3.4.3 The Product and Quotient Rule

Problem Setup

  • Objective: Differentiate a rational function that simultaneously requires the Product Rule for its numerator and the Quotient Rule for the overall fraction.

  • Implicitly-defined function (from context):
    f(x)=4x exx2+1f(x)=\frac{4x\,e^x}{x^2+1}
    • Numerator: N(x)=4x exN(x)=4x\,e^x
    • Denominator: D(x)=x2+1D(x)=x^2+1

Step 1 – Prepare for the Quotient Rule

  • Reminder of the Quotient Rule: For f(x)=N(x)D(x)f(x)=\dfrac{N(x)}{D(x)},
    f′(x)=N′(x) D(x)−N(x) D′(x)(D(x))2f'(x)=\frac{N'(x)\,D(x)-N(x)\,D'(x)}{\bigl(D(x)\bigr)^2}

  • First requirement: compute N′(x)N'(x).

Step 2 – Differentiate the Numerator with the Product Rule

  • Product Rule statement: If N(x)=u(x) v(x)N(x)=u(x)\,v(x), then N′(x)=u′(x) v(x)+u(x) v′(x)N'(x)=u'(x)\,v(x)+u(x)\,v'(x).

  • Identify parts:
    • u(x)=4x  ⇒  u′(x)=4u(x)=4x\;\Rightarrow\;u'(x)=4
    • v(x)=ex  ⇒  v′(x)=exv(x)=e^x\;\Rightarrow\;v'(x)=e^x

  • Apply the rule:
    \begin{aligned}
    N'(x)&=4\cdot e^x+4x\cdot e^x\
    &=4e^x+4x e^x
    \end{aligned}

  • Decision point: postpone simplification until later to keep algebra transparent.

Step 3 – Apply the Quotient Rule

Insert the computed pieces into the quotient-rule template:
<br>f′(x)=(4ex+4xex)(x2+1)−(4xex)⋅(2x)(x2+1)2<br><br>\boxed{f'(x)=\frac{\bigl(4e^x+4x e^x\bigr)(x^2+1)-\bigl(4x e^x\bigr)\cdot(2x)}{\bigl(x^2+1\bigr)^2}}<br>

Step 4 – Expand & Collect Like Terms (Numerator Only)

  1. Distribute 4ex+4xex4e^x+4x e^x across (x2+1)(x^2+1):
    • 4ex(x2)=4x2ex4e^x(x^2)=4x^2 e^x
    • 4ex(1)=4ex4e^x(1)=4e^x
    • 4xex(x2)=4x3ex4x e^x(x^2)=4x^3 e^x
    • 4xex(1)=4xex4x e^x(1)=4x e^x

  2. Distribute the second product: − (4xex)(2x)=−8x2ex-\,(4x e^x)(2x)= -8x^2 e^x

  3. Aggregate all five terms:
    4x3ex+(4x2ex−8x2ex)+4xex+4ex4x^3 e^x+(4x^2 e^x-8x^2 e^x)+4x e^x+4e^x
    • Combine 4x2ex−8x2ex=−4x2ex4x^2 e^x-8x^2 e^x=-4x^2 e^x

  4. Resulting unsimplified numerator:
    4x3ex−4x2ex+4xex+4ex4x^3 e^x-4x^2 e^x+4x e^x+4e^x

Step 5 – Factor for Elegance & Insight

  • Notice every term shares a common factor of 4ex4e^x.

  • Factor it out:
    <br>amp;4ex(x3−x2+x+1)<br>\begin{aligned}<br>&amp;4e^x\bigl(x^3- x^2+ x+1\bigr)<br>\end{aligned}

  • Final compact derivative:
    f′(x)=4ex(x3−x2+x+1)(x2+1)2\boxed{f'(x)=\frac{4e^x\left(x^3-x^2+x+1\right)}{\left(x^2+1\right)^2}}

Key Takeaways & Connections

  • Layered Rules: Complicated expressions often decompose into nested applications of basic rules. Preparing sub-results (here, N′(x)N'(x)) off to the side streamlines the process.

  • Factor Recognition: After heavy algebra, common factors (here 4ex4e^x) frequently emerge; factoring them:
    • Clarifies structure
    • Simplifies future work (e.g.
    • analyzing critical points
    • canceling terms if part of a larger expression).

  • Exponentials & Polynomials: The derivative preserves exe^x in every term (due to ddxex=ex\tfrac{d}{dx}e^x=e^x). Such echoes help when checking work—if an exe^x disappears entirely, an algebraic slip likely occurred.

  • Efficiency Tip: Decide strategically when to simplify—too early may obscure pattern recognition, too late can explode algebra.

  • Real-world relevance: Quotients of a polynomial and an exponential model growth processes constrained by saturation (denominator). Accurate derivatives gauge instantaneous change, crucial in control theory, pharmacokinetics, or any feedback-limited growth.

Ethical / Philosophical Reflection

  • While derivative rules look mechanical, they represent a deep philosophical idea: we can understand complex change by decomposing it into simple, local interactions (products & ratios). This mirrors problem-solving beyond math—break big issues into digestible parts.

Final Formula Recap

f(x)=4xexx2+1,f′(x)=4ex(x3−x2+x+1)(x2+1)2f(x)=\frac{4x e^x}{x^2+1},\qquad f'(x)=\frac{4e^x\left(x^3-x^2+x+1\right)}{\bigl(x^2+1\bigr)^2}