곱과 몫의 미분법

$x=3/2$

$x=0$

$f\,'(x)$ is never 0.

$x= -2,0$

$f\,''(x) = 2\cos x - x\sin x$

$f^{(4)}(x) = -4\cos x+ x\sin x$

$f\,''(x) = \cot^2x\csc x + \csc^3 x$

$f^{(8)} = 0$

(a) Since $h(x) = g(x) \cdot f(x)$, it follows $h(2) = g(2) \cdot f(2) = (5) \cdot (-3) = -15$. Further, by the product rule, we know that $h'(x) = g(x) \cdot f'(x) + f(x) \cdot g'(x)$. Using the given function and derivative values, $h'(2) = g(2) \cdot f'(2) + f(2) \cdot g'(2) = (-3)(-1/2) + (5)(2) = 23/2$. (b) The tangent line to $h$ at $a=2$ is given by $L(x) = h(2) + h'(2)(x-2)$, so using our results from (a), we have $L(x) = -15 + 23/2(x-2)$. (c) Since $r(x) = \frac{g(x)}{f(x)}$, but the quotient rule that $r'(x) = \frac{f(x) \cdot g'(x) - g(x) \cdot f'(x)}{(f(x))^2}$, and thus $r'(2) = \frac{f(2) \cdot g'(2) - g(2) \cdot f'(2)}{(f(2))^2}$. Employing the known function and derivative values, $r'(2) = \frac{(5)(2) - (-3)(-1/2)}{(5)^2} = \frac{17/2}{25} = \frac{17}{50}$. Therefore, $r'(2) > 0$, and $r$ is increasing at $a = 2$. (d) The local linearization of $r$ at $a=2$ is $L(x) = r(2) + r'(2)(x-2) = -\frac{3}{5} + \frac{17}{50}(x-2)$, and thus $r(2.06) \approx L(2.06) = -\frac{3}{5} + \frac{17}{50}(2.06-2) = -0.5796$.

(a) Using the Product Rule and the information given, $$\begin{aligned}w'(t) &= \frac{d}{dt}\left(t^t\right)\cdot\left(\arccos t\right)+t^t\cdot\frac{d}{dt}\left(\arccos t\right) \\ &= t^t\left(\ln t+1\right)\cdot\left(\arccos t\right)+t^t\cdot\frac{-1}{\sqrt{1-t^2}}\end{aligned}$$ (b) Using the local linearization formula, we have $L(t) = w\left(\frac{1}{2}\right) + w'\left(\frac{1}{2}\right) \left(t-\frac{1}{2}\right)$. Substituting $t=\frac{1}{2}$ into the formulas for $w$ and $w'$ and using a computational device gives $$w(2) = \left(\frac{1}{2}\right)^{\left(\frac{1}{2}\right)}\arccos\left(\frac{1}{2}\right) \approx 0.740$$ and $$w'(2) = \left[\left(\frac{1}{2}\right)^{\left(\frac{1}{2}\right)}\left(\ln t+1\right)\cdot\left(\arccos \left(\frac{1}{2}\right)\right)+\left(\frac{1}{2}\right)^{\left(\frac{1}{2}\right)}\cdot\frac{-1}{\sqrt{1-\left(\frac{1}{2}\right)^2}}\right] \approx -0.589\,.$$ Thus, $L(t) \approx 0.740-0.589(t-0.5)$. (c) Using the Quotient Rule, $$v'(t) = frac{(t)^{t}\left(\ln(t)+1\right)\left(\arccos(t)\right)-(t)^{t}\cdot\left(\frac{-1}{\sqrt{1-(t)^2}}\right)}{\arccos^2(t)}\,.$$ Using a computational device to evaluate $v'(2)$, it follows $v'(2) \approx 0.952$. Since this number is positive, the instantaneous rate of change of $v$ at $t=\frac{1}{2}$ is positive, which means that $v$ is increasing at that instant.

(a) If we multiply $A(t)$ acres by $Y(t)$ bushels per acre, we arrive at $C(t) = A(t)Y(t)$ bushels as the total number of bushels of corn the farmer produces in year $t$. (b) We are given that $A(0) = 7000$ acres with a yield of $Y(0) = 170$ bushels per acre, giving us $C(0) = A(0)Y(0) = (7000)(170) = 1190000$ bushels. Thus, the farmer produces $1 190 000$ bushels of corn in the present year. (c) The product rule tells us that $C'(t) = A(t)Y'(t) + A'(t)Y(t)$. (d) We know $A(0) = 7000$ and $Y(0) = 170$. We are also given that the farmer will increase his number of acres devoted to growing corn at a rate of $600$ acres/year, and he expects that right now his average yield is increasing at a rate of $8$ bushels per acre per year. This tells us that $A'(t) = 600$ and $Y'(0) = 8$. So $$C'(0) = A(0)Y'(0) + A'(0)Y(0) = (7000)(8) + (600)(170) = 158 000\,.$$ Note that the units of $A(t)$ are acres and the units of $Y'(t)$ are bushels per acre per year. So the units of $A(0)Y'(0)$ are bushels per year. Similarly, the units of $A'(t)$ are acres per year and the units of $Y(t)$ are bushels per acre, so the units of $A'(0)Y(0)$ are also bushels per acre. So the units of $C'(0)$ are bushels per acre. Therefore, $C'(0) = 158 000$ tells us that the number of bushels the farmer can produce will increase by approximately $158 000$ bushels for each year past the present year. (e) The linearization of $C$ at $t=0$ is $L(t) = C(0) + C'(0)(t-0) = 119 0000 + 158 000t$, $C(1) \approx L(1) = 119 0000 + 158 000(1) = 1 348 000$bushels.

(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.