Derivative Rules and Rates of Change

Introduction to Derivatives

• The derivative of a function ($f'(x)$) represents the slope of the tangent line to the function's graph at a specific point.

Fundamental Derivative Rules

The Constant Rule

• If a function $f(x)$ is a constant, meaning $f(x) = c$, its graph is a horizontal line.

• The slope of a horizontal line is always zero.

• Therefore, the derivative of any constant is zero.

  • Formula: If $f(x) = c$, then $f'(x) = 0$.

  • Examples:

    • If $y = 7$, then $y' = 0$.

    • If $f(x) = 0$, then $f'(x) = 0$.

    • If $s(t) = -3$, then $s'(t) = 0$.

    • If $y = k<br>pi^2$ (where $k$ is a constant), then $y' = 0$.

  • Trick Question Alert: If $f(x) = \pi^3$, the derivative $f'(x) = 0$, because $\pi^3$ is a numerical constant, not a variable raised to a power.

Linear Functions

• If a function is a linear equation $f(x) = mx + b$, its graph is a straight line.

• The derivative represents the slope, so for a linear function, the derivative is simply the slope of the line.

  • Formula: If $f(x) = mx + b$, then $f'(x) = m$.

  • Example: For $f(x) = x$ (where $m=1, b=0$), $f'(x) = 1$.

The Power Rule

• The power rule applies when a variable is raised to a power.

• Formula: If $f(x) = x^n$ (where $n$ is a rational number), then $f'(x) = nx^{n-1}$.

• Proof (briefly mentioned): Uses the binomial theorem and the limit definition of the derivative.

• Conditions for Differentiability at $x=0$:

  • If $n$ is a positive integer, the function is differentiable at $x=0$.

  • If $n$ is a negative integer (e.g., $x^{-1} = 1/x$), then the function and its derivative are undefined at $x=0$.

  • If $n$ is a fractional exponent (e.g., $x^{1/3} = \sqrt[3]{x}$), the derivative might be undefined at $x=0$ even if the original function is defined. This often indicates a vertical tangent line at that point, meaning the slope is infinite and the derivative does not exist.

• Examples:

  • If $f(x) = x^3$, then $f'(x) = 3x^2$.

  • If $f(x) = x^{1/3}$ (cube root of $x$), then $f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}} = \frac{1}{3\sqrt[3]{x^2}}$. (Note: function is defined at $x=0$, but derivative is not; this implies a vertical tangent at $x=0$).

  • If $y = \frac{1}{x^2} = x^{-2}$, then $y' = -2x^{-3} = -\frac{2}{x^3}$. (Undefined at $x=0$).

Derivatives of Trigonometric Functions

• Derivative of sine: If $f(x) = \sin x$, then $f'(x) = \cos x$.

• Derivative of cosine: If $f(x) = \cos x$, then $f'(x) = -\sin x$.

  • (Proof is similar to other basic derivative proofs but takes a few minutes, not covered in detail here).

The Constant Multiple Rule

• If a function is multiplied by a constant, the derivative of the product is the constant multiplied by the derivative of the function.

• Formula: If $g(x) = c \cdot f(x)$, then $g'(x) = c \cdot f'(x)$.

  • Alternatively: If $y = c x^n$, then $\frac{dy}{dx} = c(nx^{n-1})$.

• Examples:

  • If $y = 5x^3$, then $y' = 5(3x^2) = 15x^2$.

  • If $y = 2x^{-1}$, then $y' = 2(-1x^{-2}) = -2x^{-2} = -\frac{2}{x^2}$.

  • If $y = \frac{4t^2}{5} = \frac{4}{5}t^2$, then $y' = \frac{4}{5}(2t) = \frac{8}{5}t$.

  • If $y = 2\sqrt{x} = 2x^{1/2}$, then $y' = 2\left(\frac{1}{2}x^{-1/2}\right) = x^{-1/2} = \frac{1}{\sqrt{x}}$.

The Sum and Difference Rules

• The derivative of a sum or difference of functions is the sum or difference of their individual derivatives.

• Sum Rule: If $h(x) = f(x) + g(x)$, then $h'(x) = f'(x) + g'(x)$.

• Difference Rule: If $h(x) = f(x) - g(x)$, then $h'(x) = f'(x) - g'(x)$.

• These rules are straightforward and apply naturally.

• Example: If $h(x) = x^4 + x^3$, then $h'(x) = 4x^3 + 3x^2$.

Product Rule (Crucial Distinction)

• The derivative of a product is NOT the product of the derivatives.

  • Demonstration: If $h(x) = x^3 \cdot x^4 = x^7$, then $h'(x) = 7x^6$.

  • However, if you incorrectly take $f'(x) = (x^3)' = 3x^2$ and $g'(x) = (x^4)' = 4x^3$, then $f'(x) \cdot g'(x) = (3x^2)(4x^3) = 12x^5$, which is not $7x^6$.

• Derivation of the Product Rule (using adding zero):

  • To find the derivative of $h(x) = f(x)g(x)$, we use the limit definition:
    $h'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x)g(x + \Delta x) - f(x)g(x)}{\Delta x}$

  • Add and subtract $f(x+\Delta x)g(x)$ (a form of zero) in the numerator:
    $h'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x)g(x + \Delta x) - f(x + \Delta x)g(x) + f(x + \Delta x)g(x) - f(x)g(x)}{\Delta x}$

  • Factor $f(x + \Delta x)$ from the first two terms and $g(x)$ from the last two:
    $h'(x) = \lim_{\Delta x \to 0} \left[ f(x + \Delta x) \frac{g(x + \Delta x) - g(x)}{\Delta x} + g(x) \frac{f(x + \Delta x) - f(x)}{\Delta x} \right]$

  • As $\Delta x \to 0$:

    • $f(x + \Delta x) \to f(x)$

    • $\frac{g(x + \Delta x) - g(x)}{\Delta x} \to g'(x)$

    • $\frac{f(x + \Delta x) - f(x)}{\Delta x} \to f'(x)$

• Product Rule Formula: If $h(x) = f(x)g(x)$, then $h'(x) = f(x)g'(x) + g(x)f'(x)$.

  • Mnemonic: "First times derivative of the second, plus second times derivative of the first."

• Example application (implied, not fully worked out): To find the derivative of $x \sin x$, use $f(x)=x$ and $g(x)=\sin x$.

Quotient Rule

• The quotient rule applies to functions that are ratios of two other functions.

• Formula: If $h(x) = \frac{f(x)}{g(x)}$, then $h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}$.

  • Mnemonic (common for "low d high minus high d low over low squared"): "Low dee High minus High dee Low over Low Low (or Low Squared)."

Rates of Change and Motion

Average vs. Instantaneous Rate of Change

• Average Rate of Change: The slope of the secant line between two points on a curve. Calculated as $\frac{\Delta y}{\Delta x} = \frac{f(x<em>2) - f(x</em>1)}{x<em>2 - x</em>1}$.

  • Example: Average velocity of a car traveling $100$ miles in $2$ hours is $50$ mph, regardless of speed fluctuations.

• Instantaneous Rate of Change: The slope of the tangent line at a single point on a curve. This is given by the derivative.

  • Calculus allows us to find instantaneous rates, unlike precalculus which focuses on average rates.

• Applications of rates of change: population growth, production rates, water flow rates, velocity, and acceleration.

Motion Along a Straight Line

• Used to describe the movement of a particle or object along a horizontal or vertical line.

• Position Function: $s(t)$ gives the object's position at time $t$.

  • Delta $s$ ($\Delta s$) is the change in position over a time interval delta $t$ ($\Delta t$).

  • Average velocity over a short time interval: $\text{Average Velocity} = \frac{\Delta s}{\Delta t} = \frac{s(t + \Delta t) - s(t)}{\Delta t}$.

• Velocity Function: $v(t)$ is the instantaneous rate of change of position, or the derivative of the position function.

  • Formula: $v(t) = s'(t) = \lim_{\Delta t \to 0} \frac{s(t + \Delta t) - s(t)}{\Delta t}$.

  • Velocity can be positive (moving right/up), negative (moving left/down), or zero (at rest).

• Speed: The absolute value of velocity.

  • Formula: $\text{Speed} = |v(t)| = |s'(t)|$.

  • Speed is always non-negative.

• Acceleration Function: $a(t)$ is the rate of change of velocity, or the derivative of the velocity function (and the second derivative of the position function).

  • Formula: $a(t) = v'(t) = s''(t)$.

Free-Fall Position Function (Standard Physics Formula)

• For an object moving purely under gravity (vertical motion), the position function is given by: $s(t) = \frac{1}{2}gt^2 + v<em>0t + s</em>0$

  • Where:

    • $g$ is the acceleration due to gravity. (This will be negative in common physics applications as it pulls downwards).

    • Constants for $g$:

      • Metric system: $g = -9.8 \text{ m/s}^2$

      • English units: $g = -32 \text{ ft/s}^2$

    • $v_0$ (read as "v nought" or "v sub zero") is the initial velocity (velocity at time $t=0$).

    • $s_0$ (read as "s nought" or "s sub zero") is the initial position or initial height (position at time $t=0$).

• Example: Diver Jumping from a Platform

  • Initial height ($s_0$): $32$ feet.

  • Initial upward velocity ($v_0$): $16$ feet per second.

  • Position function (using $g = -32 \text{ ft/s}^2$):
    $s(t) = \frac{1}{2}(-32)t^2 + 16t + 32$
    $s(t) = -16t^2 + 16t + 32$

  • Question 1: When does the diver hit the water?

    • Set $s(t) = 0$ (height above water is zero).

    • $-16t^2 + 16t + 32 = 0$

    • Divide by $-16$: $t^2 - t - 2 = 0$

    • Factor: $(t-2)(t+1) = 0$

    • Solutions: $t=2$ or $t=-1$.

    • Since time cannot be negative in this context, the diver hits the water at $t=2$ seconds.

  • Question 2: What is the velocity at impact?

    • First, find the velocity function $v(t)$ by taking the derivative of $s(t)$:
      $s'(t) = v(t) = \frac{d}{dt}(-16t^2 + 16t + 32)$
      $v(t) = -32t + 16$ (constant term $32$ becomes $0$; $16t$ becomes $16$; $-16t^2$ becomes $-32t$).

    • Now, plug in the impact time $t=2$ into $v(t)$:
      $v(2) = -32(2) + 16 = -64 + 16 = -48$

    • The velocity at impact is $-48$ feet per second. The negative sign indicates the diver is moving downwards.