3.2 Basic Differentiation Rules & Rectilinear Motion

3.2 Basic Differentiation Rules & Rectilinear Motion

Quick Answer: The fundamental differentiation rules eliminate the need for computing limits directly on algebraic functions. The Power Rule $\frac{d}{dx}[x^n] = n x^{n-1}$ applies to all real exponents. In physics and particle kinematics, differentiating position $s(t)$ yields velocity $v(t) = s'(t)$, speed is $|v(t)|$, and differentiating velocity yields acceleration $a(t) = v'(t) = s''(t)$. A particle is speeding up when $v(t)$ and $a(t)$ share the same sign, and slowing down when they have opposite signs.

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1. Fundamental Differentiation Rules

Applying the limit definition to standard algebraic structures yields five primary operational rules.

The Constant Rule

The derivative of any constant value $c \in \mathbb{R}$ is zero: $\frac{d}{dx}[c] = 0$ Geometric intuition: The graph of $y = c$ is a horizontal line with slope $m = 0$ everywhere.

The General Power Rule

For any real exponent $n \in \mathbb{R}$: $\frac{d}{dx}[x^n] = n x^{n-1}$

To apply the Power Rule effectively on the CLEP exam, first rewrite radical and rational expressions into standard power forms $x^n$:

• Radical expressions: $\sqrt{x} = x^{1/2}$, $\sqrt[3]{x} = x^{1/3}$, $\sqrt[q]{x^p} = x^{p/q}$
• Reciprocal expressions: $\frac{1}{x} = x^{-1}$, $\frac{1}{x^4} = x^{-4}$, $\frac{1}{\sqrt{x}} = x^{-1/2}$
• Combined rational radicals: $\frac{7}{\sqrt[3]{x^2}} = 7x^{-2/3}$

The Constant Multiple Rule

A multiplicative constant factors out of the derivative: $\frac{d}{dx}[c \cdot f(x)] = c \cdot \frac{d}{dx}[f(x)] = c \cdot f'(x)$

The Sum and Difference Rules

The derivative of a sum or difference of differentiable functions is the sum or difference of their individual derivatives: $\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$ $\frac{d}{dx}[f(x) - g(x)] = f'(x) - g'(x)$

Combining these rules demonstrates that differentiation is a linear operator: $\frac{d}{dx}[a f(x) + b g(x)] = a f'(x) + b g'(x)$

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2. Higher-Order Derivatives

Differentiating a function $y = f(x)$ produces the first derivative $f'(x)$. Differentiating $f'(x)$ yields the second derivative, and repeated differentiation yields higher-order derivatives.

Key Property (Polynomial Termination): If $P(x)$ is a polynomial of degree $n$, then its $n$-th derivative $P^{(n)}(x)$ is a constant ($n! \cdot a_n$), and every derivative of order higher than $n$ is identically zero: $P^{(k)}(x) = 0 \quad \text{for all } k > n$

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3. Rectilinear Particle Motion Along a Line

Rectilinear motion describes the one-dimensional movement of an object along a coordinate axis (typically the $s$-axis or $x$-axis) as a function of time $t \ge 0$.

                               Rectilinear Motion Kinematics
                                              │
     Position: s(t) ──────────────────────────┼───> Location along the axis
        │ (d/dt)
        ▼
     Velocity: v(t) = s'(t) ──────────────────┼───> Directed rate of change
        │ (d/dt)                                    • v > 0: Moving right / forward
        ▼                                           • v < 0: Moving left / backward
     Acceleration: a(t) = v'(t) = s''(t) ─────┼───> • v = 0: Stopped / changing direction
                                                    
     Speed: |v(t)| = |s'(t)| ─────────────────┴───> Magnitude of velocity (scalar ≥ 0)

Kinematic Definitions

1. Position $s(t)$: The directed coordinate location of the particle relative to the origin at time $t$.
2. Displacement over $[t_1, t_2]$: The net change in position: $\Delta s = s(t_2) - s(t_1)$
3. Velocity $v(t)$: The instantaneous rate of change of position with respect to time: $v(t) = \frac{ds}{dt} = s'(t)$
   • Moving Forward / Right: $v(t) > 0$
   • Moving Backward / Left: $v(t) < 0$
   • At Rest / Stopped: $v(t) = 0$
   • Changes Direction: $v(t)$ changes sign (crosses zero from positive to negative or vice versa).
4. Speed: The absolute magnitude of velocity (always non-negative): $\text{Speed} = |v(t)| = |s'(t)|$
5. Acceleration $a(t)$: The instantaneous rate of change of velocity with respect to time: $a(t) = \frac{dv}{dt} = v'(t) = \frac{d^2 s}{dt^2} = s''(t)$

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4. Speeding Up vs. Slowing Down: The Golden Rule

A critical concept on the CLEP Calculus exam is analyzing when a moving object is speeding up or slowing down.

The Sign Rule for Speed:

• A particle is SPEEDING UP if and only if velocity $v(t)$ and acceleration $a(t)$ have the SAME SIGN ($v(t) \cdot a(t) > 0$).
• A particle is SLOWING DOWN if and only if velocity $v(t)$ and acceleration $a(t)$ have OPPOSITE SIGNS ($v(t) \cdot a(t) < 0$).

Common Student Trap: Acceleration being negative does not automatically mean the particle is slowing down! If $v(t) = -8\text{ m/s}$ and $a(t) = -4\text{ m/s}^2$, the speed $|v(t)|$ is increasing from $8\text{ m/s}$ to $12\text{ m/s}$ in the negative direction.

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5. Step-by-Step Worked Kinematic Example

Problem: A particle moves along a horizontal axis with position function $s(t) = t^3 - 6t^2 + 9t + 2, \quad t \ge 0$

Step 1: Find Velocity and Acceleration Functions

• $v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t - 1)(t - 3)$
• $a(t) = v'(t) = 6t - 12 = 6(t - 2)$

Step 2: Determine When the Particle is at Rest and Direction Intervals

• Set $v(t) = 0 \implies 3(t-1)(t-3) = 0 \implies t = 1$ and $t = 3$.
• Interval $[0, 1)$: $v(0.5) = 3(-0.5)(-2.5) > 0 \implies$ Particle moves right.
• Interval $(1, 3)$: $v(2) = 3(1)(-1) < 0 \implies$ Particle moves left.
• Interval $(3, \infty)$: $v(4) = 3(3)(1) > 0 \implies$ Particle moves right.
• Direction changes occur at $t = 1$ and $t = 3$.

Step 3: Determine Speeding Up and Slowing Down Intervals

Construct a simultaneous sign chart for $v(t)$ and $a(t)$ with key partition points $t = 1, 2, 3$:

Conclusion: The particle is speeding up on $(1, 2) \cup (3, \infty)$ and slowing down on $(0, 1) \cup (2, 3)$.

Step 4: Total Distance Traveled vs. Displacement on $[0, 4]$

• Displacement: $\Delta s = s(4) - s(0) = [(4)^3 - 6(4)^2 + 9(4) + 2] - [2] = [64 - 96 + 36 + 2] - 2 = 6 - 2 = 4$.
• Total Distance Traveled: Must split the path at turning points $t = 1$ and $t = 3$:
  • $s(0) = 2$
  • $s(1) = (1)^3 - 6(1)^2 + 9(1) + 2 = 6$
  • $s(3) = (3)^3 - 6(3)^2 + 9(3) + 2 = 2$
  • $s(4) = 6$ $\text{Total Distance} = |s(1) - s(0)| + |s(3) - s(1)| + |s(4) - s(3)| = |6 - 2| + |2 - 6| + |6 - 2| = 4 + 4 + 4 = 12$

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6. CLEP Exam Tips

• Tip 1: Always rewrite negative and fractional powers before applying the Power Rule.
• Tip 2: When asked for "speed at $t = c$", remember speed is $|v(c)|$ and can never be negative.
• Tip 3: For total distance, always find where $v(t) = 0$ in the interval and sum the absolute values of the separate segment displacements.