練習問題メモ 01

練習問題メモ 01#

$(1) y = x^{3} - 3 x^{2} + 2$

\frac{d y}{d x} = 3 x^{2} - 6 x

$(2) y = x (x - 2) (x - 4) (x - 6)$

\begin{array}{r} x (x - 2) (x - 4) (x - 6) = (x^{2} - 2 x) (x - 4) (x - 6) \\ = (x^{3} - 6 x^{2} + 8 x) (x - 6) \\ = x^{4} - 6 x^{3} - 6 x^{3} + 36 x^{2} + 8 x^{2} - 48 x \\ = x^{4} - 12 x^{3} + 44 x^{2} - 48 x \end{array}

\frac{d y}{d x} = 4 x^{3} - 36 x^{2} + 88 x - 48

import sympy as sp
x = sp.symbols("x")
y = x * (x - 2) * (x - 4) * (x - 6)
sp.expand(y)

x^{4} - 12 x^{3} + 44 x^{2} - 48 x

sp.expand(y).diff(x)

4 x^{3} - 36 x^{2} + 88 x - 48

$(3) y = \frac{5 + x}{5 - x}$

y = \frac{5 + x}{5 - x} = \frac{g (x)}{m (x)}

とおく。商の微分公式から

\begin{array}{r} \begin{aligned} \frac{d y}{d x} & = \frac{g^{'} (x) m (x) - g (x) m^{'} (x)}{m (x)^{2}} \\ = \frac{(5 + x)^{'} (5 - x) - (5 + x) (5 - x)^{'}}{(5 - x)^{2}} \\ = \frac{(5 - x) + (5 + x)}{(5 - x)^{2}} \\ = \frac{10}{(5 - x)^{2}} \end{aligned} \end{array}

$(4) y = \frac{1}{\sqrt{x^{2} + 2}}$

y = (x^{2} + 2)^{- 1 / 2}

\begin{array}{r} \frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x} = - \frac{1}{2} (x^{2} + 2)^{- 3 / 2} \times 2 x \\ = - \frac{x}{x^{2} + 2^{3 / 2}} \end{array}

import sympy as sp
x = sp.symbols("x")
y = 1/sp.sqrt(x**2 + 2)
y.diff(x)

- \frac{x}{{(x^{2} + 2)}^{\frac{3}{2}}}

$(5) y = \frac{\cos x}{x}$

商の公式を使う場合

\begin{array}{r} {(\frac{\cos x}{x})}^{'} = \frac{- \sin x \cdot x - \cos x}{x^{2}} \\ = \frac{- x \sin x + \cos x}{x^{2}} \end{array}

積の微分公式を使う

\begin{array}{r} [\cos x \cdot x^{- 1}]^{'} = - \sin x \cdot x^{- 1} + \cos x \cdot - x^{- 2} \\ = \frac{- \sin x}{x} + \frac{\cos x}{x^{2}} \\ = \frac{- x \sin x + \cos x}{x^{2}} \end{array}

$(6) y = e^{a x} (\cos b x + \sin b x)$

積の公式を使って

\begin{array}{r} \begin{aligned} [e^{a x} \cdot (\cos b x + \sin b x)]^{'} & = a e^{a x} \cdot (\cos b x + \sin b x) + e^{a x} \cdot (- b \sin b x + b \cos b x) \\ = e^{a x} \cdot (a \cos b x + a \sin b x) + e^{a x} \cdot (- b \sin b x + b \cos b x) \\ = e^{a x} [(a \cos b x) + (a \sin b x) - b \sin b x) + (b \cos b x)] \\ = e^{a x} [(a + b) \cos b x + (a - b) \sin b x] \end{aligned} \end{array}

import sympy as sp
a, b, x = sp.symbols("a b x")
y = sp.exp(a * x) * (sp.cos(b * x) + sp.sin(b * x))
y.diff(x)

a (\sin (b x) + \cos (b x)) e^{a x} + (- b \sin (b x) + b \cos (b x)) e^{a x}

$(7) y = \tan (x^{2} + 2)$

$u = x^{2} + 2$ として

\begin{array}{r} [\tan (x^{2} + 2)]^{'} = \frac{d y}{d u} \frac{d u}{d x} \\ = \frac{1}{\cos^{2} u} 2 x \\ = \frac{1}{\cos^{2} (x^{2} + 2)} 2 x \\ = \frac{2 x}{\cos^{2} (x^{2} + 2)} \end{array}

$(8) y = \log (\cos^{2} x)$

\begin{array}{r} (\cos^{2} x)^{'} = - \sin x \cos x + \cos x \cdot - \sin x \\ = - 2 \sin x \cos x \end{array}

なので

\begin{array}{r} \begin{aligned} y^{'} & = \frac{d (\log u)}{d u} \frac{d u}{d x} \\ = \frac{1}{\cos^{2} x} \times - 2 \sin x \cos x \\ = - 2 \frac{\sin x}{\cos x} \\ = - 2 \tan x \end{aligned} \end{array}

$(9) y = \log {\log (\log x)}$

\begin{array}{r} \begin{aligned} \frac{d y}{d x} & = \frac{d (\log {\log (\log x)})}{d {\log (\log x)}} \frac{d {\log (\log x)}}{d (\log x)} \frac{d (\log x)}{d x} \\ = \frac{1}{\log (\log x)} \frac{1}{\log x} \frac{1}{x} \\ = \frac{1}{\log (\log x) \cdot \log x \cdot x} \end{aligned} \end{array}

$(10) y = e^{x^{2} + 3 x} \log x$

まず $e^{x^{2} + 3 x}$ は合成関数の微分公式で

\frac{d e^{x^{2} + 3 x}}{d (x^{2} + 3 x)} \frac{d (x^{2} + 3 x)}{d x} = e^{x^{2} + 3 x} (2 x + 3)

$e^{x^{2} + 3 x} \log x$ は積の微分公式で

\begin{array}{r} \begin{aligned} \frac{d y}{d x} & = (e^{x^{2} + 3 x})^{'} \log x + e^{x^{2} + 3 x} (\log x)^{'} \\ = e^{x^{2} + 3 x} (2 x + 3) \log x + e^{x^{2} + 3 x} \frac{1}{x} \\ = e^{x^{2} + 3 x} ((2 x + 3) \log x + \frac{1}{x}) \end{aligned} \end{array}

$(11) y = 8^{x} x^{3}$

積の微分公式より

(8^{x})^{'} x^{3} + 8^{x} (x^{3})^{'}

$(8^{x})^{'}$ は対数をとって逆関数の微分として解くと $(a^{x})^{'} = x \log a$ となることから

(8^{x})^{'} = 8^{x} \log 8

よって

\begin{array}{r} (8^{x})^{'} x^{3} + 8^{x} (x^{3})^{'} = 8^{x} \log 8 \cdot x^{3} + 8^{x} 3 x^{2} \\ = 8^{x} (x^{3} \log 8 + 3 x^{2}) \end{array}

$(12) y = \frac{x}{\sqrt{a^{2} - x^{2}}} - \arcsin \frac{x}{a}$

まず第1項は

\begin{array}{r} \begin{aligned} {(\frac{x}{\sqrt{a^{2} - x^{2}}})}^{'} & = \frac{\sqrt{a^{2} - x^{2}} - x \frac{1}{2} (a^{2} - x^{2})^{- 1 / 2} \cdot (- 2 x)}{(\sqrt{a^{2} - x^{2}})^{2}} \\ = \frac{\sqrt{a^{2} - x^{2}}}{(\sqrt{a^{2} - x^{2}})^{2}} + \frac{x^{2} (a^{2} - x^{2})^{- 1 / 2}}{(\sqrt{a^{2} - x^{2}})^{2}} \\ = \frac{1}{\sqrt{a^{2} - x^{2}}} + \frac{x^{2}}{(a^{2} - x^{2})^{\frac{3}{2}}} \end{aligned} \end{array}

つづいて第2項は

\frac{d}{d x} \arcsin x = \frac{1}{\sqrt{1 - x^{2}}}

より

\begin{array}{r} \begin{aligned} \frac{d}{d x} \arcsin \frac{x}{a} & = \frac{d \arcsin u}{d u} \frac{d (1 / a) x}{d x} \\ = \frac{1}{\sqrt{1 - (\frac{x}{a})^{2}}} \cdot \frac{1}{a} \\ = \frac{1}{a \sqrt{1 - \frac{x^{2}}{a^{2}}}} \\ = \frac{1}{\sqrt{a^{2} (1 - \frac{x^{2}}{a^{2}})}} \\ = \frac{1}{\sqrt{a^{2} - x^{2}}} \end{aligned} \end{array}

よって

{(\frac{x}{\sqrt{a^{2} - x^{2}}} - \arcsin \frac{x}{a})}^{'} = \frac{1}{\sqrt{a^{2} - x^{2}}} + \frac{x^{2}}{(a^{2} - x^{2})^{\frac{3}{2}}} - \frac{1}{\sqrt{a^{2} - x^{2}}}

整理すると

\begin{array}{r} \frac{1}{\sqrt{a^{2} - x^{2}}} + \frac{x^{2}}{(a^{2} - x^{2})^{\frac{3}{2}}} - \frac{1}{\sqrt{a^{2} - x^{2}}} \\ = \frac{(a^{2} - x^{2})}{\sqrt{a^{2} - x^{2}} (a^{2} - x^{2})} + \frac{x^{2}}{\sqrt{a^{2} - x^{2}} (a^{2} - x^{2})} - \frac{(a^{2} - x^{2})}{\sqrt{a^{2} - x^{2}} (a^{2} - x^{2})} \\ = \frac{(a^{2} - x^{2}) + x^{2} - (a^{2} - x^{2})}{\sqrt{a^{2} - x^{2}} (a^{2} - x^{2})} \\ = \frac{x^{2}}{\sqrt{a^{2} - x^{2}} (a^{2} - x^{2})} \\ = \frac{x^{2}}{(a^{2} - x^{2})^{\frac{3}{2}}} \end{array}

(13) $y = \arctan (\frac{b}{a} \tan x)$

\begin{array}{r} (\arctan (x))^{'} = \frac{1}{x^{2} + 1} \\ (\tan (x))^{'} = \frac{1}{\cos^{2} x} \end{array}

より、

\begin{array}{r} {[\arctan (\frac{b}{a} \tan x)]}^{'} = \frac{1}{\frac{b^{2}}{a^{2}} \tan^{2} x + 1} \cdot \frac{b}{a} \frac{1}{\cos^{2} x} \\ = \frac{1}{\frac{b^{2}}{a^{2}} \frac{\sin^{2} x}{\cos^{2} x} + 1} \cdot \frac{b}{a \cos^{2} x} \\ = \frac{1}{\frac{b^{2}}{a^{2}} \frac{\sin^{2} x}{\cos^{2} x} + 1} \cdot \frac{a b}{a^{2} \cos^{2} x} \\ = \frac{a b}{(\frac{b^{2}}{a^{2}} \frac{\sin^{2} x}{\cos^{2} x} + 1) a^{2} \cos^{2} x} \\ = \frac{a b}{b^{2} \sin^{2} x + a^{2} \cos^{2} x} \end{array}