미분법 전체 문제

(a) We use the quotient, product, sum, and constant multiple rules to obtain $$\begin{aligned}p'(z) &= \frac{d}{dz}\left(\frac{z\tan(z)}{z^2\sec(z) + 1} + 3 e^z + 1\right) \\ &= \frac{d}{dz}\left(\frac{z\tan(z)}{z^2\sec(z) + 1}\right) + \frac{d}{dz} \left(3 e^z \right) + \frac{d}{dz}(1) \\ &= \frac{\left(z^2\sec(z) +1 \right)\frac{d}{dz}\left(z\tan(z)\right) - z\tan(z) \frac{d}{dz}\left(z^2\sec(z) + 1\right)}{\left(z^2\sec(z) + 1\right)^2} +3e^z \\ &= \frac{\left(z^2\sec(z) +1 \right)\left(z\sec^2(z)+\tan(z)\right) - z\tan(z) \left(z^2\sec(z)\tan(z)+2z\sec(z)\right)}{\left(z^2\sec(z) + 1\right)^2} +3e^z\end{aligned}$$ (b) We have $p(0) = 0+3+1 = 4$ and $p'(0) = 0+3=3$, so an equation for the tangent line to $p$ at the point where $z=0$ is $y - 4 = 3(x-0)$. (c) We found that $p'(0) = 3$ is positive, so $p$ is increasing at $t=0$.

(a) Since $h(x) = f(g(x))$, by the chain rule we know that $h'(x) = f'(g(x)) \cdot g'(x)$. Using the facts that $f(x)=x^3$ and $g(x) = \sin(x)$, it follows $f'(x) = 3x^2$ and $f'(g(x)) = 3\sin^2(x)$, and $g'(x) = \cos(x)$. Therefore, $$h'(x) = 3\sin^2(x) \cdot \cos(x)$$ and $h'\left( \frac{\pi}{4} \right) = 3 \left( \frac{1}{\sqrt{2}} \right)^2 \cdot \frac{1}{\sqrt{2}} = \frac{3}{2\sqrt{2}}$. (b) Note that $r'(x) = g'(f(x)) \cdot f'(x) = \cos(x^3) \cdot 3x^2$. Thus, we have $r'(0.25) = \cos(0.25^3) \cdot 3(0.25)^2 \approx 0.1875$, while using our work from (a), $h'(0.25) = 3\sin^2(0.25) \cdot \cos(0.25) \approx 0.1779$, so since $|h'(0.25) < r'(0.25)|$, we see that $r$ is changing more rapidly. (c) We have observed in (a) and (b) that $h'(x) = 3\sin^2(x) \cdot \cos(x)$ and $r'(x) = \cos(x^3) \cdot 3x^2$. Because both parts of the product in $h'(x)$ are periodic functions, $h'(x)$ is periodic (with period $2\pi$). However, $r'(x)$ is not periodic, since $3x^2$ grows without bound as $x$ increases.

(a) By the chain rule, $p'(x) = e^{u(x)} \cdot u'(x)$. (b) By the chain rule, $q'(x) = u'(e^x) \cdot e^x$. (c) By the chain rule, $r'(x) = -\csc^2(u(x) \cdot u'(x)$. (d) By the chain rule, $s'(x) = u'(\cot(x)) \cdot (-\csc^2(x))$. (e) By the chain rule, $a'(x) = u'(x^4) \cdot 4x^3$. (f) By the chain rule, $b'(x) = 4(u(x))^3 \cdot u'(x)$.

(a) We are asked to find $\frac{dV}{dh}$ (the derivative of $V$ with respect to the independent variable $h$) at the instant $h = 1$. If we first expand $V$ to write $$V = \frac{\pi}{3}\left(12h^2 - h^3\right)$$ we find that $$\frac{dV}{dh} = \frac{\pi}{3} \left(24h - 3h^2\right) = \pi \left(8h-h^2 \right)\,,$$ with units “cubic feet per foot”. Evaluating at $h=1$, we get $$\left. \frac{dV}{dh} \right|_{h=1} = \pi \left(8 \cdot 1 - 1^2 \right) = 7\pi$$ (b) Since $h(t) = \sin(\pi t) + 1$, the instantaneous rate of change of $h$ with respect to $t$ is $h'(t) = \pi \cos(\pi t)$, and therefore $h'(2) = \pi \cos(2\pi) = \pi$ feet per hour. (c) First we want to find $\frac{dV}{dt}$, the derivative of $V$ with respect to the independent variable $t$. Here we can write $V$ as a function of $t$ by writing $V(t) = \frac{\pi}{3}\left(12(h(t))^2 - (h(t))^3\right)$, where $h(t) = \sin(\pi t) + 1$. By the chain rule, it follows that $$\frac{dV}{dt} = \frac{\pi}{3}\left( 12 \cdot 2 h(t) \cdot h'(t) - 3 (h(t))^2 \cdot h'(t) \right)$$ and hence using $h(t) = \sin(\pi t) + 1$ and $h'(t) = \pi \cos(\pi t)$, $$\frac{dV}{dt} = \frac{\pi}{3}\left[ 12 \cdot 2 (\sin(\pi t) + 1) \cdot \pi \cos(\pi t) - 3 (\sin(\pi t) + 1)^2 \cdot \pi \cos(\pi t) \right]$$ with units “ feet per hour”. Now we want to find $\frac{dV}{dt}$ when $t=2$. Using our most recent work, we have $$\left. \frac{dV}{dt} \right|_{t=2} = \frac{\pi}{3}\left[ 12 \cdot 2 (\sin(2\pi) + 1) \cdot \pi \cos(2\pi) - 3 (\sin(2\pi) + 1)^2 \cdot \pi \cos(2\pi) \right] = 7 \pi^2$$ cubic feet per hour. (d) In (a) we are determining the instantaneous rate at which the volume changes as we increase the height of the water in the tank, while in (c) we are finding the instantaneous rate at which volume changes as we increase time. The first calculation is a direct one, while the second is more indirect, as it involves the composition of $V$ with $h$ as a function of time. In the latter case, both $h$ and $V$ are functions of time, which allows us to think not only of how $V$ is changing as a function of $h$, but also as a function of $t$.

(a) By the chain rule, $$f'(x) = \frac{1}{2\arctan(x) + 3\arcsin(x) + 5} \cdot \frac{d}{dx}\left( 2\arctan(x) + 3\arcsin(x) + 5\right)\,,$$ and by the sum and constant multiple rules, it follows that $$f'(x) = \frac{1}{2\arctan(x) + 3\arcsin(x) + 5} \cdot \left(\frac{2}{1+x^2} + \frac{3}{\sqrt{1-x^2}}\right)\,.$$ (b) By the chain rule applied to the outermost function, $$r'(z) = \frac{1}{1+\left(\ln(\arcsin(z))\right)^2} \cdot \frac{d}{dz} \left( \ln(\arcsin(z)) \right)\,.$$ Applying the chain rule again to evaluate $\frac{d}{dz} \left( \ln(\arcsin(z)) \right)$, $$r'(z) = \frac{1}{1+\left(\ln(\arcsin(z))\right)^2} \cdot \left( \frac{1}{\arcsin(z)} \right) \cdot \frac{d}{dz}\left(\arcsin(z)\right)\,.$$ Finally, using the basic rule for $\frac{d}{dz}\left(\arcsin(z)\right)$, $$r'(z) = \frac{1}{1+\left(\ln(\arcsin(z))\right)^2} \cdot \left( \frac{1}{\arcsin(z)} \right) \cdot \frac{1}{\sqrt{1-z^2}}\,.$$ (c) First, by the product rule, $$q'(t) = \arctan^2(3t) \cdot \frac{d}{dt}\left(\arcsin^4(7t)\right) + \arcsin^4(7t) \cdot \frac{d}{dt}\left(\arctan^2(3t)\right)\,.$$ Next, we recall that $\arcsin^4(7t) = \left(\arcsin(7t)\right)^4$ and $\arctan^2(3t) = \left( \arctan(3t) \right)^2$, and apply the chain rule to find the derivative of each of these functions to get $$q'(t) = \arctan^2(3t) \cdot \left[4\arcsin^3(7t) \frac{d}{dt}\left( \arcsin(7t) \right)\right] + \arcsin^4(7t) \cdot \left[2\arctan(3t) \frac{d}{dt}\left(\arctan(3t)\right) \right]\,.$$ Finally, applying the chain rule one more time to evaluate the remaining derivatives, $$q'(t) = \arctan^2(3t) \cdot \left[4\arcsin^3(7t) \left( \frac{7}{\sqrt{1-(7t)^2}} \right)\right] + \arcsin^4(7t) \cdot \left[2\arctan(3t) \left(\frac{3}{1+(3t)^2}\right) \right]\,.$$ (d) By the chain rule, $$g'(v) = \frac{1}{ \frac{\arctan(v)}{\arcsin(v) + v^2} } \cdot \frac{d}{dv}\left[ \frac{\arctan(v)}{\arcsin(v) + v^2} \right]$$ Thus, by the quotient rule, $$g'(v) = \frac{1}{\frac{\arctan(v)}{\arcsin(v) + v^2}} \cdot \frac{(\arcsin(v) + v^2) \cdot \frac{1}{1+v^2} - \arctan(v) \cdot \left(\frac{1}{\sqrt{1-v^2}} + 2v \right)}{(\arcsin(v) + v^2)^2}$$

(a) $f$ is an invertible function because its graph passes the horizontal line test. (b) To find a formula for $g = f^{-1}$, we take the equation $y = \frac{1}{4}x^3 + 4$ and solve for $x$ in terms of $y$. Doing so, we find that $y - 4 = \frac{1}{4}x^3$, so $x^3 = 4(y-4)$, and thus $$x = \sqrt[3]{4(y-4)} = f^{-1}(y)\,.$$ Writing the inverse as a function of $x$, we have found that $f^{-1}(x) = g(x) = \sqrt[3]{4x-16}$. (c) First, $f'(x) = \frac{3}{4}x^2$, and thus $f'(2) = 3$. Next, $g'(x) = \frac{1}{3}(4x-16)^{-2/3} \cdot 4$, so that $g'(6) = \frac{1}{3} \cdot 8^{-2/3} \cdot 4 = \frac{1}{3} \cdot \frac{1}{4} \cdot 4 = \frac{1}{3}$. What is special about these two derivative values is that they happen to be reciprocals. Noting that $f(2) = 3$ and $g(3) = 2$, we know that $g'(3) = \frac{1}{f'(2)}$ by the rule for the derivative of an inverse function, and this matches our finding above that $g'(3) = \frac{1}{3}$.

(a) The function $h$ is invertible because its graph passes the horizontal line test. (b) It does not seem possible to solve the equation $y = x + \sin(x)$ for $x$ in terms of $y$. (c) By the rule for the derivative of an invertible function, we have $(h^{-1})'(\frac{\pi}{2} + 1) = \frac{1}{h'\left(\frac{\pi}{2} \right)}$. Thus, we first compute $h'(x)$ and find that $h'(x) = 1 + \cos(x)$. Since $h'\left(\frac{\pi}{2} \right) = 1 + \cos\left(\frac{\pi}{2} \right) = 1$, it follows that $(h^{-1})'(\frac{\pi}{2} + 1) = \frac{1}{1} = 1$.