数学 - 解析学 - 微分と基本的な関数 - 微分係数、導関数 - 極限

解析入門原書第3版
原著

学習環境

Surface 3 (4G LTE)、Surface 3 タイプカバー、Surface ペン(端末)
Windows 10 Pro (OS)
数式入力ソフト(TeX, MathML): MathType
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解析入門原書第3版 (S.ラング(著)、松坂和夫(翻訳)、片山孝次(翻訳)、岩波書店)の第2部(微分と基本的な関数)、第3章(微分係数、導関数)、3(極限)、練習問題1-10.を取り組んでみる。

練習問題1-10.

$\begin{array}{l} f' (x) = \lim_{h \to 0} \frac{2 {(x + h)}^{2} + 3 (x + h) - (2 x^{2} + 3 x)}{h} \\ = \lim_{h \to 0} \frac{4 h x + 2 h^{2} + 3 h}{h} \\ = \lim_{h \to 0} (4 x + 2 h + 3) \\ = \lim_{h \to 0} 4 x + \lim_{h \to 0} 2 h + \lim_{h \to 0} 3 \\ = 4 x + 3 \end{array}$
$\begin{array}{l} f' (x) = \lim_{h \to 0} \frac{\frac{1}{2 (x + h) + 1} - \frac{1}{2 x + 1}}{h} \\ = \lim_{h \to 0} \frac{- 2 h}{h (2 x + 2 h + 1) (2 x + 1)} \\ = \lim_{h \to 0} \frac{- 2}{(2 x + 2 h + 1) (2 x + 1)} \\ = \frac{\lim_{h \to 0} (- 2)}{\lim_{h \to 0} (2 x + 2 h + 1) \cdot \lim_{h \to 0} (2 x + 1)} \\ = \frac{- 2}{{(2 x + 1)}^{2}} \end{array}$
$\begin{array}{l} f' (x) = \lim_{h \to 0} \frac{\frac{x + h}{x + h + 1} - \frac{x}{x + 1}}{h} \\ = \lim_{h \to 0} \frac{(x + h) (x + 1) - x (x + h + 1)}{h (x + h + 1) (x + 1)} \\ = \lim_{h \to 0} \frac{h}{h (x + h + 1) (x + 1)} \\ = \frac{1}{\lim_{h \to 0} (x + h + 1) \cdot \lim_{h \to 0} (x + 1)} \\ = \frac{1}{{(x + 1)}^{2}} \end{array}$
$\begin{array}{l} f' (x) = \lim_{h \to 0} \frac{(x + h) (x + h + 1) - x (x + 1)}{h} \\ = \lim_{h \to 0} \frac{2 h x + h}{h} \\ = \lim_{h \to 0} (2 x + 1) \\ = 2 x + 1 \end{array}$
$\begin{array}{l} f' (x) = \lim_{h \to 0} \frac{\frac{x + h}{2 (x + h) - 1} - \frac{x}{2 x - 1}}{h} \\ = \lim_{h \to 0} \frac{(x + h) (2 x - 1) - x (2 x + 2 h - 1)}{h (2 x + 2 h - 1) (2 x - 1)} \\ = \lim_{h \to 0} \frac{- h}{h (2 x + 2 h - 1) (2 x - 1)} \\ = \frac{\lim_{h \to 0} (- 1)}{\lim_{h \to 0} (2 x + 2 h - 1) \cdot \lim_{h \to 0} (2 x - 1)} \\ = \frac{- 1}{{(2 x - 1)}^{2}} \end{array}$
$\begin{array}{l} f' (x) = \lim_{h \to 0} \frac{3 {(x + h)}^{3} - 3 x^{3}}{h} \\ = \lim_{h \to 0} \frac{3 (3 h x^{2} + 3 h^{2} x + h^{3})}{h} \\ = \lim_{h \to 0} 3 \cdot (\lim_{h \to 0} 3 x^{2} + \lim_{h \to 0} 3 h x + \lim_{h \to 0} h^{2}) \\ = 3 (3 x^{2}) \\ = 9 x^{2} \end{array}$
$\begin{array}{l} f' (x) = \lim_{h \to 0} \frac{{(x + h)}^{4} - x^{4}}{h} \\ = \lim_{h \to 0} \frac{4 h x^{3} + 6 h^{2} x^{2} + 4 h^{3} x + h^{4}}{h} \\ = \lim_{h \to 0} 4 x^{3} + \lim_{h \to 0} 6 h x^{2} + \lim_{h \to 0} h^{2} x + \lim_{h \to 0} h^{3} \\ = 4 x^{3} \end{array}$
$\begin{array}{l} f' (x) = \lim_{h \to 0} \frac{{(x + h)}^{5} - x^{5}}{h} \\ = \lim_{h \to 0} \frac{5 h x^{4} + 10 h^{2} x^{3} + 10 h^{3} x^{2} + 5 h^{4} x + h^{5}}{h} \\ = \lim_{h \to 0} 5 x^{4} + \lim_{h \to 0} 10 h x^{3} + \lim_{h \to 0} 10 h^{2} x + \lim_{h \to 0} 5 h^{3} x + \lim_{h \to 0} h^{4} \\ = 5 x^{4} \end{array}$
$\begin{array}{l} f' (x) = \lim_{h \to 0} \frac{2 {(x + h)}^{3} - 2 x^{3}}{h} \\ = \lim_{h \to 0} \frac{2 (3 h x^{2} + 3 h^{2} x + h^{3})}{h} \\ = \lim_{h \to 0} 2 \cdot (\lim_{h \to 0} 3 x^{2} + \lim_{h \to 0} 3 h x + \lim_{h \to 0} h^{2}) \\ = 6 x^{2} \end{array}$
$\begin{array}{l} f' (x) = \lim_{h \to 0} \frac{(\frac{1}{2} {(x + h)}^{3} + (x + h)) - (\frac{1}{2} x^{3} + x)}{h} \\ = \lim_{h \to 0} \frac{\frac{1}{2} (3 h x^{2} + 3 h^{2} x + h^{3}) + h}{h} \\ = \lim_{h \to 0} \frac{1}{2} \cdot (\lim_{h \to 0} 3 x^{2} + \lim_{h \to 0} 3 h x + \lim_{h \to 0} h^{2}) + \lim_{h \to 0} 1 \\ = \frac{3}{2} x^{2} + 1 \end{array}$

Kamimura's blog

2017年1月27日金曜日

数学 - 解析学 - 微分と基本的な関数 - 微分係数、導関数 - 極限

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