3.2 continued Derivative Rules, Related Rates & Cost Analysis – Detailed Study Notes

3.2 – Warm-Up Algebra

• Starting quadratic (common-factor extraction)
– 0 = 3x^{2}+6x
– Factor: 3x(x+2)=0 ⇒ roots at x=0 and x=-2.

Product Rule – Symbolic & Numerical Example

• Rule (for h(x)=f(x)\,g(x)):
– h'(x)=f'(x)\,g(x)+f(x)\,g'(x).
• Given data at x=5
– f(5)=3,\;f'(5)=-2,\;g(5)=-4,\;g'(5)=7.
• Derivative at 5
– h'(5)=f'(5)\,g(5)+f(5)\,g'(5)=(-2)(-4)+3\,(7)=8+21=29.
– (Transcript shows “-29” – sign error; correct value is 29.)

Product Rule – Second Data Set (possible typo in transcript)

• Data at x=2
– f(2)=0,\;g(2)=-3,\;g'(2)=-5.
• If h(x)=f(x)\,g(x) then h'(2)=f'(2)\,g(2)+f(2)\,g'(2).
– Only g'(2) is provided; without f'(2) result cannot be completed (transcript’s “L'(2)=-5” seems to refer to another context, perhaps a quotient).
– Flag as incomplete in original source.

Related-Rates Word Problem – iPhone Price, Demand & Revenue

• Variables
– t = time (months).
– Price p(t) in dollars/phone; demand d(t) in millions of phones/month.
• Current values (at t=0)
– p(0)=380\;\text{dollars}, d(0)=4\;\text{million phones}.
– Rates: \dfrac{dp}{dt}=5\;\dfrac{\text{dollars}}{\text{month}},\qquad\dfrac{dd}{dt}=0.5\;\dfrac{\text{million phones}}{\text{month}} (500 000=0.5 million).
• Linear models implied in transcript
– p(t)=5t+380.
– d(t)=0.5t+4.
• Revenue (millions of dollars)
– R(t)=d(t)\,p(t)=\bigl(0.5t+4\bigr)\bigl(5t+380\bigr).
• Rate of change of revenue
– Product-rule derivative
\frac{dR}{dt}=0.5\bigl(5t+380\bigr)+5\bigl(0.5t+4\bigr)
=2.5t+190+2.5t+20=5t+210.
– Present rate (at t=0): \dfrac{dR}{dt}(0)=210 million dollars / month.
• Significance
– Even while price is rising, advertising causes demand to rise fast enough that revenue is currently increasing by about $210 M per month.

Quotient Rule – General Form & Mnemonic

• For h(x)=\dfrac{f(x)}{g(x)} (low = denominator g, high = numerator f)
– h'(x)=\dfrac{\text{Low}\,(\text{d High})-\text{High}\,(\text{d Low})}{(\text{Low})^{2}}.
• Classroom jingle: “Low d-High minus High d-Low, over the low squared, there you go.”

Worked Example 1 – \dfrac{5x^{2}}{3x+5}

• Identify
– High =5x^{2},\;\dfrac{d}{dx}(\text{High})=10x.
– Low =3x+5,\;\dfrac{d}{dx}(\text{Low})=3.
• Derivative
\frac{d}{dx}\Bigl(\frac{5x^{2}}{3x+5}\Bigr)=\frac{(3x+5)(10x)-5x^{2}(3)}{(3x+5)^{2}}
=\frac{30x^{2}+50x-15x^{2}}{(3x+5)^{2}}=\frac{15x^{2}+50x}{(3x+5)^{2}}.

Worked Example 2 – \dfrac{4x}{x^{2}+5} and its Tangent Line at x=1

• High/Low decomposition
– High =4x,\;\dfrac{d}{dx}(\text{High})=4.
– Low =x^{2}+5,\;\dfrac{d}{dx}(\text{Low})=2x.
• Derivative
f'(x)=\frac{(x^{2}+5)(4)-4x(2x)}{(x^{2}+5)^{2}}=\frac{4x^{2}+20-8x^{2}}{(x^{2}+5)^{2}}=\frac{-4x^{2}+20}{(x^{2}+5)^{2}}.
• Evaluate at x=1
– Slope: f'(1)=\dfrac{-4(1)^{2}+20}{(1^{2}+5)^{2}}=\dfrac{16}{36}=\dfrac{4}{9}.
– Point on curve: f(1)=\dfrac{4(1)}{1^{2}+5}=\dfrac{4}{6}=\dfrac{2}{3}.
• Tangent line (point–slope form)
y-\frac{2}{3}=\frac{4}{9}\bigl(x-1\bigr).
– Can be rearranged to y=\frac{4}{9}x-\frac{2}{9}.

Quotient Rule – Numerical Substitution Exercises

Exercise A

• Data at x=2: f(2)=9,\;f'(2)=6,\;g(2)=4,\;g'(2)=-3.
• h(x)=\dfrac{f(x)}{g(x)} ⇒
h'(2)=\frac{6\cdot4-9\cdot(-3)}{4^{2}}=\frac{24+27}{16}=\frac{51}{16}.

Exercise B

• Data at x=3: f(3)=1,\;f'(3)=0,\;g(3)=-3,\;g'(3)=-5.
• h'(3)=\frac{0\cdot(-3)-1\cdot(-5)}{(-3)^{2}}=\frac{5}{9}.

Cost, Average Cost & Marginal Average Cost

Model

• Total cost (example pulled from transcript context)
– High (numerator) =3x^{2}+120 dollars.
– Low (denominator) =2x^{2}+x units produced.
– Hence average cost
\overline{C}(x)=\frac{3x^{2}+120}{2x^{2}+x}.

Marginal Average Cost

• Differentiate using quotient rule
– High: H=3x^{2}+120,\;H'=6x.
– Low: L=2x^{2}+x,\;L'=4x+1.
• \overline{C}\,'(x)=\frac{H'\,L-H\,L'}{L^{2}}
=\frac{6x(2x^{2}+x)-(3x^{2}+120)(4x+1)}{(2x^{2}+x)^{2}}.
• Simplify numerator
– Expand first term: 12x^{3}+6x^{2}.
– Expand second term: 12x^{3}+480x+3x^{2}+120.
– Combine: \bigl(12x^{3}+6x^{2}\bigr)-\bigl(12x^{3}+480x+3x^{2}+120\bigr)=3x^{2}-480x-120.
– Factor: 3\bigl(x^{2}-160x-40\bigr).
• Critical points (set numerator = 0)
x^{2}-160x-40=0 \;\Longrightarrow\;x=\frac{160\pm\sqrt{160^{2}+4\cdot40}}{2}\approx160.124\;\text{or}\;-0.124.
– Negative quantity discarded if x counts items produced.
– Thus marginal average cost is zero at about x\approx160.12 units (transcript labels this “hundred”, perhaps meaning 160 (×100)=16 000 units).

Key Connections & Remarks

• Product and quotient rules extend directly to real-world rates (revenue, cost).
• Careful sign tracking is crucial; one transcription slip flipped the sign of h'(5).
• For linear approximations (price and demand) the product rule delivers instantaneous revenue growth without explicit multiplication before differentiating.
• Tangent-line problems reinforce that slope = derivative and point on curve = original function value.
• In cost analysis, zeros of marginal average cost often indicate most efficient production levels.