1 Problem 1: Basic Chain Rule (x)

Differentiate the following functions using the chain rule:

\(f(x) = (3x + 7)^6\)
\(g(x) = \sqrt{5x - 2}\)
\(h(x) = (x^2 - 4x + 1)^{10}\)
\(k(x) = \frac{1}{(2x + 3)^4}\)

Solution

a) \(f(x) = (3x + 7)^6\)

Outer: \(u^6\), derivative: \(6u^5\)
Inner: \(3x + 7\), derivative: \(3\)

\[f'(x) = 6(3x + 7)^5 \cdot 3 = 18(3x + 7)^5\]

b) \(g(x) = \sqrt{5x - 2} = (5x - 2)^{1/2}\)

Outer: \(u^{1/2}\), derivative: \(\frac{1}{2}u^{-1/2}\)
Inner: \(5x - 2\), derivative: \(5\)

\[g'(x) = \frac{1}{2}(5x - 2)^{-1/2} \cdot 5 = \frac{5}{2\sqrt{5x - 2}}\]

c) \(h(x) = (x^2 - 4x + 1)^{10}\)

Outer: \(u^{10}\), derivative: \(10u^9\)
Inner: \(x^2 - 4x + 1\), derivative: \(2x - 4\)

\[h'(x) = 10(x^2 - 4x + 1)^9 \cdot (2x - 4) = 20(x - 2)(x^2 - 4x + 1)^9\]

d) \(k(x) = \frac{1}{(2x + 3)^4} = (2x + 3)^{-4}\)

Outer: \(u^{-4}\), derivative: \(-4u^{-5}\)
Inner: \(2x + 3\), derivative: \(2\)

\[k'(x) = -4(2x + 3)^{-5} \cdot 2 = \frac{-8}{(2x + 3)^5}\]

2 Problem 2: Chain Rule with Product Rule (xx)

Differentiate the following functions:

\(f(x) = x^3(2x - 1)^4\)
\(g(x) = (x^2 + 1)^2(3x - 5)^3\)
\(h(x) = \frac{x^2}{(x + 1)^3}\)

Solution

a) \(f(x) = x^3(2x - 1)^4\) — Product rule needed

\(u = x^3\), \(u' = 3x^2\)
\(v = (2x - 1)^4\), \(v' = 4(2x - 1)^3 \cdot 2 = 8(2x - 1)^3\) (chain rule!)

\[f'(x) = 3x^2(2x - 1)^4 + x^3 \cdot 8(2x - 1)^3\]

Factor out \(x^2(2x - 1)^3\): \[= x^2(2x - 1)^3[3(2x - 1) + 8x] = x^2(2x - 1)^3(14x - 3)\]

b) \(g(x) = (x^2 + 1)^2(3x - 5)^3\) — Product of composites

\(u = (x^2 + 1)^2\), \(u' = 2(x^2 + 1) \cdot 2x = 4x(x^2 + 1)\)
\(v = (3x - 5)^3\), \(v' = 3(3x - 5)^2 \cdot 3 = 9(3x - 5)^2\)

\[h'(x) = 4x(x^2 + 1)(3x - 5)^3 + (x^2 + 1)^2 \cdot 9(3x - 5)^2\]

Factor \((x^2 + 1)(3x - 5)^2\): \[= (x^2 + 1)(3x - 5)^2[4x(3x - 5) + 9(x^2 + 1)]\] \[= (x^2 + 1)(3x - 5)^2(21x^2 - 20x + 9)\]

c) \(g(x) = \frac{x^2}{(x + 1)^3}\) — Quotient rule needed

\(u = x^2\), \(u' = 2x\)
\(v = (x + 1)^3\), \(v' = 3(x + 1)^2\) (chain rule!)

\[g'(x) = \frac{2x(x + 1)^3 - x^2 \cdot 3(x + 1)^2}{(x + 1)^6}\]

Factor \((x + 1)^2\) from numerator: \[= \frac{(x + 1)^2[2x(x + 1) - 3x^2]}{(x + 1)^6} = \frac{-x^2 + 2x}{(x + 1)^4} = \frac{x(2 - x)}{(x + 1)^4}\]

3 Problem 3: Simplifying Before Differentiating (x)

Differentiate by simplifying first:

\(f(x) = \sqrt{(2x + 1)^3}\)
\(g(x) = \frac{1}{\sqrt{x^2 + 1}}\)

Solution

a) \(f(x) = \sqrt{(2x + 1)^3} = (2x + 1)^{3/2}\)

\[f'(x) = \frac{3}{2}(2x + 1)^{1/2} \cdot 2 = 3\sqrt{2x + 1}\]

b) \(g(x) = \frac{1}{\sqrt{x^2 + 1}} = (x^2 + 1)^{-1/2}\)

\[g'(x) = -\frac{1}{2}(x^2 + 1)^{-3/2} \cdot 2x = \frac{-x}{(x^2 + 1)^{3/2}}\]

4 Problem 4: Implicit Differentiation - Business Contexts (x)

Find the derivative for each business relationship:

A company’s price \(p\) and quantity \(q\) satisfy a constant revenue constraint: \(pq = 5000\). Find \(\frac{dq}{dp}\).
Marketing spend \(M\) and sales \(S\) follow: \(MS = 12000\). Find \(\frac{dS}{dM}\).
Budget constraint: \(30L + 50K = 9000\) where \(L\) = labor hours and \(K\) = capital units. Find \(\frac{dK}{dL}\).

Solution

a) \(pq = 5000\)

Differentiate both sides with respect to \(p\): \[q + p\frac{dq}{dp} = 0\] \[\frac{dq}{dp} = -\frac{q}{p}\]

Interpretation: If price doubles, quantity must halve to maintain revenue.

b) \(MS = 12000\)

Differentiate with respect to \(M\): \[S + M\frac{dS}{dM} = 0\] \[\frac{dS}{dM} = -\frac{S}{M}\]

Interpretation: Increasing marketing requires proportionally less sales to maintain the same product.

c) \(30L + 50K = 9000\)

Differentiate with respect to \(L\): \[30 + 50\frac{dK}{dL} = 0\] \[\frac{dK}{dL} = -\frac{30}{50} = -\frac{3}{5}\]

Interpretation: For every 5 additional labor hours, capital must decrease by 3 units to stay on budget.

5 Problem 5: Related Rates - Customer Growth (xx)

A company’s revenue \(R\) (in €1000) depends on its customer base \(C\) (in thousands): \[R = 80\sqrt{C}\]

The company is gaining 2,000 new customers per month.

Find \(\frac{dR}{dC}\) (marginal revenue per customer).
Find \(\frac{dR}{dt}\) in terms of \(C\) and \(\frac{dC}{dt}\).
How fast is revenue growing when \(C = 100\) (100,000 customers)?
How fast is revenue growing when \(C = 400\)?
Explain why revenue growth slows as the customer base grows.

Solution

a) Marginal revenue per customer: \[R = 80C^{1/2}\] \[\frac{dR}{dC} = 80 \cdot \frac{1}{2}C^{-1/2} = \frac{40}{\sqrt{C}}\]

b) Revenue growth rate: \[\frac{dR}{dt} = \frac{dR}{dC} \cdot \frac{dC}{dt} = \frac{40}{\sqrt{C}} \cdot \frac{dC}{dt}\]

c) When \(C = 100\): Given: \(\frac{dC}{dt} = 2\) thousand/month \[\frac{dR}{dt} = \frac{40}{\sqrt{100}} \cdot 2 = \frac{40}{10} \cdot 2 = 8\]

Revenue is growing at €8,000 per month.

d) When \(C = 400\): \[\frac{dR}{dt} = \frac{40}{\sqrt{400}} \cdot 2 = \frac{40}{20} \cdot 2 = 4\]

Revenue is growing at €4,000 per month (half as fast!).

e) Explanation:

The formula \(\frac{dR}{dt} = \frac{40}{\sqrt{C}} \cdot \frac{dC}{dt}\) contains \(\sqrt{C}\) in the denominator. As \(C\) increases:

Each new customer contributes less marginal revenue
This reflects diminishing returns: early customers are easier/cheaper to monetize
Eventually, market saturation means new customers add minimal value

6 Problem 6: Related Rates - Production and Profit (xx)

A retail chain’s annual profit \(P\) (in €1000) and number of stores \(n\) are related by: \[P = 150\sqrt{n} - 4n\]

The company opens 3 new stores per year.

Find \(\frac{dP}{dn}\) and interpret its meaning.
Find \(\frac{dP}{dt}\) when \(n = 25\) stores.
Find \(\frac{dP}{dt}\) when \(n = 100\) stores.
At how many stores does profit stop growing (i.e., \(\frac{dP}{dn} = 0\))?

Solution

a) Marginal profit per store: \[\frac{dP}{dn} = \frac{75}{\sqrt{n}} - 4\]

Interpretation: This is the additional profit from opening one more store. It decreases as \(n\) grows.

b) When \(n = 25\): \[\frac{dP}{dt} = \frac{dP}{dn} \cdot \frac{dn}{dt} = \left(\frac{75}{\sqrt{25}} - 4\right) \cdot 3\] \[= \left(\frac{75}{5} - 4\right) \cdot 3 = (15 - 4) \cdot 3 = 33\]

Profit is growing at €33,000 per year.

c) When \(n = 100\): \[\frac{dP}{dt} = \left(\frac{75}{\sqrt{100}} - 4\right) \cdot 3 = (7.5 - 4) \cdot 3 = 10.5\]

Profit is growing at €10,500 per year (much slower!).

d) Profit stops growing when: \[\frac{dP}{dn} = 0\] \[\frac{75}{\sqrt{n}} - 4 = 0\] \[\frac{75}{\sqrt{n}} = 4\] \[\sqrt{n} = \frac{75}{4} = 18.75\] \[n = 351.6\]

At approximately 352 stores, profit is maximized. Beyond this, each new store actually reduces total profit due to cannibalization and overhead costs.

7 Problem 7: Related Rates - Market Share (xxx)

A company’s profit \(P\) (in €1000) and market share \(m\) (as a percentage) are related by: \[P = 800m - 15m^2\]

Market share is increasing at 1.5% per month.

Find \(\frac{dP}{dt}\) in terms of \(m\).
How fast is profit changing when \(m = 10\%\)?
How fast is profit changing when \(m = 25\%\)?
At what market share is profit maximized?
If the company currently has \(m = 30\%\), should they continue trying to grow market share? Why or why not?

Solution

a) Profit growth rate: \[\frac{dP}{dt} = \frac{dP}{dm} \cdot \frac{dm}{dt}\] \[= (800 - 30m) \cdot \frac{dm}{dt}\]

b) When \(m = 10\): \[\frac{dP}{dt} = (800 - 30 \cdot 10) \cdot 1.5 = 500 \cdot 1.5 = 750\]

Profit is increasing at €750,000 per month.

c) When \(m = 25\): \[\frac{dP}{dt} = (800 - 30 \cdot 25) \cdot 1.5 = 50 \cdot 1.5 = 75\]

Profit is increasing at only €75,000 per month.

d) Maximum profit when \(\frac{dP}{dm} = 0\): \[800 - 30m = 0\] \[m = \frac{800}{30} \approx 26.7\%\]

e) At \(m = 30\%\): \[\frac{dP}{dm} = 800 - 30(30) = 800 - 900 = -100\] \[\frac{dP}{dt} = -100 \cdot 1.5 = -150\]

No! At 30% market share, profit is decreasing at €150,000 per month. The company has passed optimal size. Aggressive expansion is destroying value (perhaps due to price wars, excessive marketing costs, or diseconomies of scale).