練習問題メモ 01 - データサイエンス関連+αのメモ

次の関数を微分せよ。

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

\frac{dy}{dx} = 3x^2 - 6x

(1)

$\text { (2) } y=x(x-2)(x-4)(x-6)$

x(x-2)(x-4)(x-6) = (x^2 - 2x)(x-4)(x-6)\\ = (x^3 - 6x^2 + 8x)(x-6)\\ = x^4 - 6x^3 - 6x^3 + 36x^2 + 8x^2 - 48x\\ = x^4 - 12x^3 + 44x^2 - 48x\\

(2)

\frac{dy}{dx} = 4 x^3 - 36 x^2 + 88x - 48

(3)

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

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sp.expand(y).diff(x)

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

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

(4)

とおく。商の微分公式から

\begin{align} \frac{dy}{dx} &= \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{align}

(5)

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

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

(6)

\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = - \frac{1}{2} (x^2 + 2)^{-3/2} \times 2x\\ = - \frac{ x }{x^2 + 2^{3/2}}

(7)

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

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

商の公式を使う場合

\left(\frac{\cos x}{x}\right)' = \frac{-\sin x \cdot x - \cos x}{x^2}\\ = \frac{-x \sin x + \cos x}{x^2}\\

(8)

積の微分公式を使う

[\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}\\

(9)

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

積の公式を使って

\begin{align} [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 bx + b \cos b x)\\ &= e^{a x} \cdot (a \cos b x + a \sin b x) + e^{a x} \cdot (-b \sin bx + b \cos b x)\\ &= e^{a x} [(a \cos b x) + (a \sin b x) -b \sin bx) + (b \cos b x)]\\ &= e^{a x} [(a +b )\cos b x + (a - b) \sin b x]\\ \end{align}

(10)

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)

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

$u=x^2+2$ として

[\tan (x^2+2)]' = \frac{dy}{du} \frac{du}{dx}\\ = \frac{1}{\cos^2 u} 2 x\\ = \frac{1}{\cos^2 (x^2+2)} 2 x\\ = \frac{ 2 x}{\cos^2 (x^2+2)}\\

(13)

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

(\cos^2 x)' = -\sin x \cos x + \cos x \cdot -\sin x\\ = -2 \sin x \cos x

(14)

なので

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

(15)

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

\begin{align} \frac{dy}{dx} &= \frac{d (\log \{\log (\log x)\})}{d \{\log (\log x)\}} \frac{d \{\log (\log x)\}}{d (\log x)} \frac{d (\log x)}{dx}\\ &= \frac{1}{\log (\log x)} \frac{1}{\log x} \frac{1}{x}\\ &= \frac{1}{\log (\log x) \cdot \log x \cdot x} \end{align}

(16)

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

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)}{dx} = e^{x^2+3 x} ( 2x+3)

(17)

$e^{x^2+3 x} \log x$ は積の微分公式で

\begin{align} \frac{dy}{dx} &= (e^{x^2+3 x})' \log x + e^{x^2+3 x} (\log x)' \\ &= e^{x^2+3 x} ( 2x+3)\log x + e^{x^2+3 x} \frac{1}{x}\\ &= e^{x^2+3 x} \left( (2x+3)\log x + \frac{1}{x} \right)\\ \end{align}

(18)

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

積の微分公式より

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

(19)

$(8^x)'$ 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

(20)

よって

(8^x)' x^3 + 8^x (x^3)' = 8^x \log 8 \cdot x^3 + 8^x 3x^2\\ = 8^x (x^3 \log 8 + 3x^2)\\

(21)

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

まず第1項は

\begin{align} \left( \frac{x}{\sqrt{a^2-x^2}} \right)' &= \frac{\sqrt{a^2-x^2} - x \frac{1}{2} (a^2-x^2)^{-1/2} \cdot (-2x) }{(\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{align}

(22)

つづいて第2項は

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

(23)

より

\begin{align} \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{align}

(24)

よって

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

(25)

整理すると

\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}}}

(26)

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

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

(27)

より、

\left[ \arctan \left(\frac{b}{a} \tan x\right)\right]' = \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{ab}{a^2 \cos^2 x}\\ = \frac{ab}{(\frac{b^2}{a^2} \frac{\sin^2 x}{\cos^2 x} + 1) a^2 \cos^2 x}\\ = \frac{ab}{ b^2 \sin^2 x + a^2 \cos^2 x }\\

(28)