• これは Silvanus P. Thompson 著 Calculus Made Easy (1914) の翻訳です。

• この翻訳の PDF 版を BOOTH で販売しています (学生版は無料)。

# 練習問題の解答

## 練習問題 I

$$\text{(1)}$$ $$\dfrac{dy}{dx} = 13x^{12}$$ $$\$$
$$\text{(2)}$$ $$\dfrac{dy}{dx} = - \dfrac{3}{2} x^{-\frac{5}{2}}$$ $$\$$
$$\text{(3)}$$ $$\dfrac{dy}{dx} = 2ax^{(2a-1)}$$ $$\$$
$$\text{(4)}$$ $$\dfrac{du}{dt} = 2.4t^{1.4}$$ $$\$$
$$\text{(5)}$$ $$\dfrac{dz}{du} = \dfrac{1}{3} u^{-\frac{2}{3}}$$ $$\$$
$$\text{(6)}$$ $$\dfrac{dy}{dx} = -\dfrac{5}{3}x^{-\frac{8}{5}}$$ $$\$$
$$\text{(7)}$$ $$\dfrac{du}{dx} = -\dfrac{8}{5}x^{-\frac{13}{5}}$$ $$\$$
$$\text{(8)}$$ $$\dfrac{dy}{dx} = 2ax^{a-1}$$ $$\$$
$$\text{(9)}$$ $$\dfrac{dy}{dx} = \dfrac{3}{q} x^{\frac{3-q}{q}}$$ $$\$$
$$\text{(10)}$$ $$\dfrac{dy}{dx} = -\dfrac{m}{n} x^{-\frac{m+n}{n}}$$ $$\$$

## 練習問題 II

$$\text{(1)}$$ $$\dfrac{dy}{dx} = 3ax^{2}$$ $$\$$
$$\text{(2)}$$ $$\dfrac{dy}{dx} = 13 \times \frac{3}{2}x^{\frac{1}{2}}$$ $$\$$
$$\text{(3)}$$ $$\dfrac{dy}{dx} = 6x^{-\frac{1}{2}}$$ $$\$$
$$\text{(4)}$$ $$\dfrac{dy}{dx} = \dfrac{1}{2}c^{\frac{1}{2}} x^{-\frac{1}{2}}$$ $$\$$
$$\text{(5)}$$ $$\dfrac{du}{dz} = \dfrac{an}{c} z^{n-1}$$ $$\$$
$$\text{(6)}$$ $$\dfrac{dy}{dt} = 2.36t$$ $$\$$

$$\text{(7)}$$ $$\dfrac{dl_t}{dt} = 0.000012×l_0$$

$$\text{(8)}$$ $$\dfrac{dC}{dV} = abV^{b-1}$$ および $$0.98,\$$ $$3.00,\$$ $$7.47$$ 燭毎ボルト

$$\text{(9)}$$ \begin{aligned} \dfrac{dn}{dD} &= -\dfrac{1}{LD^{2}} \sqrt{\dfrac{gT}{\pi \sigma}}, \dfrac{dn}{dL} = -\dfrac{1}{DL^{2}} \sqrt{\dfrac{gT}{\pi \sigma}}, \\ \dfrac{dn}{d \sigma} &= -\dfrac{1}{2DL} \sqrt{\dfrac{gT}{\pi \sigma^{3}}}, \dfrac{dn}{dT} = \dfrac{1}{2DL} \sqrt{\dfrac{g}{\pi \sigma T}} \end{aligned}

$$\text{(10)}$$ \begin{aligned} \dfrac{t\ {\footnotesize \text{が変化するときの}}\ P\ {\footnotesize \text{の変化率}}}{D\ {\footnotesize \text{が変化するときの}}\ P\ {\footnotesize \text{の変化率}}} = -\dfrac{D}{t} \end{aligned}

$$\text{(11)}$$ $$2\pi,\$$ $$2\pi r,\$$ $$\pi l,\$$ $$\dfrac{2}{3}\pi rh,\$$ $$8\pi r,\$$ $$4\pi r^{2}$$

$$\text{(12)}$$ $$\dfrac{dD}{dT} = \dfrac{0.000012l_t}{\pi}$$

## 練習問題 III

$$\text{(1)}$$ \begin{aligned} \text{(a)}\ & 1 + x + \dfrac{x^{2}}{2} + \dfrac{x^{3}}{6} + \dfrac{x^{4}}{24} + \cdots \\ \text{(b)}\ & 2ax + b \\ \text{(c)}\ & 2x + 2a \\ \text{(d)}\ & 3x^{2} + 6ax + 3a^{2} \\ \end{aligned}

$$\text{(2)}$$ $$\dfrac{dw}{dt} = a - bt$$ $$\$$
$$\text{(3)}$$ $$\dfrac{dy}{dx} = 2x$$ $$\$$
$$\text{(4)}$$ $$14110x^{4} - 65404x^{3} - 2244x^{2} + 8192x + 1379$$
$$\text{(5)}$$ $$\dfrac{dx}{dy} = 2y + 8$$ $$\$$
$$\text{(6)}$$ $$185.9022654x^{2} + 154.36334$$ $$\$$
$$\text{(7)}$$ $$\dfrac{-5}{(3x + 2)^{2}}$$ $$\$$
$$\text{(8)}$$ $$\dfrac{6x^{4} + 6x^{3} + 9x^{2}}{(1 + x + 2x^{2})^{2}}$$ $$\$$
$$\text{(9)}$$ $$\dfrac{ad - bc}{(cx + d)^{2}}$$ $$\$$
$$\text{(10)}$$ $$\dfrac{anx^{-n-1} + bnx^{n-1} + 2nx^{-1}}{(x^{-n} + b)^{2}}$$ $$\$$
$$\text{(11)}$$ $$b + 2ct$$ $$\$$

$$\text{(12)}$$ $$R_0(a + 2bt),\$$ $$R_0 \left(a + \dfrac{b}{2\sqrt{t}}\right),\$$ $$-\dfrac{R_0(a + 2bt)}{(1 + at + bt^{2})^{2}} \left(= \dfrac{R^{2} (a + 2bt)}{R_0}\right)$$

$$\text{(13)}$$ $$1.4340(0.000014t - 0.001024),\$$ $$-0.00117,\$$ $$-0.00107,\$$ $$-0.00097$$

$$\text{(14)}$$ $$\dfrac{dE}{dl} = b + \dfrac{k}{i},\$$ $$\dfrac{dE}{di} = -\dfrac{c + kl}{i^{2}}$$

## 練習問題 IV

$$\text{(1)}$$ $$17 + 24x,\$$ $$24$$

$$\text{(2)}$$ $$\dfrac{x^{2} + 2ax - a}{(x + a)^{2}},\$$ $$\dfrac{2a(a + 1)}{(x + a)^{3}}$$

$$\text{(3)}$$ $$1 + x + \dfrac{x^{2}}{1 \times 2} + \dfrac{x^{3}}{1 \times 2 \times 3},\$$ $$1 + x + \dfrac{x^{2}}{1 \times 2}$$

$$\text{(1)}$$ \begin{aligned} \text{(a)}\ & \dfrac{d^{2}y}{dx^{2}} = \dfrac{d^{3} y}{dx^{3}} = 1 + x + \frac{1}{2}x^{2} + \frac{1}{6} x^{3} + \cdots\\ \text{(b)}\ & 2a,\ 0 \\ \text{(c)}\ & 2,\ 0 \\ \text{(d)}\ & 6x + 6a,\ 6 \\ \end{aligned}

$$\text{(2)}$$ $$-b,\ 0$$ $$\$$
$$\text{(3)}$$ $$2,\ 0$$ $$\$$

$$\text{(4)}$$ $$56440x^{3} - 196212x^{2} - 4488x + 8192,\ 169320x^{2} - 392424x - 4488$$

$$\text{(5)}$$ $$2,\ 0$$ $$\$$
$$\text{(6)}$$ $$371.80453x,\ 371.80453$$ $$\$$
$$\text{(7)}$$ $$\dfrac{30}{(3x + 2)^{3}},\ -\dfrac{270}{(3x + 2)^{4}}$$ $$\$$

$$\text{(1)}$$ $$\dfrac{6a}{b^{2}} x,\ \dfrac{6a}{b^{2}}$$ $$\$$
$$\text{(2)}$$ $$\dfrac{3a \sqrt{b}} {2 \sqrt{x}} - \dfrac{6b \sqrt[3]{a}}{x^{3}},\ \dfrac{18b \sqrt[3]{a}}{x^{4}} - \dfrac{3a \sqrt{b}}{4 \sqrt{x^{3}} }$$ $$\$$

$$\text{(3)}$$ $$\dfrac{2}{\sqrt[3]{\theta^{8}}} - \dfrac{1.056}{\sqrt[5]{\theta^{11}}},\$$ $$\dfrac{2.3232}{\sqrt[5]{\theta^{16}}} - \dfrac{16}{3 \sqrt[3]{\theta^{11}}}$$

$$\text{(4)}$$ \ \begin{aligned} & 810t^{4} - 648t^{3} + 479.52t^{2} - 139.968t + 26.64, \\ & 3240t^{3} - 1944t^{2} + 959.04t - 139.968 \end{aligned}

$$\text{(5)}$$ $$12x + 2,\ 12$$ $$\$$
$$\text{(6)}$$ $$6x^{2} - 9x,\ 12x - 9$$ $$\$$

$$\text{(7)}$$ \ \begin{aligned} &\dfrac{3}{4} \left(\dfrac{1}{\sqrt{\theta}} + \dfrac{1}{\sqrt{\theta^{5}}}\right) +\dfrac{1}{4} \left(\dfrac{15}{\sqrt{\theta^{7}}} - \dfrac{1}{\sqrt{\theta^{3}}}\right), \\ &\dfrac{3}{8} \left(\dfrac{1}{\sqrt{\theta^{5}}} - \dfrac{1}{\sqrt{\theta^{3}}}\right) -\dfrac{15}{8}\left(\dfrac{7}{\sqrt{\theta^{9}}} + \dfrac{1}{\sqrt{\theta^{7}}}\right) \end{aligned}

## 練習問題 V

$$\text{(2)}$$ $$64,\ 147.2,\ 0.32 \, {\footnotesize\text{フィート毎秒}}$$

$$\text{(3)}$$ $$x = a - gt,\$$ $$\ddot{x} = -g$$

$$\text{(4)}$$ $$45.1 \, {\footnotesize\text{フィート毎秒}}$$

$$\text{(5)}$$ $$12.4 \, {\footnotesize\text{フィート毎秒}},\$$ 一定である。

$$\text{(6)}$$ $${\footnotesize \text{角速度}} = 11.2 \, {\footnotesize\text{ラジアン毎秒}}$$ $${\footnotesize \text{角加速度}}= 9.6 \, {\footnotesize\text{ラジアン毎秒毎秒}}$$

$$\text{(7)}$$ $$v = 20.4t^{2} - 10.8,\$$ $$a = 40.8t,\$$ $$172.8 \, {\footnotesize\text{インチ毎秒}},\$$ $$122.4 \, {\footnotesize\text{インチ毎秒毎秒}}$$

$$\text{(8)}$$ $$v = \dfrac{1}{30 \sqrt[3]{(t - 125)^{2}}},\$$ $$a = - \dfrac{1}{45 \sqrt[3]{(t - 125)^{5}}}$$

$$\text{(9)}$$ $$v = 0.8 - \dfrac{8t}{(4 + t^{2})^{2}},$$ $$a = \dfrac{24t^{2} - 32}{(4 + t^{2})^{3}},\$$ $$0.7926,\$$ $$0.00211$$

$$\text{(10)}$$ $$n = 2,\$$ $$n = 11$$

## 練習問題 VI

$$\text{(1)}$$ $$\dfrac{x}{\sqrt{ x^{2} + 1}}$$ $$\$$
$$\text{(2)}$$ $$\dfrac{x}{\sqrt{ x^{2} + a^{2}} }$$ $$\$$
$$\text{(3)}$$ $$- \dfrac{1}{2 \sqrt{(a + x)^{3}} }$$ $$\$$
$$\text{(4)}$$ $$\dfrac{ax}{\sqrt{(a - x^{2})^{3}} }$$ $$\$$
$$\text{(5)}$$ $$\dfrac{2a^{2} - x^{2}}{x^{3} \sqrt{ x^{2} - a^{2}} }$$ $$\$$
$$\text{(6)}$$ $$\dfrac{\frac{3}{2} x^{2} \left[ \frac{8}{9} x \left( x^{3} + a \right) - \left( x^{4} + a \right) \right]}{(x^{4} + a)^{\frac{2}{3}} (x^{3} + a)^{\frac{3}{2}} }$$ $$\$$
$$\text{(7)}$$ $$\dfrac{2a \left(x - a \right)}{(x + a)^{3}}$$ $$\$$
$$\text{(8)}$$ $$\dfrac{5}{2} y^{3}$$ $$\$$
$$\text{(9)}$$ $$\dfrac{1}{(1 - \theta) \sqrt{1 - \theta^{2}} }$$ $$\$$

## 練習問題 VII

$$\text{(1)}$$ $$\dfrac{dw}{dx} = \dfrac{3x^{2} \left( 3 + 3x^{3} \right)} {27 \left(\frac{1}{2} x^{3} + \frac{1}{4} x^{6} \right)^{3}}$$

$$\text{(2)}$$ $$\dfrac{dv}{dx} = - \dfrac{12x}{\sqrt{1 + \sqrt{2} + 3x^{2}} \left(\sqrt{3} + 4 \sqrt{1 + \sqrt{2} + 3x^{2}}\right)^{2}}$$

$$\text{(3)}$$ $$\dfrac{du}{dx} = - \dfrac{x^{2} \left(\sqrt{3} + x^{3} \right)} {\sqrt{ \left[ 1 + \left( 1 + \dfrac{x^{3}}{\sqrt{3}} \right) ^{2} \right]^{3}}}$$

## 練習問題 VIII

$$\text{(2)}$$ $$1.44$$

$$\text{(4)}$$ 導関数は $$\dfrac{dy}{dx} = 3x^{2} + 3,\$$ で、対応する値は $$3,\$$ $$3 \dfrac{3}{4},\$$ $$6,\$$ $$15$$

$$\text{(5)}$$ $$± \sqrt{2}$$

$$\text{(6)}$$ $$\dfrac{dy}{dx} = - \dfrac{4}{9} \dfrac{x}{y},\$$ $$x = 0$$ で傾きは $$0,\$$ $$x = 1$$ で傾きは $$\mp \dfrac{1}{3 \sqrt{2}}$$

$$\text{(7)}$$ $$m = 4,\$$ $$n = -3$$

$$\text{(8)}$$ 交点は $$x = 1$$ と $$x = -3$$ で、角度はそれぞれ $$153^{\circ}\;26',\$$ $$2^{\circ}\;28'$$

$$\text{(9)}$$ 交点は $$x = 3.57, y = 3.50$$ で、角度は $$16^{\circ}\;16'$$

$$\text{(10)}$$ $$x = \dfrac{1}{3},$$ $$y = 2 \dfrac{1}{3},$$ $$b = -\dfrac{5}{3}$$

## 練習問題 IX

$$\text{(1)}$$ 極小値: $$x = 0,\ y = 0\qquad$$ 極大値: $$x = -2,\ y = -4$$

$$\text{(2)}$$ $$x = a$$.

$$\text{(4)}$$ $$25 \sqrt{3} \, {\footnotesize\text{平方インチ}}$$

$$\text{(5)}$$ $$\dfrac{dy}{dx} = - \dfrac{10}{x^{2}} + \dfrac{10}{(8 - x)^{2}}$$; $$x = 4,\$$ $$y = 5$$

$$\text{(6)}$$ 極大値 $$x = -1\qquad$$ 極小値 $$x = 1$$

$$\text{(7)}$$ 各辺の中点を結んだ四角形

$$\text{(8)}$$ $$r = \dfrac{2}{3} R,\ r = \dfrac{R}{2},\$$ 最大値は存在しない

$$\text{(9)}$$ $$r = R \sqrt{\dfrac{2}{3}},\ r = \dfrac{R}{\sqrt{2}},\ r = 0.8506R$$

$$\text{(10)}$$ $$\dfrac{8}{r} \, {\footnotesize\text{平方フィート毎秒}}$$

$$\text{(11)}$$ $$r = \dfrac{R \sqrt{8}}{3}$$

$$\text{(12)}$$ $$n = \sqrt{\dfrac{NR}{r}}$$

## 練習問題 X

$$\text{(1)}$$ 極大値: $$x = -2.19,\ y = 24.19\quad$$ 極小値: $$x = 1.52,\ y = -1.38$$

$$\text{(2)}$$ $$\dfrac{dy}{dx} = \dfrac{b}{a} - 2cx,\$$ $$\dfrac{d^{2} y}{dx^{2}} = -2c,\$$ $$x = \dfrac{b}{2ac}$$ (極大値)

$$\text{(3)}$$ $$\text{(a)}$$ 一つの極大値と二つの極小値 $$\$$ $$\text{(b)}$$ $$x = 0$$ に一つの極大値 (もう一つの解は虚数)

$$\text{(4)}$$ 極小値: $$x = 1.71,\ y = 6.14$$

$$\text{(5)}$$ 極大値: $$x = -0.5,\ y = 4$$

$$\text{(6)}$$ 極大値: $$x = 1.414,\ y = 1.7675,\$$ 極小値: $$x = -1.414,\ y = 1.7675$$

$$\text{(7)}$$ 極大値: $$x = -3.565,\ y = 2.12,\$$ 極小値: $$x = +3.565,\ y = 7.88$$

$$\text{(8)}$$ $$0.4N,\ 0.6N$$

$$\text{(9)}$$ $$x = \sqrt{\dfrac{a}{c}}$$

$$\text{(10)}$$ スピード: 毎時 $$8.66$$ 海里, $$\$$ 航海時間: $$115.47$$ 時間, $$\$$ 最小コスト: $$112$$ ポンド $$12$$ シリング

$$\text{(11)}$$ $$x = 7.5,\ y = ±5.414$$ で極大値および極小値を取る。

$$\text{(12)}$$ 極小値: $$x = \dfrac{1}{2},\ y= 0.25$$ $$\$$ 極大値: $$x = - \dfrac{1}{3},\ y= 1.408$$

## 練習問題 XI

$$\text{(1)}$$ $$\dfrac{2}{ x - 3} + \dfrac{1}{ x + 4}$$ $$\$$
$$\text{(2)}$$ $$\dfrac{1}{ x - 1} + \dfrac{2}{ x - 2}$$ $$\$$
$$\text{(3)}$$ $$\dfrac{2}{ x - 3} + \dfrac{1}{ x + 4}$$ $$\$$
$$\text{(4)}$$ $$\dfrac{5}{ x - 4} - \dfrac{4}{ x - 3}$$ $$\$$
$$\text{(5)}$$ $$\dfrac{19}{13(2x + 3)} - \dfrac{22}{13(3x - 2)}$$ $$\$$
$$\text{(6)}$$ $$\dfrac{2}{ x - 2} + \dfrac{4}{ x - 3} - \dfrac{5}{ x - 4}$$ $$\$$

$$\text{(7)}$$ $$\dfrac{1}{6(x - 1)} + \dfrac{11}{15(x + 2)} + \dfrac{1}{10(x - 3)}$$

$$\text{(8)}$$ $$\dfrac{7}{9(3x + 1)} + \dfrac{71}{63(3x - 2)} - \dfrac{5}{7(2x + 1)}$$

$$\text{(9)}$$ $$\dfrac{1}{3(x - 1)} + \dfrac{2x + 1}{3(x^{2} + x + 1)}$$ $$\$$
$$\text{(10)}$$ $$x + \dfrac{2}{3(x + 1)} + \dfrac{1 - 2x}{3(x^{2} - x + 1)}$$ $$\$$
$$\text{(11)}$$ $$\dfrac{3}{(x + 1)} + \dfrac{2x + 1}{x^{2} + x + 1}$$ $$\$$
$$\text{(12)}$$ $$\dfrac{1}{ x - 1} - \dfrac{1}{ x - 2} + \dfrac{2}{(x - 2)^{2}}$$ $$\$$

$$\text{(13)}$$ $$\dfrac{1}{4(x - 1)} - \dfrac{1}{4(x + 1)} + \dfrac{1}{2(x + 1)^{2}}$$

$$\text{(14)}$$ $$\dfrac{4}{9(x - 1)} - \dfrac{4}{9(x + 2)} - \dfrac{1}{3(x + 2)^{2}}$$

$$\text{(15)}$$ $$\dfrac{1}{ x + 2} - \dfrac{x - 1}{ x^{2} + x + 1} - \dfrac{1}{(x^{2} + x + 1)^{2}}$$

$$\text{(16)}$$ $$\dfrac{5}{ x + 4} -\dfrac{32}{(x + 4)^{2}} + \dfrac{36}{(x + 4)^{3}}$$

$$\text{(17)}$$ $$\dfrac{7}{9(3x - 2)^{2}} + \dfrac{55}{9(3x - 2)^{3}} + \dfrac{73}{9(3x - 2)^{4}}$$

$$\text{(18)}$$ $$\dfrac{1}{6(x - 2)} + \dfrac{1}{3(x - 2)^{2}} - \dfrac{x}{6(x^{2} + 2x + 4)}$$

## 練習問題 XII

$$\text{(1)}$$ $$ab(e^{ax} + e^{-ax})$$ $$\$$
$$\text{(2)}$$ $$2at + \dfrac{2}{t}$$ $$\$$
$$\text{(3)}$$ $$\log_e n$$ $$\$$
$$\text{(5)}$$ $$npv^{n-1}$$ $$\$$
$$\text{(6)}$$ $$\dfrac{n}{x}$$ $$\$$
$$\text{(7)}$$ $$\dfrac{3e^{- \frac{x}{x-1}} }{(x - 1)^{2}}$$ $$\$$
$$\text{(8)}$$ $$6x e^{-5x} - 5(3x^{2} + 1)e^{-5x}$$ $$\$$
$$\text{(9)}$$ $$\dfrac{ax^{a-1}}{x^a + a}$$ $$\$$

$$\text{(10)}$$ $$\left(\dfrac{6x}{3x^{2}-1} + \dfrac{1}{2\left(\sqrt x + x\right)}\right) \left(3x^{2}-1\right)\left(\sqrt x + 1\right)$$

$$\text{(11)}$$ $$\dfrac{1 - \log_e \left(x + 3\right)}{\left(x + 3\right)^{2}}$$ $$\$$
$$\text{(12)}$$ $$a^x\left(ax^{a-1} + x^a \log_e a\right)$$ $$\$$
$$\text{(14)}$$ 極小値: $$x = 0.694,\ y = 0.7$$ $$\$$
$$\text{(15)}$$ $$\dfrac{1 + x}{x}$$ $$\$$
$$\text{(16)}$$ $$\dfrac{3}{x} (\log_e ax)^{2}$$ $$\$$

## 練習問題 XIII

$$\text{(1)}$$ $$\dfrac{t}{T} = x$$ ($$t = 8x$$) として本文中に示した数表を使う。

$$\text{(2)}$$ $${\footnotesize \text{時定数}} = 34.627,\$$ $$1 {\footnotesize \text{{\footnotesize \text{\%}} となるまでの時間}} = 159.46$$ 分

$$\text{(3)}$$ $$2t = x$$ として数表を使う。

$$\text{(5)}$$ $$\text{(a) } x^x \left(1 + \log_e x\right)\$$ $$\text{(b) } 2x(e^x)^x\$$ $$\text{(c) } e^{x^x} \times x^x \left(1 + \log_e x\right)$$

$$\text{(6)}$$ $$0.14$$ 秒

$$\text{(7)}$$ $$\text{(a) }$$ $$1.642\$$ $$\text{(b) }$$ $$15.58$$

$$\text{(8)}$$ $$\mu = 0.00037,\$$ $$31.2 \, {\footnotesize\text{分}}$$

$$\text{(9)}$$ $$63.4 {\footnotesize \text{\%}},\$$ $$220 \, {\footnotesize\text{キロメートル}}$$

$$\text{(10)}$$ $$k = 0.133,\ 0.145,\ 0.155$$, $${\footnotesize \text{平均}} = 0.144,\$$ $${\footnotesize \text{誤差}} = -10.2 {\footnotesize \text{\%}},\ -0.9 {\footnotesize \text{\%}},\ +77.2{\footnotesize \text{\%}}$$

$$\text{(11)}$$ $$x = \dfrac{1}{e}$$ で極小値

$$\text{(12)}$$ $$x = e$$ で極小値

$$\text{(13)}$$ $$x = \log_e a$$ で極小値

## 練習問題 XIV

$$\text{(1)}$$ \begin{aligned} \text{(i)} \ \ & \dfrac{dy}{d\theta} = A \cos \left( \theta - \dfrac{\pi}{2} \right) \\ \text{(ii)} \ \ & \dfrac{dy}{d\theta} = 2\sin\theta \cos\theta = \sin2\theta,\ \dfrac{dy}{d\theta} = 2\cos2\theta \\ \text{(iii)} \ \ & \dfrac{dy}{d\theta} = 3\sin^{2} \theta \cos\theta,\ \dfrac{dy}{d\theta} = 3\cos3\theta \end{aligned}

$$\text{(2)}$$ $$\theta = 45^{\circ} = \dfrac{\pi}{4} \, {\footnotesize\text{ラジアン}}$$

$$\text{(3)}$$ $$\dfrac{dy}{dt} = -n \sin 2\pi nt$$

$$\text{(4)}$$ $$a^x \log_e a \cos a^x$$

$$\text{(5)}$$ $$\dfrac{\cos x}{\sin x} = \operatorname{cotan} x$$

$$\text{(6)}$$ $$18.2 \cos \left(x + 26^{\circ} \right)$$

$$\text{(7)}$$ 傾きは $$\dfrac{dy}{d\theta} = 100\cos\left(\theta - 15^{\circ} \right)$$ であり、$$(\theta -15^{\circ}) = 0$$ つまり $$\theta = 15^{\circ}$$ で極大値 $$100$$ を取る。$$\theta = 75^{\circ}$$ で傾きは $$100\cos(75^{\circ} - 15^{\circ}) = 100\cos 60^{\circ} = 100 \times \dfrac{1}{2} = 50$$ となる。

$$\text{(8)}$$ $$\small \cos\theta \sin2\theta + 2\cos2\theta \sin\theta = 2\sin\theta\left(\cos^{2} \theta + \cos2\theta\right) = 2\sin\theta\left(3\cos^{2} \theta - 1\right)$$

$$\text{(9)}$$ $$amn\theta^{n-1} \tan^{m-1}\left(\theta^n\right)\sec^{2} \theta^n$$

$$\text{(10)}$$ $$e^x \left(\sin^{2} x + \sin2x\right),\$$ $$e^x \left(\sin^{2} x + 2\sin2x + 2\cos2x\right)$$

$$\text{(11)}$$ $$\text{(i) } \dfrac{dy}{dx} = \dfrac{ab}{\left(x + b\right)^{2}}\ \$$ $$\text{(ii) }\ \dfrac{a}{b} e^{-\frac{x}{b}}\ \$$ $$\text{(iii) }\ \dfrac{1}{90}^{\circ} \times \dfrac{ab}{\left(b^{2} + x^{2}\right)}$$

(\text{(12)}\) \begin{aligned} & \text{(i)} & & \dfrac{dy}{dx} = \sec x \tan x \\ & \text{(ii)} & & \dfrac{dy}{dx} = - \dfrac{1}{\sqrt{ 1 - x^{2}}} \\ & \text{(iii)} & & \dfrac{dy}{dx} = \dfrac{1}{ 1 + x^{2}} \\ & \text{(iv)} & & \dfrac{dy}{dx} = \dfrac{1}{x \sqrt{ x^{2} - 1}} \\ & \text{(v)} & & \dfrac{dy}{dx} = \dfrac{\sqrt{ 3\sec x} \left(3\sec^{2} x - 1\right)}{2} \end{aligned}

$$\text{(13)}$$ $$\dfrac{dy}{d\theta} = 4.6\left(2\theta + 3\right)^{1.3} \cos\left(2\theta + 3\right)^{2.3}$$

$$\text{(14)}$$ $$\dfrac{dy}{d\theta} = 3\theta^{2} + 3\cos \left( \theta + 3 \right) - \log_e 3 \left( \cos\theta \times 3^{\sin\theta} + 3\theta \right)$$

$$\text{(15)}$$ $$\theta = \cot\theta,\ \theta = ±0.86$$; $$+0.86$$ で極大値、$$-0.86$$ で極小値

## 練習問題 XV

$$\text{(1)}$$ $$x^{3} - 6x^{2} y - 2y^{2},\ \dfrac{1}{3} - 2x^{3} - 4xy$$

$$\text{(2)}$$ \begin{aligned} 2xyz & + y^{2} z + z^{2} y + 2xy^{2} z^{2}, \\ 2xyz & + x^{2} z + xz^{2} + 2x^{2} yz^{2}, \\ 2xyz & + x^{2} y + xy^{2} + 2x^{2} y^{2} z \end{aligned}

$$\text{(3)}$$ $$\dfrac{1}{r} \{ \left(x - a\right) + \left( y - b \right) + \left( z - c \right) \} = \dfrac{ \left( x + y + z \right) - \left( a + b + c \right) }{r},\$$ $$\dfrac{3}{r}$$

$$\text{(4)}$$ $$dy = vu^{v-1}\, du + u^v \log_e u\, dv$$

$$\text{(5)}$$ \begin{aligned} dy & = 3\sin v u^{2}\, du + u^{3} \cos v\, dv, \\ dy & = u \sin x^{u-1} \cos x\, dx + (\sin x)^u \log_e \sin x du, \\ dy & = \dfrac{1}{v}\, \dfrac{1}{u}\, du - \log_e u \dfrac{1}{v^{2}}\, dv \end{aligned}

$$\text{(7)}$$ $$x = y = -\dfrac{1}{2}$$ で極小値

$$\text{(8)}$$ \begin{aligned} \text{(a) } & {\footnotesize \text{高さ}} = 2 \, {\footnotesize\text{フィート}}, {\footnotesize \text{幅}} = {\footnotesize \text{奥行き}} = 1 \, {\footnotesize\text{フィート}},\ {\footnotesize \text{体積}} = 2 \, {\footnotesize\text{立体フィート}} \\ \text{(b) } & {\footnotesize \text{半径}} = \dfrac{2}{\pi} \, {\footnotesize\text{フィート}},\ {\footnotesize \text{高さ}} = 2 \, {\footnotesize\text{フィート}}, {\footnotesize \text{体積}} = 2.54 \, {\footnotesize\text{立方フィート}} \end{aligned}

$$\text{(9)}$$ 三つの角度が等しいとき、積は極大になる。

$$\text{(10)}$$ $$x = y = 1$$ で極小

$$\text{(11)}$$ $$x = \dfrac{1}{2},\$$ $$y = 2$$ で極小

$$\text{(12)}$$ $${\footnotesize \text{頂点の角度}} = 90^{\circ},\$$ $${\footnotesize \text{三角形の長さの等しい辺の長さ}} = {\footnotesize \text{奥行き}} = \sqrt[3]{2V}$$

## 練習問題 XVI

$$\text{(1)}$$ $$\frac{4}{3}$$ $$\$$
$$\text{(2)}$$ $$0.6344$$ $$\$$
$$\text{(3)}$$ $$0.2624$$ $$\$$
$$\text{(4)}$$ $$\text{(a) } y = \frac{1}{8} x^{2} + C$$ $$\$$
$$\text{(4)}$$ $$\text{(b) } y = \sin x + C$$ $$\$$
$$\text{(5)}$$ $$y = x^{2} + 3x + C$$ $$\$$

## 練習問題 XVII

$$\text{(1)}$$ $$\dfrac{4\sqrt{a} x^{\frac{3}{2}} }{3} + C$$ $$\$$
$$\text{(2)}$$ $$-\dfrac{1}{x^{3}} + C$$ $$\$$
$$\text{(3)}$$ $$\dfrac{x^{4}}{4a} + C$$ $$\$$
$$\text{(4)}$$ $$\dfrac{1}{3} x^{3} + ax + C$$ $$\$$
$$\text{(5)}$$ $$-2x^{-\frac{5}{2}} + C$$ $$\$$
$$\text{(6)}$$ $$x^{4} + x^{3} + x^{2} + x + C$$ $$\$$
$$\text{(7)}$$ $$\dfrac{ax^{2}}{4} + \dfrac{bx^{3}}{9} + \dfrac{cx^{4}}{16} + C$$ $$\$$

$$\text{(8) }$$ $$\dfrac{x^{2} + a}{x + a} = x - a + \dfrac{a^{2} + a}{x + a} = \dfrac{x^{2}}{2} - ax + (a^{2} + a)\log_e (x + a) + C$$

$$\text{(9)}$$ $$\dfrac{x^{4}}{4} + 3x^{3} + \dfrac{27}{2} x^{2} + 27x + C$$ $$\$$
$$\text{(10)}$$ $$\dfrac{x^{3}}{3} + \dfrac{2 - a}{2} x^{2} - 2ax + C$$ $$\$$
$$\text{(11)}$$ $$a^{2} \left(2x^{\frac{3}{2}} + \dfrac{9}{4} x^{\frac{4}{3}}\right) + C$$ $$\$$
$$\text{(12)}$$ $$-\dfrac{1}{3} \cos\theta - \dfrac{1}{6} \theta + C$$ $$\$$
$$\text{(13)}$$ $$\dfrac{\theta}{2} + \dfrac{\sin 2a\theta}{4a} + C$$ $$\$$
$$\text{(14)}$$ $$\dfrac{\theta}{2} - \dfrac{\sin 2\theta}{4} + C$$ $$\$$
$$\text{(15)}$$ $$\dfrac{\theta}{2} - \dfrac{\sin 2a\theta}{4a} + C$$ $$\$$
$$\text{(16)}$$ $$\dfrac{1}{3} e^{3x} + C$$ $$\$$
$$\text{(17)}$$ $$\log(1 + x) + C$$ $$\$$
$$\text{(18)}$$ $$-\log_e (1 - x) + C$$ $$\$$

## 練習問題 XVIII

$$\text{(1)}$$ $${\footnotesize \text{面積}} = 60,\$$ $${\footnotesize \text{二次平均}} = 10$$ $$\$$
$$\text{(2)}$$ $${\footnotesize \text{面積}} = \dfrac{4}{3} a^{2}\sqrt{a}$$ $$\$$
$$\text{(3)}$$ $${\footnotesize \text{面積}} = 2,\$$ $${\footnotesize \text{二次平均}} = \dfrac{2}{\pi}$$ $$\$$
$$\text{(4)}$$ $${\footnotesize \text{面積}} = 1.57,\$$ $${\footnotesize \text{二次平均}} = 0.5$$ $$\$$
$$\text{(5)}$$ $$0.572,\$$ $$0.0476$$ $$\$$
$$\text{(6)}$$ $${\footnotesize \text{体積}} = \pi r^{2} \dfrac{h}{3}$$ $$\$$
$$\text{(7)}$$ $$1.25$$ $$\$$
$$\text{(8)}$$ $$79.4$$ $$\$$
$$\text{(9)}$$ $${\footnotesize \text{体積}} = 4.9348$$; $${\footnotesize \text{表面積}} = 12.57$$ $$\$$
$$\text{(10)}$$ $$a\log_e a,\$$ $$\dfrac{a}{a - 1} \log_e a$$ $$\$$

$$\text{(12)}$$ $${\footnotesize \text{算術平均}} = 9.5,\$$ $${\footnotesize \text{二次平均}} = 10.85$$

$$\text{(13)}$$ $${\footnotesize \text{算術平均}} = 0,\$$ $${\footnotesize \text{二次平均}} = \dfrac{1}{\sqrt{2}} \sqrt{A_1^{2} + A_3^{2}}$$

$$x$$ に $$0$$ を代入すると全ての項が $$0$$ となり、$$2\pi$$ を代入すると $$A_1^{2} \pi + A_3^{2} \pi$$ となる。

$$\text{(14)}$$ $${\footnotesize \text{面積}} = 62.6,\$$ $${\footnotesize \text{二次平均}} = 10.42$$

$$\text{(16)}$$ $$436.3$$

## 練習問題 XIX

$$\text{(1)}$$ $$\dfrac{x\sqrt{a^{2} - x^{2}} }{2} + \dfrac{a^{2}}{2} \sin^{-1} \dfrac{x}{a} + C$$ $$\$$
$$\text{(2)}$$ $$\dfrac{x^{2}}{2}(\log_e x - \frac{1}{2}) + C$$ $$\$$
$$\text{(3)}$$ $$\dfrac{x^{a+1}}{a + 1} \left(\log_e x - \dfrac{1}{a + 1}\right) + C$$ $$\$$
$$\text{(4)}$$ $$\sin e^x + C$$ $$\$$
$$\text{(5)}$$ $$\sin(\log_e x) + C$$ $$\$$
$$\text{(6)}$$ $$e^x (x^{2} - 2x + 2) + C$$ $$\$$
$$\text{(7)}$$ $$\dfrac{1}{a + 1} (\log_e x)^{a+1} + C$$ $$\$$
$$\text{(8)}$$ $$\log_e(\log_e x) + C$$ $$\$$

$$\text{(9)}$$ $$2\log_e(x - 1) + 3\log_e(x + 2) + C$$

$$\text{(10)}$$ $$\dfrac{1}{2} \log_e(x - 1) + \dfrac{1}{5} \log_e(x - 2) + \dfrac{3}{10} \log_e(x + 3) + C$$

$$\text{(11)}$$ $$\dfrac{b}{2a} \log_e \dfrac{x - a}{x + a} + C$$ $$\$$
$$\text{(12)}$$ $$\log_e \dfrac{x^{2} - 1}{x^{2} + 1} + C$$ $$\$$
$$\text{(13)}$$ $$\dfrac{1}{4} \log_e \dfrac{1 + x}{1 - x} + \dfrac{1}{2} \arctan x + C$$ $$\$$

$$\text{(14)}$$ $$\dfrac{1}{\sqrt{a}} \log_e \dfrac{\sqrt{a} - \sqrt{a - bx^{2}}}{x\sqrt{a}}$$

$$\text{(15)}$$ $$\dfrac{1}{\sqrt{a}} \log_e \dfrac{\sqrt{a} - \sqrt{a - bx^{2}}}{x\sqrt{a}}$$