1st term as $x^2$ and the 2nd term as $e^x$.
  • Compute the derivative of the 1st term: $\frac{d}{dx}(x^2) = 2x$.
  • Compute the derivative of the 2nd term: $\frac{d}{dx}(e^x) = e^x$.
  • Apply the criss-cross cross-multiplication:
  • $\frac{d}{dx} (x^2 e^x) = (2x) e^x + e^x (x^2)$
• Worked Example 2: Finding the derivative of $x^3 (2\sqrt{x} + 1)$
  • Identify the 1st term as $x^3$ and the 2nd term as $2\sqrt{x} + 1$.
  • Derivative of the 1st term is $3x^2$.
  • Derivative of the 2nd term is $\frac{1}{\sqrt{x}}$.
  • Cross-multiply and sum to complete the derivative application using the Product Rule.

The Quotient Rule for Differentiation

• The Quotient Rule is used to differentiate functions that are expressed as the ratio of two functions, $\frac{f(x)}{g(x)}$.
• Standard formula representations:
  • $\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2}$
  • Equivalent form: $\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2}$
• Terminology and Verbal Mnemonic:
  • Label the numerator $f(x)$ as $\text{High}$ and the denominator $g(x)$ as $\text{Low}$.
  • Verbal memory mnemonic: "Low times d-high minus high times d-low all over low squared."
• Criss-Cross Diagrammatic Technique:
  • Place the top function $f(x)$ and bottom function $g(x)$ in the top row.
  • Place their respective derivatives $f'(x)$ and $g'(x)$ in the bottom row.
  • Multiply diagonally $f'(x) \cdot g(x)$ and subtract the product $f(x) \cdot g'(x)$.
  • Divide the entire difference by the denominator squared, [g(x)]^2$.\n* Worked Example 1: Finding the derivative of \frac{x^4}{x^3} using the Quotient Rule\n * Identify \text{High} = x^4$and$\text{Low} = x^3.\n * Derivatives: \text{d-high} = \frac{d}{dx}(x^4) = 4x^3$, and$\text{d-low} = \frac{d}{dx}(x^3) = 3x^2.\n * Substitute into the formula:\n * \frac{d}{dx} \left( \frac{x^4}{x^3} \right) = \frac{x^3 \cdot \frac{d}{dx}(x^4) - x^4 \cdot \frac{d}{dx}(x^3)}{[x^3]^2}\n * = \frac{x^3(4x^3) - x^4(3x^2)}{x^6}\n * Algebraic simplification:\n * = \frac{(4x^3) \cdot x^3 - (3x^2) \cdot x^4}{x^6}\n * = \frac{4x^6 - 3x^6}{x^6}\n * = \frac{x^6}{x^6} = 1\n* Worked Example 2: Finding the derivative of \frac{w^2 + 3w + 4}{w^2 - 1}\n * Identify numerator: w^2 + 3w + 4$with derivative$2w + 3.\n * Identify denominator: w^2 - 1$with derivative$2w.\n * Apply criss-cross setup:\n * Multiply derivative of numerator by denominator: (2w + 3)(w^2 - 1).\n * Subtract derivative of denominator times numerator: (2w)(w^2 + 3w + 4).\n * Square the original denominator: (w^2 - 1)^2$.
  • Resulting expression:
  • $\frac{d}{dw} \left( \frac{w^2 + 3w + 4}{w^2 - 1} \right) = \frac{(2w + 3)(w^2 - 1) - (2w)(w^2 + 3w + 4)}{(w^2 - 1)^2}$