곱과 몫의 미분법

(a) Observe that the units on $g(v)$ are “kilometers per liter”. Since the units on $f(v)$ are “liters per kilometer”, it follows that $g$ is the reciprocal of $f$: $g(v) = \frac{1}{f(v)}$. By the quotient rule, we then know that $$g'(v) = \frac{f(v) \cdot 0 - 1 \cdot f'(v)}{f(v)^2} = -\frac{f'(v)}{f(v)^2}\,.$$ Further, using the given values we can say that $g(80) = \frac{1}{f(80)} = \frac{1}{0.05} = 20$ kilometers per liter, and $g'(80) = -\frac{f'(80)}{f(80)^2} = -\frac{0.0004}{0.05^2} = -0.16$. Note that the units on $g'(80)$ are “kilometers per liter per kilometer per hour”. (b) Since the units on $h(v)$ are “liters per hour”, and the units on $f(v)$ are “liters per kilometer”, we note that if we multiply $f(v)$ by $v$ (whose units are “kilometers per hour”), we get that the units on $v \cdot f(v)$ match the units of $h(v)$. Thus, $$h(v) = v \cdot f(v)\,.$$ By the product rule, it follows that $$h'(v) = v \cdot \frac{d}{dv}[f(v)] + \frac{d}{dv}[v] \cdot f(v) = v \cdot f'(v) + 1 \cdot f(v)\,.$$ Using the known function and derivative values of $f$, we determine that $h(80) = 80 \cdot 0.05 = 4$ liters per hour and $h'(80) = 80 \cdot 0.0004 + 0.05 = 0.082$ liters per hour per kilometer per hour. (c) To explain the meaning of these function and derivative values to someone who doesn't know calculus, we could say something like the following:

• The given values of $f(80) = 0.05$ and $f'(80) = 0.0004$ tell us that when the car is going $80$ kilometers per hour, it is using $0.05$ liters of fuel for every kilometer it travels. In addition, for each additional kilometer per hour of velocity above $80$ kph, we expect that the car will use about an additional $0.0004$ liters of fuel per kilometer.
• The values of $g(80) = 20$ and $g'(80) = -0.16$ tell us that when the car is going $80$ kilometers per hour, it is able to travel $20$ kilometers for each liter of fuel used. In addition, for each additional kilometer per hour of velocity above $80$ kph, we expect that the car will be able to travel about $0.16$ fewer kilometers per liter of fuel.
• Finally, the values of $h(80) = 4$ and $h'(80) = 0.82$ tell us that when the car is going $80$ kilometers per hour, it is using $4$ liters of fuel for every hour it travels. In addition, for each additional kilometer per hour of velocity above $80$ kph, we expect that the car will ues about $0.082$ more liters of fuel per hour. Note that all three of these observations are saying fundamentally the same thing, just from different perspectives: the first regards fuel consumption measured in liters per kilometer; the second regards fuel consumption in kilometers per liter; and the last regards fuel consumption in liters per hour.