Session 07 - Differentiation Rules: Basics, Powers, Products and Quotients

Apply the Power Rule, Constant Multiple Rule, and Constant Rule term-by-term:

1. Pre-Process and differentiate $f(x)$:
   • Convert radicals and denominators to power form:
     $$f(x) = 4x^{-\frac{1}{2}} - 3x^{\frac{2}{3}} + 2x^{-3}$$
   • Apply the Power Rule term-by-term:
     $$f'(x) = 4\left(-\frac{1}{2}x^{-\frac{3}{2}}\right) - 3\left(\frac{2}{3}x^{-\frac{1}{3}}\right) + 2\left(-3x^{-4}\right) = -2x^{-\frac{3}{2}} - 2x^{-\frac{1}{3}} - 6x^{-4}$$
   • Post-process into standard radical/fraction notation:
     $$f'(x) = -\frac{2}{\sqrt{x^3}} - \frac{2}{\sqrt[3]{x}} - \frac{6}{x^4}$$
2. Post-Process $g'(x)$ by factoring out the lowest power: The powers are $-\frac{1}{3}$ and $\frac{2}{3}$. The smallest power is $-\frac{1}{3}$, and the common coefficient is $2$:
   • Factoring out $2x^{-\frac{1}{3}}$ subtracts $-\frac{1}{3}$ from the second power: $\frac{2}{3} - \left(-\frac{1}{3}\right) = \frac{3}{3} = 1$.
     $$g'(x) = 2x^{-\frac{1}{3}}\left(1 + 2x^1\right) = \frac{2(1 + 2x)}{\sqrt[3]{x}}$$

• (a) The derivative measures instantaneous slope. The constant function $f(x) = c$ produces a horizontal line with zero vertical change for any horizontal step ($\Delta y / \Delta x = 0$), so $\frac{d}{dx}[c] = 0$. The linear function $f(x) = cx$ is a line through the origin with constant steepness $c$, so $\frac{d}{dx}[cx] = c$.
• (b) The Quotient Rule is designed for fractions where both the numerator and denominator contain variables. When the denominator is a constant scalar, rewrite $f(x) = \frac{1}{3}(x^4 - 6x^2 + 5)$. By the Constant Multiple Rule, $f'(x) = \frac{1}{3}(4x^3 - 12x) = \frac{4}{3}x^3 - 4x$, completing the derivative in one step.
• (c) In Beam Deflection and Shearing Force (Civil Engineering), the bending moment equation contains $\frac{16}{\sqrt{x+4}}$. Converting this term to $16(x+4)^{-1/2}$ transforms a fractional radical into power form, allowing the engineer to directly apply the Power Rule to compute shear force $V(x) = \frac{dM}{dx} = 48 + 8(x+4)^{-3/2} - 4x$ and evaluate load values at the bridge abutments.

Apply the Power Rule and Constant Multiple Rule term-by-term:

1. Factor out the constant denominator $\frac{1}{2}$:
   $$g(t) = \frac{1}{2}(t^4 - 6t^2 + 8)$$
2. Differentiate using the Constant Multiple and Power Rules:
   $$g'(t) = \frac{1}{2}(4t^3 - 12t + 0) = 2t^3 - 6t$$

1. Pre-process into rational and negative exponents:
   $$h(x) = 3x^{\frac{1}{2}} + 5x^{-2}$$
2. Differentiate term-by-term:
   $$h'(x) = 3\left(\frac{1}{2}x^{-\frac{1}{2}}\right) + 5\left(-2x^{-3}\right) = \frac{3}{2}x^{-\frac{1}{2}} - 10x^{-3}$$
3. Format in standard radical and fraction notation:
   $$h'(x) = \frac{3}{2\sqrt{x}} - \frac{10}{x^3}$$

1. Pre-Process and Rewrite using rational exponents: Notice that $\frac{3}{\sqrt[3]{t}} = 3t^{-\frac{1}{3}}$. Substitute into $f(t)$:
   $$f(t) = (7 + t^7)(3t^{-\frac{1}{3}} + 3t)$$
2. Distribute (FOIL) before differentiating:
   $$f(t) = 7(3t^{-\frac{1}{3}}) + 7(3t) + t^7(3t^{-\frac{1}{3}}) + t^7(3t)$$
   Simplify the exponents ($7 - \frac{1}{3} = \frac{21}{3} - \frac{1}{3} = \frac{20}{3}$ and $7 + 1 = 8$):
   $$f(t) = 21t^{-\frac{1}{3}} + 21t + 3t^{\frac{20}{3}} + 3t^8$$
3. Differentiate term-by-term using the Power Rule:
   $$f'(t) = 21\left(-\frac{1}{3}t^{-\frac{4}{3}}\right) + 21(1) + 3\left(\frac{20}{3}t^{\frac{17}{3}}\right) + 3\left(8t^7\right)$$
   $$f'(t) = -7t^{-\frac{4}{3}} + 21 + 20t^{\frac{17}{3}} + 24t^7$$
4. Format final answer:
   $$f'(t) = -\frac{7}{\sqrt[3]{t^4}} + 21 + 20\sqrt[3]{t^{17}} + 24t^7$$

1. Pre-Process to clear the compound trigonometric fraction: Recall that $\csc \theta = \frac{1}{\sin \theta}$. Substitute this into $r(\theta)$:
   $$r(\theta) = \frac{\sin \theta}{1 + \frac{1}{\sin \theta}} = \frac{\sin \theta \cdot \sin \theta}{\left(1 + \frac{1}{\sin \theta}\right)\sin \theta} = \frac{\sin^2 \theta}{\sin \theta + 1}$$
2. Build Component Dictionary for Quotient Rule:
   • $u = \sin^2 \theta = (\sin \theta)(\sin \theta) \implies u' = (\cos \theta)(\sin \theta) + (\sin \theta)(\cos \theta) = 2\sin \theta \cos \theta$
   • $v = \sin \theta + 1 \implies v' = \cos \theta$
3. Apply the Quotient Rule $\frac{u'v - uv'}{v^2}$:
   $$r'(\theta) = \frac{(2\sin \theta \cos \theta)(\sin \theta + 1) - (\sin^2 \theta)(\cos \theta)}{(\sin \theta + 1)^2}$$
4. Expand numerator and combine like terms:
   $$r'(\theta) = \frac{2\sin^2 \theta \cos \theta + 2\sin \theta \cos \theta - \sin^2 \theta \cos \theta}{(\sin \theta + 1)^2} = \frac{\sin^2 \theta \cos \theta + 2\sin \theta \cos \theta}{(\sin \theta + 1)^2}$$
5. Factor out the common term $\sin \theta \cos \theta$:
   $$r'(\theta) = \frac{\sin \theta \cos \theta (\sin \theta + 2)}{(\sin \theta + 1)^2}$$

1. Build Component Dictionary:
   • High ($u$): $x^2 - 1 \implies u' = 2x$
   • Low ($v$): $x^3 + x \implies v' = 3x^2 + 1$
2. Assemble using "Low dHigh minus High dLow over Low Low":
   $$y' = \frac{(x^3 + x)(2x) - (x^2 - 1)(3x^2 + 1)}{(x^3 + x)^2}$$
3. Distribute numerator terms with protective brackets:
   $$y' = \frac{(2x^4 + 2x^2) - [3x^4 + x^2 - 3x^2 - 1]}{(x^3 + x)^2} = \frac{2x^4 + 2x^2 - [3x^4 - 2x^2 - 1]}{(x^3 + x)^2}$$
4. Distribute negative sign and combine like terms:
   $$y' = \frac{2x^4 + 2x^2 - 3x^4 + 2x^2 + 1}{(x^3 + x)^2} = \frac{-x^4 + 4x^2 + 1}{(x^3 + x)^2}$$

1. Build Component Dictionary:
   • $u = t^3 \implies u' = 3t^2$
   • $v = \cos(t) \implies v' = -\sin(t)$
2. Assemble into $u'v + uv'$:
   $$h'(t) = (3t^2)(\cos t) + (t^3)(-\sin t) = 3t^2 \cos t - t^3 \sin t = t^2(3\cos t - t\sin t)$$

• (a) As outlined in the Conceptual Framework, the master rule is chosen by asking what the final arithmetic operation would be when evaluating $f(x)$ for a scalar. In $f(x) = \frac{x^2 \sin x}{x+1}$, one would compute $x^2 \sin x$ in the top, $x+1$ in the bottom, and perform division last. Thus, the Quotient Rule is the primary structural frame, and the Product Rule is executed as a sub-routine when finding $u'$.
• (b) In the Quotient Rule formula $\frac{u'v - uv'}{v^2}$, the subtraction operator applies to the entire product $uv'$. In Worked Example A, $uv' = (x^2 - 1)(3x^2 + 1) = 3x^4 - 2x^2 - 1$. Without protective brackets, students typically only negate the first term ($-3x^4 - 2x^2 - 1$), causing catastrophic sign errors.
• (c) In RLC Circuit Dynamic Filter Transfer (Electrical Engineering), the frequency gain function $H(\omega) = \frac{10\omega}{\omega^2 + 25}$ is differentiated via the Quotient Rule to find $\frac{dH}{d\omega} = \frac{10(25 - \omega^2)}{(\omega^2 + 25)^2}$. Setting $\frac{dH}{d\omega} = 0$ yields the resonant peak $\omega = 5\text{ krad/s}$, which allows engineers to tune bandpass filters for maximum telemetry gain.

• Part (a): Algebraic $\times$ Trigonometric
  1. Build Component Dictionary:
     • $u = x^2 \implies u' = 2x$
     • $v = \sin(x) \implies v' = \cos(x)$
  2. Apply Product Rule ($u'v + uv'$):
     $$f'(x) = (2x)(\sin x) + (x^2)(\cos x) = 2x\sin(x) + x^2\cos(x)$$
  3. Factor out common factor $x$:
     $$f'(x) = x(2\sin x + x\cos x)$$
• Part (b): Algebraic $\times$ Exponential
  1. Build Component Dictionary:
     • $u = x^2 \implies u' = 2x$
     • $v = e^x \implies v' = e^x$
  2. Apply Product Rule ($u'v + uv'$):
     $$g'(x) = (2x)(e^x) + (x^2)(e^x)$$
  3. Factor out common factor $x e^x$:
     $$g'(x) = x e^x (2 + x) = x(x + 2)e^x$$

1. Build Component Dictionary:
   • High ($u$): $x^2 - 4 \implies u' = 2x$
   • Low ($v$): $x^2 + 1 \implies v' = 2x$
2. Apply Quotient Rule $\frac{u'v - uv'}{v^2}$:
   $$f'(x) = \frac{(2x)(x^2 + 1) - (x^2 - 4)(2x)}{(x^2 + 1)^2}$$
3. Expand the numerator with protective brackets:
   $$f'(x) = \frac{(2x^3 + 2x) - [2x^3 - 8x]}{(x^2 + 1)^2}$$
4. Distribute the negative sign and combine like terms:
   $$f'(x) = \frac{2x^3 + 2x - 2x^3 + 8x}{(x^2 + 1)^2} = \frac{10x}{(x^2 + 1)^2}$$

1. Build Component Dictionary:
   • High ($u$): $e^x \implies u' = e^x$
   • Low ($v$): $x^2 + 1 \implies v' = 2x$
2. Apply Quotient Rule $\frac{u'v - uv'}{v^2}$:
   $$h'(x) = \frac{(e^x)(x^2 + 1) - (e^x)(2x)}{(x^2 + 1)^2}$$
3. Factor out $e^x$ in the numerator:
   $$h'(x) = \frac{e^x(x^2 + 1 - 2x)}{(x^2 + 1)^2} = \frac{e^x(x^2 - 2x + 1)}{(x^2 + 1)^2}$$
4. Factor the quadratic into a perfect square:
   $$h'(x) = \frac{e^x(x - 1)^2}{(x^2 + 1)^2}$$

1. Rewrite $\cot(x)$ in terms of sine and cosine:
   $$\cot(x) = \frac{\cos(x)}{\sin(x)}$$
2. Build Component Dictionary for Quotient Rule:
   • Numerator: $u = \cos(x) \implies u' = -\sin(x)$
   • Denominator: $v = \sin(x) \implies v' = \cos(x)$
3. Apply the Quotient Rule $\frac{u'v - uv'}{v^2}$:
   $$\frac{d}{dx}[\cot(x)] = \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{(\sin x)^2} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}$$
4. Factor out $-1$ and apply the Pythagorean Identity ($\sin^2 x + \cos^2 x = 1$):
   $$\frac{d}{dx}[\cot(x)] = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = \frac{-(1)}{\sin^2 x} = -\frac{1}{\sin^2 x}$$
5. Use the reciprocal identity ($\csc x = \frac{1}{\sin x}$):
   $$\frac{d}{dx}[\cot(x)] = -\csc^2(x)$$

1. Establish the 4-step cycle for $f(x) = \sin(x)$:
   • $f^{(0)}(x) = \sin x$
   • $f^{(1)}(x) = \cos x$
   • $f^{(2)}(x) = -\sin x$
   • $f^{(3)}(x) = -\cos x$
2. Compute $86 \bmod 4$:
   $$86 \div 4 = 21 \text{ with a remainder of } 2 \implies 86 \equiv 2 \pmod 4$$
3. Determine the derivative:
   $$f^{(86)}(x) = f^{(2)}(x) = -\sin(x)$$

1. Find point of tangency:
   $$y_1 = f(2)g(2) = (3)(-4) = -12 \implies (2, -12)$$
2. Differentiate structurally via Product Rule:
   $$y'(x) = f'(x)g(x) + f(x)g'(x)$$
3. Evaluate slope at $x = 2$:
   $$m = y'(2) = f'(2)g(2) + f(2)g'(2) = (5)(-4) + (3)(2) = -20 + 6 = -14$$
4. Construct tangent line equation in point-slope form:
   $$y - (-12) = -14(x - 2) \implies y + 12 = -14x + 28 \implies y = -14x + 16$$

• (a) Under the Power Rule $\frac{d}{dx}[x^n] = n x^{n-1}$, each successive derivative reduces the degree of each monomial by 1. For a degree $n$ polynomial, taking $n$ derivatives reduces all terms to constants ($n! a_n$), and taking the $(n+1)^{\text{th}}$ derivative differentiates constants to $0$. In contrast, $\sin(x)$ and $\cos(x)$ alternate functions and signs in a 4-step periodic loop ($\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x$) and never extinguish.
• (b) Because the sequence of derivatives repeats with period 4, completing 4 derivatives returns you to the original function ($f^{(4k)}(x) = f(x)$). Dividing $86$ by $4$ yields $21$ full cycles plus a remainder of $2$ ($86 = 4 \times 21 + 2$). Therefore, $f^{(86)}(x) = f^{(2)}(x) = -\sin(x)$.
• (c) In Automated High-Speed Rail Jerk Analysis (Robotics & Computer Engineering), calculating velocity ($s'(t)$) and acceleration ($s''(t)$) only confirms that the vehicle speeds up smoothly. The third derivative $j(t) = s'''(t)$ measures the rate of change of acceleration; if $|j(t)| > 1.5\text{ m/s}^3$, passengers experience abrupt forces that cause nausea, while mechanical guideway actuators experience destructive dynamic shocks.

1. Pre-Process the function:
   $$f(t) = 4t^3 - 3t^{-5}$$
2. Compute First Derivative $f'(t)$:
   $$f'(t) = 4(3t^2) - 3(-5t^{-6}) = 12t^2 + 15t^{-6}$$
3. Compute Second Derivative $f''(t)$:
   $$f''(t) = 12(2t^1) + 15(-6t^{-7}) = 24t - 90t^{-7}$$
4. Compute Third Derivative $f'''(t)$:
   $$f'''(t) = 24(1) - 90(-7t^{-8}) = 24 + 630t^{-8}$$
5. Post-Process into fraction notation:
   $$f'''(t) = 24 + \frac{630}{t^8}$$

1. Establish the 4-step cyclic derivative pattern for $f(x) = \cos x$:
   • $f^{(0)}(x) = \cos x$ (remainder 0)
   • $f^{(1)}(x) = -\sin x$ (remainder 1)
   • $f^{(2)}(x) = -\cos x$ (remainder 2)
   • $f^{(3)}(x) = \sin x$ (remainder 3)
   • $f^{(4)}(x) = \cos x$ (remainder 0)
2. Find the $42^{\text{nd}}$ derivative ($f^{(42)}(x)$):
   $$42 \div 4 = 10 \text{ R } 2 \implies 42 \equiv 2 \pmod 4$$
   $$f^{(42)}(x) = f^{(2)}(x) = -\cos x$$
3. Find the $55^{\text{th}}$ derivative ($f^{(55)}(x)$):
   $$55 \div 4 = 13 \text{ R } 3 \implies 55 \equiv 3 \pmod 4$$
   $$f^{(55)}(x) = f^{(3)}(x) = \sin x$$

• Part (a): Constant Multiple Rule
  1. Differentiate structurally:
     $$\frac{d}{dx}[-2f(x)] = -2f'(x)$$
  2. Evaluate at $x = 5$ using the table ($f'(5) = -2$):
     $$-2f'(5) = -2(-2) = 4$$
• Part (b): Product Rule
  1. Differentiate $x^2 f(x)$ structurally via Product Rule:
     $$\frac{d}{dx}[x^2 f(x)] = \frac{d}{dx}[x^2]f(x) + x^2 f'(x) = 2x f(x) + x^2 f'(x)$$
  2. Evaluate at $x = 3$ using table values ($f(3) = 3$ and $f'(3) = 7$):
     $$\left.\frac{d}{dx}[x^2 f(x)]\right|_{x=3} = 2(3)f(3) + (3)^2 f'(3) = 6(3) + 9(7) = 18 + 63 = 81$$
• Part (c): Tangent Line via Quotient Rule
  1. Find point of tangency $(4, y(4))$:
     $$y(4) = \frac{g(4)}{f(4)} = \frac{-1}{2} = -\frac{1}{2} \implies \left(4, -\frac{1}{2}\right)$$
  2. Differentiate $y = \frac{g(x)}{f(x)}$ structurally via Quotient Rule:
     $$y'(x) = \frac{g'(x)f(x) - g(x)f'(x)}{[f(x)]^2}$$
  3. Evaluate slope at $x = 4$ using table values ($g'(4) = 1, f(4) = 2, g(4) = -1, f'(4) = -1$):
     $$m = y'(4) = \frac{(1)(2) - (-1)(-1)}{(2)^2} = \frac{2 - 1}{4} = \frac{1}{4}$$
  4. Construct tangent line equation in point-slope form:
     $$y - \left(-\frac{1}{2}\right) = \frac{1}{4}(x - 4)$$
     $$y + \frac{1}{2} = \frac{1}{4}x - 1 \implies y = \frac{1}{4}x - \frac{3}{2}$$