3.4.1 The Product and Quotient Rule

Recap – Sum & Difference Rule

• Previously established: for any differentiable f(x) and g(x)
  • \frac{d}{dx}[f(x)+g(x)] = f'(x) + g'(x)
  • \frac{d}{dx}[f(x)-g(x)] = f'(x) - g'(x)
• Temptation to extend this linear behavior to products/quotients is incorrect; new rules are required.

Product Rule

• Conditions: f(x) and g(x) must be differentiable at the point of interest.
• Rule ("first ∙ derivative of second + second ∙ derivative of first"):
  \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)
• Symmetry: either factor may be differentiated first; both orders appear in the final sum.
• Proof available in the textbook; skipped in lecture.

Product-Rule Example 1

Target expression: (v^2)\bigl(2\sqrt v + 1\bigr) (second factor rewritten as 2v^{1/2}+1).
Step-by-step:

• Differentiate the first factor: \frac{d}{dv}[v^2] = 2v.
• Multiply by the untouched second factor: 2v\,\bigl(2v^{1/2}+1\bigr) = 4v^{3/2}+2v.
• Add the untouched first factor times the derivative of the second:
  • \frac{d}{dv}[2v^{1/2}+1] = 2\cdot\frac12 v^{-1/2} + 0 = v^{-1/2}
  • Product: v^2\,v^{-1/2}=v^{3/2}.
• Combine like terms:
  4v^{3/2}+v^{3/2}+2v = 5v^{3/2}+2v.
• Final simplified derivative: 5v^{3/2}+2v.

Product-Rule Example 2

Target expression: x^2 e^x.

• Derivative: 2x e^x + x^2 e^x.
• Optional factorization (textbook’s preference):
  x e^x\,(2+x).
  Either form is accepted unless a specific style is requested.

Quotient Rule

• Conditions: f(x), g(x) differentiable and g(x)\neq0 at the point.
• Rule ("low ∙ d-high – high ∙ d-low over low²"):
  \frac{d}{dx}\Bigl[\frac{f(x)}{g(x)}\Bigr] = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^{2}}
• Key contrasts with product rule:
  • Numerator involves subtraction, not addition.
  • Entire expression divided by the square of the denominator.
• Proof again omitted but available in the text.

Quotient-Rule Example 1

Target expression: \dfrac{x^2+3x+4}{x^2-1}.

• Identify f(x)=x^2+3x+4, g(x)=x^2-1.
• Compute derivatives: f'(x)=2x+3, g'(x)=2x.
• Apply the rule:
  \frac{d}{dx}\Bigl[\frac{f}{g}\Bigr]=\frac{(x^2-1)(2x+3)-(x^2+3x+4)(2x)}{(x^2-1)^2}.
• Expand & simplify numerator:
  • First product: 2x(x^2-1)+3(x^2-1)=2x^3-2x+3x^2-3.
  • Second product (to be subtracted): 2x(x^2+3x+4)=2x^3+6x^2+8x.
  • Combine:
    [2x^3-2x+3x^2-3] - [2x^3+6x^2+8x] = -3x^2-10x-3.
• Factor out the negative sign for aesthetics:
  \frac{-(3x^2+10x+3)}{(x^2-1)^2}.
• Final derivative:
  -\dfrac{3x^2+10x+3}{(x^2-1)^2}.

Quotient-Rule Example 2

Target: e^{-x}, rewritten as \dfrac{1}{e^x} to highlight quotient form.

• Treat as f(x)=1, g(x)=e^x.
• Derivatives: f'(x)=0, g'(x)=e^x.
• Apply rule:
  \frac{g\,f' - f\,g'}{g^2} = \frac{e^x \cdot 0 - 1\cdot e^x}{(e^x)^2} = -\frac{e^x}{e^{2x}} = -e^{-x}.
• Therefore \frac{d}{dx}[e^{-x}] = -e^{-x} (an identity worth memorizing).

Key Takeaways & Connections

• Linear rules (sum/difference) do NOT extend to products & quotients.
• Product rule uses addition; quotient rule uses subtraction and a squared denominator.
• Mnemonics:
  • Product: "(d of 1)·2 + 1·(d of 2)".
  • Quotient: "low d-high minus high d-low over low-squared".
• Familiar derivatives reused throughout:
  • Power rule \frac{d}{dx}[x^n]=nx^{n-1}.
  • \frac{d}{dx}[e^x]=e^x, giving \frac{d}{dx}[e^{-x}]=-e^{-x} via chain or quotient rule.
• Algebraic simplification (factoring, combining like terms) is expected in final answers unless instructions state otherwise.
• Domain considerations: quotient rule derivatives undefined where g(x)=0 (division by zero).
• Proofs connect to limit definitions of the derivative; consult textbook if formal justification is required.