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.