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

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

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

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

練習問題 III

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

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

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

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

第六章の例

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

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

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

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

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

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

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

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

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

二次平均の計算には難しい積分が関係する。定義から \[ {\footnotesize \text{二次平均}} = \sqrt{\dfrac{1}{2\pi} \int_0^{2\pi} (A_1 \sin x + A_3 \sin 3x)^{2}\, dx} \] が分かる。積分 \[ \int (A_1^{2} \sin^{2} x + 2A_1 A_3 \sin x \sin 3x + A_3^{2} \sin^{2} 3x)\, dx \] の計算では、\(\sin^{2} x = \dfrac{1 - \cos 2x}{2}\) から分かる \[ \sin^{2} 3x = \dfrac{1 - \cos 6x}{2} \] と加法定理から分かる \(2\sin x \sin 3x = \cos 2x - \cos 4x\) を使うと計算でき、次の結果となる: \[ \dfrac{A_1^{2}}{2} \left( x - \dfrac{\sin 2x}{2} \right) + A_1 A_3 \left( \dfrac{\sin 2x}{2} - \dfrac{\sin 4x}{4} \right) + \dfrac{A_3^{2}}{2} \left( x - \dfrac{\sin 6x}{6} \right) \]

\(x\) 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

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

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

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

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

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

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

答を微分すると元の式に戻ることを確認するとよい。

熱心な学生には自身の力を試すために他にも練習問題を作って解いてみることを勧める。積分結果を微分して元に戻るかを見れば、計算が正しいかを確認できる。

理解を深めるための練習問題が載っている本はいくつもある。例えば R. G. Blaine 著 The Calculus and its Applications や F. M. Saxelby 著 A Course in Practical Mathematics がある。



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