Chain Rule - Business Calculus

1. A factory's daily production cost is $C(n)$ dollars, where $n$ is the number of units produced. The number of units produced $t$ hours into the workday is given by $n(t)$. The derivative $C'(n)$ represents the marginal cost per unit, and $n'(t)$ is the production rate in units per hour. What is the practical interpretation of the composite derivative $(C \circ n)'(3)$?
   Answer: The rate at which production costs are changing, in dollars per hour, three hours into the workday. By the chain rule, $(C \circ n)'(t) = \frac{d}{dt}C(n(t)) = C'(n(t)) \cdot n'(t)$. Let's analyze the units. $C'(n)$ has units of dollars per unit. $n'(t)$ has units of units per hour. Their product, $C'(n(t)) \cdot n'(t)$, has units of (dollars/unit) $\times$ (units/hour) = dollars/hour. This represents the rate of change of cost with respect to time. Evaluating at $t=3$ gives this rate at the specific moment three hours into the workday.
2. The value of a particular asset is given by the function $V(x) = (x^2+3)\sqrt{4x-3}$, where $x$ is the number of years since its acquisition. Find the rate at which the asset's value is changing after 3 years.
   Answer: $26$ We need to find $V'(3)$. The function is a product $V(x) = u(x)v(x)$ where $u(x)=x^2+3$ and $v(x)=(4x-3)^{1/2}$. We use the product rule, $V'(x) = u'(x)v(x) + u(x)v'(x)$. The derivatives are $u'(x)=2x$ and, using the chain rule, $v'(x) = \frac{1}{2}(4x-3)^{-1/2} \cdot 4 = \frac{2}{\sqrt{4x-3}}$. So, $V'(x) = (2x)\sqrt{4x-3} + (x^2+3)\frac{2}{\sqrt{4x-3}}$. Now, we evaluate at $x=3$: $V'(3) = (2\cdot3)\sqrt{4\cdot3-3} + (3^2+3)\frac{2}{\sqrt{4\cdot3-3}} = 6\sqrt{9} + (12)\frac{2}{\sqrt{9}} = 6(3) + \frac{24}{3} = 18 + 8 = 26$.
3. If $y = e^{x \ln(x^2 + 1)}$, then $\frac{dy}{dx}$ equals:
   Answer: $e^{x \ln(x^2 + 1)} \left(\ln(x^2 + 1) + \frac{2x^2}{x^2 + 1}\right)$ Using chain rule on $y = e^{u}$ where $u = x\ln(x^2 + 1)$: $\frac{dy}{dx} = e^u \cdot \frac{du}{dx}$. For $\frac{du}{dx}$, use product rule: $\frac{du}{dx} = 1 \cdot \ln(x^2 + 1) + x \cdot \frac{2x}{x^2 + 1} = \ln(x^2 + 1) + \frac{2x^2}{x^2 + 1}$. Choice B has wrong coefficient in the fraction. Choice C has $2x$ instead of $2x^2$ in numerator. Choice D omits the $\ln(x^2 + 1)$ term completely.
4. The weekly revenue $R$ from selling a new software is given by $R(x) = 5000e^{-0.01x}$ dollars, where $x$ is the number of units sold. The number of units sold depends on the advertising budget $a$, in thousands of dollars, according to $x(a) = a^2 + 9a$. Find the rate of change of revenue with respect to the advertising budget when the budget is $1,000.
   Answer: $-550e^{-0.1}$ We need to find $dR/da$ at $a=1$ (since $a$ is in thousands of dollars). By the chain rule, $dR/da = (dR/dx) \cdot (dx/da)$. First, find the derivatives: $dR/dx = 5000e^{-0.01x} \cdot (-0.01) = -50e^{-0.01x}$ and $dx/da = 2a+9$. When $a=1$, the number of units sold is $x(1) = 1^2 + 9(1) = 10$. The rate of change of sales is $x'(1) = 2(1)+9=11$. Now evaluate $dR/dx$ at $x=10$: $dR/dx|_{x=10} = -50e^{-0.01(10)} = -50e^{-0.1}$. Finally, multiply the rates: $dR/da|_{a=1} = (-50e^{-0.1}) \cdot (11) = -550e^{-0.1}$.
5. A company finds that its daily cost $C$ is a function of the natural logarithm of its production output $x$, given by the formula $C(x) = (\ln(x^3 + 1))^2$. Find the marginal cost $C'(x)$ when production output is $x=1$.
   Answer: $3\ln(2)$ To find the marginal cost $C'(x)$, we must differentiate $C(x)$ using the chain rule twice. Let $u = \ln(v)$ and $v = x^3+1$, so $C=u^2$. The derivative is $C'(x) = \frac{dC}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$. The component derivatives are $\frac{dC}{du} = 2u = 2\ln(x^3+1)$, $\frac{du}{dv} = \frac{1}{v} = \frac{1}{x^3+1}$, and $\frac{dv}{dx} = 3x^2$. Combining these gives $C'(x) = 2\ln(x^3+1) \cdot \frac{1}{x^3+1} \cdot 3x^2 = \frac{6x^2 \ln(x^3+1)}{x^3+1}$. Evaluating at $x=1$: $C'(1) = \frac{6(1)^2 \ln(1^3+1)}{1^3+1} = \frac{6\ln(2)}{2} = 3\ln(2)$.
6. The profit $P(x)$, in hundreds of dollars, from a concert is modeled by $P(x) = 200\ln(S(x))$, where $S(x) = 0.5x^2 + 10$ is an audience satisfaction index and $x$ is the number of tickets sold in thousands. Find the marginal profit, $P'(x)$, when 10,000 tickets are sold.
   Answer: $100/3$ The marginal profit is the derivative of the profit function, $P'(x)$. We use the chain rule: $P'(x) = 200 \cdot \frac{1}{S(x)} \cdot S'(x)$. The derivative of the inner function is $S'(x) = 2(0.5)x = x$. Substituting this in gives $P'(x) = 200 \cdot \frac{1}{0.5x^2 + 10} \cdot x = \frac{200x}{0.5x^2 + 10}$. We need to evaluate this at $x=10$ (since $x$ is in thousands). $P'(10) = \frac{200(10)}{0.5(10)^2 + 10} = \frac{2000}{0.5(100) + 10} = \frac{2000}{50 + 10} = \frac{2000}{60} = \frac{100}{3}$.
7. A particle moves along a path where its position is given by $s(t) = \tan^{-1}(t^3 - 2t)$. What is the particle's velocity when $t = 1$?
   Answer: $\frac{1}{2}$ Using chain rule: $s'(t) = \frac{1}{1 + (t^3 - 2t)^2} \cdot (3t^2 - 2)$. At $t = 1$: the inner function $t^3 - 2t = 1 - 2 = -1$, and its derivative is $3(1)^2 - 2 = 1$. So $s'(1) = \frac{1}{1 + (-1)^2} \cdot 1 = \frac{1}{2}$. Choice B incorrectly calculates $1 + (-1)^2 = 4$. Choice C uses the wrong derivative of the inner function. Choice D forgets to evaluate the denominator correctly.
8. Let $h(x) = \sin(\cos(x^3))$. Which expression represents $h'(x)$?
   Answer: $\cos(\cos(x^3)) \cdot (-\sin(x^3)) \cdot 3x^2$ Using chain rule with three layers: $h'(x) = \cos(\cos(x^3)) \cdot \frac{d}{dx}[\cos(x^3)] \cdot \frac{d}{dx}[x^3] = \cos(\cos(x^3)) \cdot (-\sin(x^3)) \cdot 3x^2$. Choice A forgets the negative sign when differentiating cosine. Choice C incorrectly switches sine and cosine in the outer function. Choice D has the wrong coefficient for $x^2$.
9. The number of subscribers $S$ to a streaming service, in millions, t months after launch is modeled by $S(t) = (3t^2 + 4)^{3/2}$. What is the rate of growth of subscribers 2 months after launch, in millions per month?
   Answer: 72 The rate of growth is the derivative, $S'(t)$. We use the chain rule: $S'(t) = \frac{3}{2}(3t^2 + 4)^{(3/2 - 1)} \cdot \frac{d}{dt}(3t^2+4)$. This gives $S'(t) = \frac{3}{2}(3t^2 + 4)^{1/2} \cdot (6t)$. Simplifying, $S'(t) = 9t\sqrt{3t^2+4}$. We need to evaluate this at $t=2$: $S'(2) = 9(2)\sqrt{3(2^2)+4} = 18\sqrt{3(4)+4} = 18\sqrt{12+4} = 18\sqrt{16} = 18(4) = 72$. So the growth rate is 72 million subscribers per month.
10. Let $f(x) = \ln(\sin(e^x))$. Which of the following represents $f'(x)$?
    Answer: $\frac{\cos(e^x) \cdot e^x}{\sin(e^x)}$ When you encounter a composite function like $f(x) = \ln(\sin(e^x))$, you need to apply the chain rule systematically. This function has three layers: the natural logarithm on the outside, sine in the middle, and an exponential function on the inside. To find $f'(x)$, work from the outside in. Start with the derivative of $\ln(u)$, which is $\frac{1}{u}$. Here, $u = \sin(e^x)$, so the first step gives us $\frac{1}{\sin(e^x)}$. Next, multiply by the derivative of the middle layer: $\sin(e^x)$. The derivative of $\sin(v)$ is $\cos(v)$, where $v = e^x$, giving us $\cos(e^x)$. Finally, multiply by the derivative of the innermost function $e^x$, which is $e^x$. Combining all parts: $f'(x) = \frac{1}{\sin(e^x)} \cdot \cos(e^x) \cdot e^x = \frac{\cos(e^x) \cdot e^x}{\sin(e^x)}$ Choice A ($\cos(e^x) \cdot e^x$) represents only the derivative of $\sin(e^x)$ but ignores the outer logarithm layer. Choice B ($\frac{\cos(e^x)}{\sin(e^x)}$) correctly handles the logarithm and sine layers but forgets to differentiate $e^x$. Choice C ($\frac{e^x}{\sin(e^x)}$) incorrectly treats the sine function as if it weren't there during differentiation. Remember: with nested functions, the chain rule requires you to differentiate every layer and multiply all those derivatives together. Missing any layer leads to an incomplete answer.
11. The population of a city t years after 2020 is modeled by $P(t) = (t^2 + 100)^{3/2}$. Find the rate at which the population growth is accelerating in 2030.
    Answer: $45\sqrt{2}$ The rate of acceleration is the second derivative, $P''(t)$. First, find the first derivative, $P'(t)$, using the chain rule: $P'(t) = \frac{3}{2}(t^2+100)^{1/2} \cdot (2t) = 3t(t^2+100)^{1/2}$. Next, find the second derivative, $P''(t)$, using the product rule and the chain rule: $P''(t) = (3)(t^2+100)^{1/2} + (3t) \left[ \frac{1}{2}(t^2+100)^{-1/2} \cdot (2t) \right]$. This simplifies to $P''(t) = 3\sqrt{t^2+100} + \frac{3t^2}{\sqrt{t^2+100}}$. The year 2030 corresponds to $t=10$. Evaluating at $t=10$: $P''(10) = 3\sqrt{10^2+100} + \frac{3(10^2)}{\sqrt{10^2+100}} = 3\sqrt{200} + \frac{300}{\sqrt{200}} = 3(10\sqrt{2}) + \frac{300}{10\sqrt{2}} = 30\sqrt{2} + \frac{30}{\sqrt{2}}$. Rationalizing the second term gives $30\sqrt{2} + \frac{30\sqrt{2}}{2} = 30\sqrt{2} + 15\sqrt{2} = 45\sqrt{2}$.
12. A company's manufacturing cost, $C$, in thousands of dollars, is a function of the number of units produced, $q$, according to $C(q) = \sqrt{q^2 + 567}$. The number of units produced is a function of time, $t$, in hours, given by $q(t) = t^3$. What is the rate of change of the manufacturing cost with respect to time, in thousands of dollars per hour, when $t=3$?
    Answer: $20.25 The problem asks for $dC/dt$ at $t=3$. Using the chain rule, $dC/dt = (dC/dq) \cdot (dq/dt)$. First, find the derivatives: $C'(q) = \frac{1}{2}(q^2+567)^{-1/2}(2q) = \frac{q}{\sqrt{q^2+567}}$ and $q'(t) = 3t^2$. Next, evaluate the functions at $t=3$. When $t=3$, the production level is $q(3) = 3^3 = 27$ units. The rate of production is $q'(3) = 3(3^2) = 27$ units/hour. The marginal cost at $q=27$ is $C'(27) = \frac{27}{\sqrt{27^2 + 567}} = \frac{27}{\sqrt{729+567}} = \frac{27}{\sqrt{1296}} = \frac{27}{36} = 0.75$ thousand dollars/unit. Finally, $dC/dt|_{t=3} = C'(27) \cdot q'(3) = 0.75 \cdot 27 = 20.25$ thousand dollars/hour.
13. A company's market share $M$ (as a percentage) $t$ years from now is given by $M(t) = \frac{100t}{(t^2+9)^{3/2}}$. At what rate will the company's market share be changing in 4 years?
    Answer: $-0.736\%$ per year We need to find $M'(4)$. Using the quotient rule, $M'(t) = \frac{100 \cdot (t^2+9)^{3/2} - 100t \cdot \frac{d}{dt}(t^2+9)^{3/2}}{((t^2+9)^{3/2})^2}$. The derivative of the denominator requires the chain rule: $\frac{d}{dt}(t^2+9)^{3/2} = \frac{3}{2}(t^2+9)^{1/2}(2t) = 3t(t^2+9)^{1/2}$. Substituting this in: $M'(t) = \frac{100(t^2+9)^{3/2} - 100t(3t(t^2+9)^{1/2})}{(t^2+9)^3}$. Factor out $100(t^2+9)^{1/2}$ from the numerator: $M'(t) = \frac{100(t^2+9)^{1/2}[(t^2+9)-3t^2]}{(t^2+9)^3} = \frac{100(9-2t^2)}{(t^2+9)^{5/2}}$. Now evaluate at $t=4$: $M'(4) = \frac{100(9-2(4^2))}{(4^2+9)^{5/2}} = \frac{100(9-32)}{(16+9)^{5/2}} = \frac{100(-23)}{25^{5/2}} = \frac{-2300}{(\sqrt{25})^5} = \frac{-2300}{5^5} = \frac{-2300}{3125} = -0.736$.
14. Let $h(x) = f(e^{x^2})$, where $f$ is a differentiable function. If it is known that $f'(e^4) = 10$, what is the value of $h'(2)$?
    Answer: $40e^4$ This question tests the chain rule with composite functions, which is essential when dealing with nested business functions like compound interest or layered cost structures. To find $h'(2)$, you need to apply the chain rule to $h(x) = f(e^{x^2})$. The chain rule states that if you have a composition $f(g(x))$, then the derivative is $f'(g(x)) \cdot g'(x)$. Here, the outer function is $f$ and the inner function is $g(x) = e^{x^2}$. So $h'(x) = f'(e^{x^2}) \cdot \frac{d}{dx}[e^{x^2}]$. First, find the derivative of the inner function: $\frac{d}{dx}[e^{x^2}] = e^{x^2} \cdot 2x$ (using the chain rule again). Therefore: $h'(x) = f'(e^{x^2}) \cdot e^{x^2} \cdot 2x$ At $x = 2$: $h'(2) = f'(e^{2^2}) \cdot e^{2^2} \cdot 2(2) = f'(e^4) \cdot e^4 \cdot 4$ Since $f'(e^4) = 10$: $h'(2) = 10 \cdot e^4 \cdot 4 = 40e^4$ Choice A ($20e^4$) results from forgetting to apply the chain rule to $e^{x^2}$, missing the $2x$ factor. Choice C ($10e^2$) incorrectly uses $e^2$ instead of $e^4$ and omits the $2x$ factor. Choice D ($80e^2$) has the wrong exponential term but accidentally gets close by doubling somewhere. Strategy tip: When seeing nested functions, identify each layer clearly and apply the chain rule systematically. Always multiply by the derivative of each inner function as you work outward.
15. A company's cost function is $C(q) = 50\sqrt{q^2 + 4q + 13}$ dollars for producing $q$ units. What is the marginal cost when $q = 3$ units?
    Answer: $\frac{25(2q + 4)}{\sqrt{q^2 + 4q + 13}}$ evaluated at $q = 3$ Using chain rule: $C'(q) = 50 \cdot \frac{1}{2\sqrt{q^2 + 4q + 13}} \cdot (2q + 4) = \frac{25(2q + 4)}{\sqrt{q^2 + 4q + 13}}$. At $q = 3$: $C'(3) = \frac{25(10)}{\sqrt{34}} = \frac{250}{\sqrt{34}}$. Choice A incorrectly factors the derivative of the inner function. Choice B forgets to apply the $\frac{1}{2}$ factor from differentiating the square root. Choice D doubles the coefficient incorrectly.

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