Comprehensive Calculus Notes: Limits, Differentiability, Differentiation Rules, and Kinematics

The Limit Definition of the Derivative and Differentiability

• Definition of the Derivative:

  • The general definition of the derivative of a function $f(x)$ with respect to the independent variable $x$ is defined as the limit of the difference quotient:   $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$   where $h$ represents the change in the independent variable.
  • When calculating the derivative at a specific point $x = c$, the limit definition takes the form:   $f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$
  • Alternatively, the derivative at a specific point $x = c$ can be expressed using the limit of the average rate of change as $x$ approaches $c$:   $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$
  • Geometric and physical interpretations:
  • The derivative at $x = c$ yields the exact slope of the tangent line to the curve $y = f(x)$ at the point $(c, f(c))$.
  • The derivative represents the instantaneous rate of change of the function value with respect to the independent variable at that specific point.

• Differentiability and Its Relationship to Continuity:

  • If the limit defining $f'(c)$ exists, the function $f$ is said to be differentiable at $c$.
  • Differentiability Implies Continuity Theorem: If a function $f$ is differentiable at $x = c$, then $f$ is guaranteed to be continuous at $x = c$.
  • Non-Reversibility (One-Way Theorem): The converse of this theorem is false. A function being continuous at $x = c$ does not guarantee that it is differentiable at $x = c$.
  • Existence of the Derivative Limit: For $f'(x)$ to exist, the specific limit of the difference quotient must exist:   $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$   The existence of the function's limit $\lim_{x \to c} f(x)$ alone is insufficient to guarantee differentiability.

One-Sided Derivatives

• Right-Hand and Left-Hand Derivatives:

  • Analogous to one-sided limits, one-sided derivatives evaluate differentiability at boundary points of an interval or points of non-smoothness.
  • On a closed interval $[a, b]$, the right-hand derivative of $f$ at the left endpoint $a$ is defined as:   $f'(a^+) = \lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$
  • On a closed interval $[a, b]$, the left-hand derivative of $f$ at the right endpoint $b$ is defined as:   $f'(b^-) = \lim_{h \to 0^-} \frac{f(b+h) - f(b)}{h}$
  • One-sided derivatives are required when assessing differentiability at the endpoints of a closed domain.

• Theorem on the Existence of the Two-Sided Derivative:

  • A function $f$ is differentiable at an interior point $x = c$ if and only if both the left-hand derivative $f'(c^-)$ and the right-hand derivative $f'(c^+)$ exist and are equal:   $f'(c^+) = f'(c^-) = f'(c)$
  • If both $f'(c^+)$ and $f'(c^-)$ exist but are not equal ($f'(c^+) \neq f'(c^-)$), then the two-sided derivative $f'(c)$ does not exist, and $f$ is not differentiable at $x = c$.

Discontinuity, Corner Points, and Non-Differentiability

• Non-Differentiability due to Removable Discontinuity:

  • Consider the piecewise function $g(x)$ defined as:   $g(x) = \begin{cases} x^2 & \text{if } x \neq 1 \\ 2 & \text{if } x = 1 \end{cases}$
  • Continuity Analysis:
  • $g(1) = 2$ exists.
  • $\lim_{x \to 1} g(x) = 1^2 = 1$.
  • Because $\lim_{x \to 1} g(x) = 1 \neq g(1) = 2$, the function has a removable discontinuity at $x = 1$ and is not continuous on all real numbers $\mathbb{R}$.
  • Differentiability Analysis at $x = 1$:
  • Setting up the limit definition of the derivative at $x = 1$:     $g'(1) = \lim_{h \to 0} \frac{g(1+h) - g(1)}{h} = \lim_{h \to 0} \frac{(1+h)^2 - 2}{h}$
  • Expanding the algebraic terms:     $g'(1) = \lim_{h \to 0} \frac{1 + 2h + h^2 - 2}{h} = \lim_{h \to 0} \frac{h^2 + 2h - 1}{h} = \lim_{h \to 0} \left(2 + h - \frac{1}{h}\right)$
  • Plotting or taking the one-sided limits of $2 + h - \frac{1}{h}$ as $h \to 0$ reveals vertical asymptotes along the $h = 0$ axis:
    • As $h \to 0^+$, $\frac{1}{h} \to \infty$, so the expression approaches $-\infty$.
    • As $h \to 0^-$, $\frac{1}{h} \to -\infty$, so the expression approaches $+\infty$.
  • Because the limit is unbounded and the one-sided limits fail to converge to a finite real number, $g'(1)$ does not exist. Removable discontinuities directly prevent differentiability.

• Non-Differentiability due to Sharp Corners (Continuous Function Example):

  • Consider the absolute value function $f(x) = |x|$ on $\mathbb{R}$.
  • Continuity: $f(x) = |x|$ is continuous for all real numbers $\mathbb{R}$, including $x = 0$
  • Differentiability Analysis at $x = 0$ using one-sided derivatives:
  • Right-hand derivative at $x = 0$:     $f'(0^+) = \lim_{h \to 0^+} \frac{|0+h| - |0|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$
  • Left-hand derivative at $x = 0$:     $f'(0^-) = \lim_{h \to 0^-} \frac{|0+h| - |0|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$
  • Since $f'(0^+) = 1 \neq -1 = f'(0^-)$, the overall limit does not exist, so $f(x) = |x|$ is not differentiable at $x = 0$.
  • Sharp points, corners, or cusps on a graph visually indicate that left-hand and right-hand slopes disagree, creating a point of non-differentiability.
  • At points away from the corner (e.g., $x > 0$ or $x < 0$), the absolute value function is smooth and differentiable.

• Non-Differentiability due to Unbounded Limits / Vertical Tangents:

  • Consider the cube root function $y = f(x) = x^{1/3} = \sqrt[3]{x}$ at $x = 0$
  • Evaluating the limit definition of the derivative at $x = 0$:   $f'(0) = \lim_{h \to 0} \frac{(0+h)^{1/3} - 0^{1/3}}{h} = \lim_{h \to 0} \frac{h^{1/3}}{h} = \lim_{h \to 0} \frac{1}{h^{2/3}}$
  • As $h \to 0$, $h^{2/3} \to 0^+$ through positive values, causing $\frac{1}{h^{2/3}} \to +\infty$
  • Because the limit approaches infinity, it is unbounded and lacks a finite limiting value. Consequently, the derivative does not exist at $x = 0$.
  • Geometrically, an unbounded derivative limit indicates the presence of a vertical tangent line at that point on the curve.

Basic Differentiation Rules

• The Power Rule:

  • If $a$ is any real number exponent, the derivative of $x^a$ is given by:   $\frac{d}{dx}[x^a] = a x^{a-1}$
  • Example 1: For $f(x) = x^4$:   $f'(x) = 4 x^{4-1} = 4 x^3$
  • Example 2: For $g(x) = \frac{1}{x} = x^{-1}$, applying the power rule yields:   $g'(x) = -1 x^{-1-1} = -1 x^{-2} = -\frac{1}{x^2}$

• The Sum Rule for Derivatives:

  • If two functions $f$ and $g$ are differentiable at $x = c$, then their sum $(f + g)$ is differentiable at $c$, and:   $(f + g)'(c) = f'(c) + g'(c)$
  • In general function notation:   $\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$

• The Constant Multiple Rule for Derivatives:

  • If $f$ is a differentiable function at $x = c$ and $k \in \mathbb{R}$ is any constant scalar:   $(k \cdot f)'(c) = k \cdot f'(c)$
  • In general function notation:   $\frac{d}{dx}[k \cdot f(x)] = k \cdot f'(x)$
  • The derivative of a pure constant term $C$ is always zero:   $\frac{d}{dx}[C] = 0$

• Comprehensive Computational Examples:

  • Example 1: Constant Multiple with Power Rule
  • Function: $y = 4 x^5$
  • Derivative computation: $y' = 4 \cdot \frac{d}{dx}[x^5] = 4 \cdot (5 x^4) = 20 x^4$
  • Example 2: Sum Rule with Polynomial Kinematics
  • Position function: $s(t) = 7 t^3 + 5 t^2$
  • Velocity function computation:     $v(t) = s'(t) = 7(3 t^2) + 5(2 t) = 21 t^2 + 10 t$
  • Example 3: Fractional and Negative Exponents
  • Function: $y = 4 \sqrt{x} - \frac{2}{x^2} = 4 x^{1/2} - 2 x^{-2}$
  • Term-by-term differentiation:     $y' = 4 \left(\frac{1}{2} x^{-1/2}\right) - 2 \left(-2 x^{-3}\right) = 2 x^{-1/2} + 4 x^{-3} = \frac{2}{\sqrt{x}} + \frac{4}{x^3}$
  • Example 4: Basic Power Rule Instances
  • $\frac{d}{dx}[x^2] = 2 x$
  • $\frac{d}{dx}[x^{100}] = 100 x^{99}$
  • Exponential Function Rule:
  • The derivative of the natural exponential function $e^x$ is uniquely equal to itself:     $\frac{d}{dx}[e^x] = e^x$

Equations of Tangent Lines

• Procedure for Finding the Equation of a Tangent Line:

  1. Calculate the derivative function $g'(x)$ to establish the general slope formula.
  2. Evaluate the derivative at the given target $x$-coordinate ($x = x_0$) to compute the specific numerical slope $m = g'(x_0)$.
  3. Determine the point of tangency $(x_0, g(x_0))$ by evaluating the original function at $x_0$.
  4. Substitute slope $m$ and point $(x_0, g(x_0))$ into point-slope form:   $y - g(x_0) = m(x - x_0)$

• Worked Example:

  • Given a derivative function $g'(x) = 16 x - 5$ evaluated at $x = 1$:   $m = g'(1) = 16(1) - 5 = 11$
  • The resulting slope of the tangent line at $x = 1$ is $11$

The Product Rule

• Definition and Formula:

  • The derivative of a product of two differentiable functions is not equal to the product of their individual derivatives.
  • Let $f$ and $g$ be functions differentiable on a common interval $I$ containing $c$. The derivative of their product is defined as:   $(f \cdot g)'(c) = f'(c) g(c) + f(c) g'(c)$
  • In general operator notation:   $\frac{d}{dx}[f(x) g(x)] = f'(x) g(x) + f(x) g'(x)$

• Comparative Examples:

  • Example 1: Function $f(x) = (3 x^2)(4 x + 1)$
  • Method A (Algebraic Expansion without Product Rule):
    • Expand original expression: $f(x) = 12 x^3 + 3 x^2$
    • Apply power rule: $f'(x) = 36 x^2 + 6 x$
  • Method B (Using Product Rule):
    • Assign components: Let $g(x) = 3 x^2 \implies g'(x) = 6 x$, and let $h(x) = 4 x + 1 \implies h'(x) = 4$
    • Apply formula $f'(x) = g'(x) h(x) + g(x) h'(x)$: $f'(x) = (6 x)(4 x + 1) + (3 x^2)(4) = 24 x^2 + 6 x + 12 x^2 = 36 x^2 + 6 x$
    • Both methods yield the identical result: $36 x^2 + 6 x$
  • Example 2: Function $h(x) = (2 x^5 - 1)(x^3 + 1)$
  • Assign components: Let $u(x) = 2 x^5 - 1 \implies u'(x) = 10 x^4$, and let $v(x) = x^3 + 1 \implies v'(x) = 3 x^2$
  • Apply Product Rule:     $h'(x) = u'(x) v(x) + u(x) v'(x) = (10 x^4)(x^3 + 1) + (2 x^5 - 1)(3 x^2)$
  • Expand terms: $h'(x) = 10 x^7 + 10 x^4 + 6 x^7 - 3 x^2 = 16 x^7 + 10 x^4 - 3 x^2$

The Quotient Rule

• Definition and Formula:

  • Let $f$ and $g$ be differentiable functions on an open interval containing $c$, provided $g(c) \neq 0$.
  • The quotient $\frac{f}{g}$ is differentiable at $c$, and its derivative is:   $\left(\frac{f}{g}\right)'(c) = \frac{f'(c) g(c) - f(c) g'(c)}{[g(c)]^2}$
  • In general operator notation:   $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x) g(x) - f(x) g'(x)}{[g(x)]^2}$

• Reciprocal Special Case Trick:

  • For a fraction with a constant numerator $1$ over a non-zero differentiable function $f(x)$, where $f(x) \neq 0$:   $\frac{d}{dx}\left[\frac{1}{f(x)}\right] = \frac{-f'(x)}{[f(x)]^2}$
  • If the numerator contains a scalar constant $k$:   $\frac{d}{dx}\left[\frac{k}{f(x)}\right] = \frac{-k \cdot f'(x)}{[f(x)]^2}$

• Comparative Examples:

  • Example 1: Function $g(x) = \frac{3 x^2 + x}{x}$
  • Method A (Algebraic Simplification):
    • Simplify fraction for $x \neq 0$: $g(x) = 3 x + 1$
    • Take derivative: $g'(x) = 3$
  • Method B (Applying Quotient Rule):
    • Numerator $f(x) = 3 x^2 + x \implies f'(x) = 6 x + 1$
    • Denominator $h(x) = x \implies h'(x) = 1$
    • Apply rule:       $g'(x) = \frac{(6 x + 1)(x) - (3 x^2 + x)(1)}{x^2} = \frac{6 x^2 + x - 3 x^2 - x}{x^2} = \frac{3 x^2}{x^2} = 3$
  • Example 2: Non-simplifying Rational Function
  • Function: $y = \frac{\sqrt{x}}{-3 x + 4}$
  • Numerator component: $f(x) = x^{1/2} \implies f'(x) = \frac{1}{2 \sqrt{x}}$
  • Denominator component: $g(x) = -3 x + 4 \implies g'(x) = -3$
  • Applying Quotient Rule:     $y' = \frac{\left(\frac{1}{2 \sqrt{x}}\right)(-3 x + 4) - (\sqrt{x})(-3)}{(-3 x + 4)^2}$

Higher-Order Derivatives and Kinematics

• Definition of Higher-Order Derivatives:

  • Taking the derivative of a derivative function $f'(x)$ yields the second derivative, denoted as $f''(x)$.
  • Notations for the second derivative:
  • Prime notation: $f''(x)$ or $y''$
  • Operator notation: $\frac{d}{dx}\left[\frac{df}{dx}\right] = \frac{d^2 f}{dx^2}$ or $\frac{d^2 y}{dx^2}$
  • Third-order derivatives $f'''(x) = \frac{d^3 f}{dx^3}$ and higher are similarly defined by repeated differentiation.

• Physical Applications: Position, Velocity, and Acceleration:

  • Position Function: Denoted $s(t)$, represents spatial location at time $t$ (measured in meters, $\text{m}$).
  • Velocity Function: Denoted $v(t)$, represents instantaneous rate of change of position with respect to time (measured in meters per second, $\text{m/s}$):   $v(t) = s'(t) = \frac{ds}{dt}$
  • Acceleration Function: Denoted $a(t)$, represents instantaneous rate of change of velocity with respect to time (measured in meters per second squared, $\text{m/s}^2$):   $a(t) = v'(t) = s''(t) = \frac{d^2 s}{dt^2}$

• Physical Interpretation of Direction and Speeding Up / Slowing Down:

  • Direction of Motion (Velocity Sign):
  • Positive velocity ($v(t) > 0$) indicates motion in the positive coordinate direction (rightward/upward).
  • Negative velocity ($v(t) < 0$) indicates motion in the negative coordinate direction (leftward/downward).
  • Speeding Up vs. Slowing Down (Relative Signs of $v(t)$ and $a(t)$):
  • Speeding Up: Occurs when velocity $v(t)$ and acceleration $a(t)$ have the same sign (both positive or both negative).
  • Slowing Down (Decelerating): Occurs when velocity $v(t)$ and acceleration $a(t)$ have opposite signs (one positive, one negative).
  • Real-World Analogy: Driving a car in reverse produces negative velocity. Applying the gas pedal in reverse increases speed in the negative direction (negative velocity and negative acceleration = speeding up in reverse). Applying the brakes while in reverse produces positive acceleration against the negative velocity (opposite signs = slowing down).

• Kinematic Worked Problems:

  • Problem 1: Constant Acceleration Motion
  • Given position $s(t) = 16 + 24 t - 16 t^2$ with $s$ in meters ($\text{m}$) and $t$ in seconds ($\text{s}$).
  • Velocity function derivation: $v(t) = s'(t) = 24 - 32 t$
  • Acceleration function derivation: $a(t) = v'(t) = s''(t) = -32\,\text{m/s}^2$
  • Evaluation at $t = 2\,\text{s}$:
    • Velocity: $v(2) = 24 - 32(2) = 24 - 64 = -40\,\text{m/s}$. The negative value indicates downward/leftward directional movement.
    • Acceleration: $a(2) = -32\,\text{m/s}^2$
    • Analysis: Because both $v(2) = -40\,\text{m/s}$ and $a(2) = -32\,\text{m/s}^2$ are negative, the object is speeding up in the downward direction.
  • Problem 2: Time-Dependent Acceleration
  • Consider a higher-degree polynomial position function yielding a time-dependent acceleration function $a(t) = -96 t$
  • At time $t = 2\,\text{s}$:
    • Acceleration evaluation: $a(2) = -96(2) = -192\,\text{m/s}^2$
    • Acceleration varies continuously over time rather than remaining constant.

Questions and Discussion

• Question: If a derivative limit evaluates to positive infinity ($\infty$) or negative infinity ($-\infty$), does the derivative exist?

  • Answer: No. For a derivative to exist, the difference quotient limit must converge to a single, finite real number. If a limit approaches $\pm \infty$, the limit is unbounded, meaning there is no limiting real value. Consequently, the function is not differentiable at that point (which geometrically corresponds to a vertical tangent line).

• Question: When evaluating the alternative limit form of the derivative, what does the variable approach?

  • Answer: In the form $\lim_{x \to c} \frac{f(x) - f(c)}{x - c}$, the variable $x$ approaches the fixed point $c$. In contrast, in the $h$-form $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$, the variable $h$ (representing the change in the independent variable) always approaches $0$.

• Question: If both velocity and acceleration are negative, is the object decelerating or accelerating?

  • Answer: The object is speeding up (accelerating in the negative direction). Deceleration (slowing down) requires velocity and acceleration to have opposite signs. When both vectors share the same sign, acceleration reinforces velocity, causing the object's speed to increase.