3.3.2 Rules of Differentiation

Slope of a Tangent Line (Finding where the derivative equals a prescribed value)

  • Problem statement

    • Given f(x)=2x315x2+24xf(x)=2x^3-15x^2+24x, find all xx-values where the tangent line to the graph has slope 66.
    • "Slope of the tangent" = value of the first derivative f(x)f'(x) at the desired point.
  • Differentiate with basic rules

    • Power Rule (d/dx)[xn]=nxn1(d/dx)[x^n]=nx^{n-1} and Constant Multiple Rule apply.
    • Compute
      \begin{aligned}
      f'(x) &= (d/dx)(2x^3) - (d/dx)(15x^2) + (d/dx)(24x)\
      &= 3\cdot 2x^{3-1} - 2\cdot 15x^{2-1} + 24\
      &= 6x^2 - 30x + 24.
      \end{aligned}
  • Set derivative equal to target slope
    6x230x+24=6.6x^2 - 30x + 24 = 6.

    • Subtract 66 from both sides: 6x230x+18=0.6x^2 - 30x + 18 = 0.
    • Factor out common 66: 6(x25x+3)=0x25x+3=0.6(x^2 - 5x + 3)=0 \Rightarrow x^2 - 5x + 3 = 0.
  • Quadratic Formula

    • General form ax2+bx+c=0x=b±b24ac2a.ax^2+bx+c=0 \Rightarrow x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}.
    • Here a=1,  b=5,  c=3a=1,\;b=-5,\;c=3:
      x = \frac{-(-5) \pm \sqrt{(-5)^2-4(1)(3)}}{2(1)}
      = \frac{5 \pm \sqrt{25-12}}{2}
      = \frac{5 \pm \sqrt{13}}{2}.
    • Decimal approximations (useful for graphing or intuition):
      51320.697,5+1324.303.\frac{5-\sqrt{13}}{2} \approx 0.697, \qquad \frac{5+\sqrt{13}}{2} \approx 4.303.
  • Interpretation/Significance

    • At x0.697x \approx 0.697 and x4.303x \approx 4.303, the curve is rising with slope 66.
    • In applications this process finds where a system meets a given rate-of-change requirement (e.g.
      economics: marginal cost = given value, physics: velocity equals specific value, etc.).

Higher-Order Derivatives (Concept & Notation)

  • Idea

    • "Higher order" means differentiating repeatedly.
    • Notation:
    • First derivative: f(x)f'(x) or dydx.\frac{dy}{dx}.
    • Second derivative: f(x)=d2ydx2f''(x) = \frac{d^2y}{dx^2} — captures concavity/acceleration.
    • Third derivative: f(x)=d3ydx3f'''(x) = \frac{d^3y}{dx^3} — appears in physics as "jerk" (rate of change of acceleration).
    • nn-th derivative: f(n)(x).f^{(n)}(x).
  • Why care?

    • Analyze curvature, inflection points, motion (position→velocity→acceleration→jerk→…).
    • Used in Taylor series, differential equations, control systems.

Example 1 — Polynomial (Find the third derivative)

  • Reconstructed original function (deduced from derivative steps in the video):
    g(x)=3x35x+12.g(x)=3x^3-5x+12.

  • Step-by-step differentiation

    • First derivative
      g(x)=9x25g'(x)=9x^2-5 (since (d/dx)(3x3)=9x2,(d/dx)(3x^3)=9x^2, (d/dx)(5x)=5,(d/dx)(-5x)=-5, constant 1212 vanishes).
    • Second derivative
      g(x)=(d/dx)(9x25)=18x.g''(x)=(d/dx)(9x^2-5)=18x.
    • Third derivative
      g(x)=(d/dx)(18x)=18.g'''(x)=(d/dx)(18x)=18.
  • Observations

    • The third derivative of a cubic polynomial is always constant.
    • Physical analogy: a constant jerk implies linearly increasing acceleration.

Example 2 — Mixed Linear & Exponential (Find up to third derivative)

  • Function
    y(t)=3t+2et.y(t)=3t+2e^{t}.

  • Notation reminder

    • Using Leibniz form dydt,  d2ydt2,  d3ydt3.\frac{dy}{dt},\;\frac{d^2y}{dt^2},\;\frac{d^3y}{dt^3}.
  • Derivatives

    • First derivative
      dydt=3+2et,\frac{dy}{dt}=3+2e^{t}, because (d/dt)(3t)=3(d/dt)(3t)=3 and (d/dt)(et)=et.(d/dt)(e^{t})=e^{t}.
    • Second derivative
      d2ydt2=2et\frac{d^2y}{dt^2}=2e^{t} (derivative of constant 33 is 00; 2et2e^{t} remains unchanged).
    • Third derivative
      d3ydt3=2et\frac{d^3y}{dt^3}=2e^{t} (exponential function differentiates to itself indefinitely).
  • Pattern insight

    • For aetae^{t}, every derivative is still aetae^{t}.
    • Linear terms vanish after first differentiation.

Quick Reference: Rules Employed & Their Significance

  • Power Rule: ddx(xn)=nxn1\frac{d}{dx}(x^{n}) = nx^{n-1} (foundation for differentiating polynomials).
  • Constant Multiple Rule: ddx[cf(x)]=cf(x).\frac{d}{dx}[c\,f(x)] = c\,f'(x).
  • Sum/Difference Rule: differentiate term-by-term.
  • Derivative of exe^{x}: remains exe^{x} — makes exponentials pivotal in modeling continuous growth/decay.
  • Quadratic Formula: universal method when factoring fails; discriminant b24acb^2-4ac signals number/nature of roots.

Broader Connections & Practical Takeaways

  • Tangent‐slope problems illustrate how derivatives convert geometric questions into algebraic ones.
  • Higher-order derivatives underpin concavity testing, optimization (via second derivative test), and series approximations (Taylor/Maclaurin).
  • Exponentials’ self-replicating derivatives explain their ubiquity in differential equations (e.g.
    Newton’s law of cooling, population models).
  • Recognizing when to switch from factoring to quadratic formula saves time and prevents errors.

Suggested Practice / Next Steps

  • Re-work the tangent‐slope example with a different target slope (e.g.
    m=10m=10) to reinforce the derivative-to-algebra transition.
  • Compute 4th and 5th derivatives of the exponential example to verify the pattern.
  • Sketch f(x)=2x315x2+24xf(x)=2x^3-15x^2+24x and mark the points where the slope equals 66 to internalize the geometric meaning.
  • Read ahead on the second derivative test for local maxima/minima to see higher-order derivatives in action.