Calculus: Power Rule, Sum Rule, Derivative Notations, and Higher-Order Derivatives

Definition and Notations of the Derivative

• Review of the Limit Definition of the Derivative:

  • The derivative of a function $f$ at a point $x$ is defined as:     $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

  • This definition holds provided that the limit exists.

  • Geometrically, the value $f'(x)$ represents the slope of the tangent line to the curve $y = f(x)$ at the point $x$.

• Notation Conventions:

  • When $y$ is a dependent variable determined by the independent variable $x$, there are two primary notations used for the derivative function:

  • Leibniz Notation: $\frac{dy}{dx}$

  • Lagrange Notation: $f'(x)$

  • Equivalence between notations:     $\text{Leibniz Notation} \longrightarrow \frac{dy}{dx} = f'(x) \longleftarrow \text{Lagrange Notation}$

• Geometric Interpretation of Leibniz Notation:

  • $dy$ represents the "infinitesimal rise", given by $f(x+h) - f(x)$.

  • $dx$ represents the "infinitesimal run", given by $(x+h) - x = h$.

  • The term "infinitesimal" signifies taking the limit as $h \to 0$.

  • Thus, the derivative expression $\frac{dy}{dx}$ directly represents:     $\frac{dy}{dx} = \frac{\text{rise}}{\text{run}} = \text{slope}$

• Trade-offs of Leibniz Notation:

  • Advantage: Clearly identifies both the dependent variable ($y$) and the independent variable ($x$).

  • Disadvantage: Not as tidy or compact as Lagrange notation ($f'(x)$).

Example 1: Differentiating a Cubic Function

• Problem Statement:

  • Find the derivative of the cubic function $y = x^3$ using the limit definition.

  • Note that the operator notation $\frac{d}{dx}[x^3]$ is simply another way of writing "differentiate this function".

• Step-by-Step Derivation:

  • Apply the limit definition of the derivative:     $\frac{dy}{dx} = \frac{d}{dx}[x^3] = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h}$

  • Expand $(x+h)^3$ using the Binomial Expansion with coefficients from Pascal's Triangle ($1$, $3$, $3$, $1$):     $(x+h)^3 = x^3 + 3x^2 h + 3xh^2 + h^3$

  • Substitute the expansion back into the limit expression:     $\frac{dy}{dx} = \lim_{h \to 0} \frac{(x^3 + 3x^2 h + 3xh^2 + h^3) - x^3}{h}$

  • Cancel the $x^3$ and $-x^3$ terms in the numerator:     $\frac{dy}{dx} = \lim_{h \to 0} \frac{3x^2 h + 3xh^2 + h^3}{h}$

  • Divide each term in the numerator by $h$:     $\frac{dy}{dx} = \lim_{h \to 0} (3x^2 + 3xh + h^2)$

  • Evaluate the limit as $h \to 0$:     $\frac{dy}{dx} = 3x^2 + 3x(0) + (0)^2 = 3x^2$

• Sanity Checks:

  • Positive Derivative: The derivative $\frac{dy}{dx} = 3x^2$ is positive for all $x \neq 0$, which aligns with $y = x^3$ being a strictly increasing function everywhere except at the origin.

  • Zero Derivative at Origin: The derivative is zero at $x = 0$ because the graph of $y = x^3$ flattens out for that single instantaneous moment.

The Power Rule

• Observed Algorithm:

  • Comparing the original function $y = x^3$ with its derivative $\frac{dy}{dx} = 3x^2$ reveals a two-step pattern:

  1. Use the original power as the multiplier/coefficient ($3$ brought down in front).

  2. Lower the original exponent power by one ($3 - 1 = 2$).

• Formal Statement of the Power Rule:

  • If $y = x^p$, where $p$ is any constant, then:     $\frac{dy}{dx} = p x^{p-1}$

• Proof for Positive Integer Exponents ($p \in \mathbb{Z}^+$):

  • Using the Binomial Expansion Theorem for $(x+h)^p$:     $(x+h)^p = x^p + p x^{p-1} h + \binom{p}{2} x^{p-2} h^2 + \dots + h^p$

  • Substitute into the limit definition of the derivative:     $\frac{dy}{dx} = \lim_{h \to 0} \frac{(x+h)^p - x^p}{h}$     $\frac{dy}{dx} = \lim_{h \to 0} \frac{\left( x^p + p x^{p-1} h + \binom{p}{2} x^{p-2} h^2 + \dots + h^p \right) - x^p}{h}$

  • Cancel the $x^p$ terms:     $\frac{dy}{dx} = \lim_{h \to 0} \frac{p x^{p-1} h + \binom{p}{2} x^{p-2} h^2 + \dots + h^p}{h}$

  • Divide through by $h$:     $\frac{dy}{dx} = \lim_{h \to 0} \left( p x^{p-1} + \binom{p}{2} x^{p-2} h + \dots + h^{p-1} \right)$

  • Evaluate as $h \to 0$: every term after the first term contains at least one factor of $h$ and thus vanishes to $0$

  • The only term that survives is $p x^{p-1}$.

  • Proving the Power Rule when $p$ is not a positive integer requires additional mathematical tools that are developed in subsequent lessons.

• Caution Regarding Domains:

  • If $p$ is not a positive integer, the domain of $f(x) = x^p$ may not be all real numbers ($(-\infty, \infty)$).

  • Examples of Power Functions and Their Domains:

  • Function $x^4$: Domain is $(-\infty, \infty)$

  • Function $x^{-4}$: Domain is $(-\infty, 0) \cup (0, \infty)$

  • Function $x^{1/3}$: Domain is $(-\infty, \infty)$

  • Function $x^{-1/3}$: Domain is $(-\infty, 0) \cup (0, \infty)$

  • Function $x^{1/2} = \sqrt{x}$: Domain is $[0, \infty)$

  • Function $x^{-1/2} = \frac{1}{\sqrt{x}}$: Domain is $(0, \infty)$

Differentiation Rules for Sums and Constant Multiples

• Utility:

  • The Power Rule allows calculation of derivatives without using the limit definition, which becomes extremely efficient when combined with the Sum Rule and Constant Multiple Rule.

• Sum Rule:

  • If $f$ and $g$ are both differentiable at $x$, then the derivative of their sum is equal to the sum of their individual derivatives:     $\frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)] = f'(x) + g'(x)$

• Constant Multiple Rule:

  • For any constant $c$, constant coefficients remain unaffected during differentiation:     $\frac{d}{dx}[c f(x)] = c \frac{d}{dx}[f(x)] = c f'(x)$

• Proof of the Sum Rule:

  • Apply the limit definition to the sum function $f(x) + g(x)$:     $\frac{d}{dx}[f(x) + g(x)] = \lim_{h \to 0} \frac{(f(x+h) + g(x+h)) - (f(x) + g(x))}{h}$

  • Rearrange terms in the numerator:     $= \lim_{h \to 0} \frac{(f(x+h) - f(x)) + (g(x+h) - g(x))}{h}$

  • Split into two limits:     $= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} + \lim_{h \to 0} \frac{g(x+h) - g(x)}{h}$

  • Apply definitions of $f'(x)$ and $g'(x)$:     $= f'(x) + g'(x)$

• Proof of the Constant Multiple Rule:

  • Apply the limit definition to $c f(x)$:     $\frac{d}{dx}[c f(x)] = \lim_{h \to 0} \frac{c f(x+h) - c f(x)}{h}$

  • Factor out constant $c$:     $= \lim_{h \to 0} c \left( \frac{f(x+h) - f(x)}{h} \right)$

  • Pull constant $c$ outside the limit:     $= c \cdot \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = c f'(x)$

Example 2: Differentiating Polynomials and Interval Analysis

• Problem Statement:

  • Given the polynomial function:     $y = 4x^3 - 3x^2 - 18x + 11$

  • Part (a): Calculate the derivative $\frac{dy}{dx}$.

  • Part (b): Determine intervals where the original function $y$ is increasing or decreasing.

• Part (a) Solution:

  • Apply the Sum Rule, Constant Multiple Rule, and Power Rule:     $\frac{dy}{dx} = \frac{d}{dx}[4x^3 - 3x^2 - 18x + 11]$     $\frac{dy}{dx} = 4 \frac{d}{dx}[x^3] - 3 \frac{d}{dx}[x^2] - 18 \frac{d}{dx}[x^1] + \frac{d}{dx}[11]$     $\frac{dy}{dx} = 4(3x^2) - 3(2x) - 18(1) + 0$     $\frac{dy}{dx} = 12x^2 - 6x - 18$

• Part (b) Solution:

  • To determine where the function is increasing or decreasing, solve for where the derivative is positive or negative.

  • Factor out $6$ from the derivative expression:     $\frac{dy}{dx} = 6(2x^2 - x - 3)$

  • Factoring quadratic $2x^2 - x - 3$: find two numbers that multiply to $2 \times (-3) = -6$ and add up to $-1$. The numbers are $-3$ and $2$.

  • Factored derivative:     $\frac{dy}{dx} = 6(2x - 3)(x + 1)$

  • Critical points occur where $\frac{dy}{dx} = 0$:     $x = \frac{3}{2} \quad \text{and} \quad x = -1$

  • Interval Sign Analysis:

  • Interval 1 ($x < -1$):

    • Factors $(2x-3)$ and $(x+1)$ are both negative.

    • Derivative sign: (\text{neg}) \times (\text{neg}) = \text{positive} > 0$.\n * Function behavior: **Increasing**.\n * **Interval 2 (-1 < x < \frac{3}{2})**:\n * Factor (2x-3)$is negative; factor$(x+1) is positive.\n * Derivative sign: (\text{neg}) \times (\text{pos}) = \text{negative} < 0$.

    • Function behavior: Decreasing.

  • Interval 3 ($x > \frac{3}{2}$):

    • Factors $(2x-3)$ and $(x+1)$ are both positive.

    • Derivative sign: (\text{pos}) \times (\text{pos}) = \text{positive} > 0$.\n * Function behavior: **Increasing**.\n\n\n\n# Higher-Order Derivatives\n\n* **Concept**:\n * Since the derivative f'$is itself a function, taking its derivative yields a new function$f'', called the **second derivative**.\n * Repeated differentiation produces the third derivative, fourth derivative, and so forth up to the n^{\text{th}} derivative.\n\n* **Notation Summary Table**:\n * **1st Derivative**:\n * Leibniz Notation: \frac{dy}{dx}\n * Lagrange Notation: f'\n * **2nd Derivative**:\n * Leibniz Notation: \frac{d^2 y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]\n * Lagrange Notation: f''\n * **n^{\text{th}} Derivative**:\n * Leibniz Notation: \frac{d^n y}{dx^n} = \frac{d}{dx}\left[\frac{d^{n-1}y}{dx^{n-1}}\right]\n * Lagrange Notation: f^{(n)}\n\n\n\n# Kinematics Example: Position, Velocity, and Acceleration\n\n* **Scenario**:\n * A basketball is tossed upward in the air (like a tipoff) and allowed to fall back to the ground.\n * Its height f(t)$at time$t is given by:\n    f(t) = -5t^2 + 3t + 2 \quad (\text{meters})\n * Parameter breakdown:\n * Initial height at t = 0$:$2\,\text{meters}\n * Initial upward velocity: 3\,\text{meters/sec}\n * Time t$measured in seconds. Ground contact occurs when$f(t) = 0$at$t = 1\,\text{second}.\n\n* **First Derivative (Velocity)**:\n * The first derivative of position with respect to time represents velocity:\n    f'(t) = -10t + 3 \quad (\text{meters/sec})\n\n* **Second Derivative (Acceleration)**:\n * The second derivative of position represents the rate of change of velocity, which is acceleration:\n    f''(t) = -10 \quad (\text{meters/sec}^2)\n\n* **Physical Interpretation**:\n * The ball's velocity decreases by 10\,\text{meters/sec} every second.\n * Acceleration is constant (-10\,\text{meters/sec}^2) because gravity is the sole force acting on the ball (ignoring air resistance for simplicity).\n\n\n\n# Core Summary\n\n* **Dual Notations**: Derivatives can be written in Leibniz notation (\frac{dy}{dx}$) or Lagrange notation ($f'(x)), and fluency in both is required.\n* **Algebraic Shortcuts**: The Power Rule (\frac{d}{dx}[x^p] = p x^{p-1}$) combined with the Sum Rule ($\frac{d}{dx}[f+g] = f' + g') allows rapid differentiation of polynomials without resorting to limit evaluations.\n* **Physical Applications**: Derivatives model rates of change; in kinematics, the first derivative of position yields velocity (f'(t)$), and the second derivative yields acceleration ($f''(t)$$).