数学 - 解析学 - 微分と基本的な関数 - 微分係数、導関数 - 導関数(ニュートン商、極限、直線の方程式)

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
• MathML対応ブラウザ: Firefox、Safari
• MathML非対応ブラウザ(Internet Explorer, Google Chrome...)用JavaScript Library: MathJax

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)、2(導関数)、練習問題1-11を取り組んでみる。

練習問題1-10

1. 
   $\begin{array}{l}\underset{h\to 0}{\lim }\frac{{\left(x+h\right)}^2+1-\left(x^2+1\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{2hx+h^2}{h}\\ =\underset{h\to 0}{\lim }\left(2x+h\right)\\ =2x\end{array}$
2. 
   $\begin{array}{l}\underset{h\to 0}{\lim }\frac{{\left(x+h\right)}^3-x^3}{h}\\ =\underset{h\to 0}{\lim }\frac{3x^{2}h+3x{h}^2+h^3}{h}\\ =\underset{h\to 0}{\lim }\left(3x^2+3xh+h^2\right)\\ =3x^2\end{array}$
3. 
   $\begin{array}{l}\underset{h\to 0}{\lim }\frac{2{\left(x+h\right)}^3-2x^3}{h}\\ =\underset{h\to 0}{\lim }\frac{6x^{2}h+6x{h}^2+2h^3}{h}\\ =\underset{h\to 0}{\lim }\left(6x^2+6xh+2h^2\right)\\ =6x^2\end{array}$
4. 
   $\begin{array}{l}\underset{h\to 0}{\lim }\frac{3{\left(x+h\right)}^2-3x^2}{h}\\ =\underset{h\to 0}{\lim }\frac{6xh+3h^2}{h}\\ =\underset{h\to 0}{\lim }\left(6x+3h\right)\\ =6x\end{array}$
5. 
   $\begin{array}{l}\underset{h\to 0}{\lim }\frac{{\left(x+h\right)}^2-5-\left(x^2-5\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{2xh+h^2}{h}\\ =\underset{h\to 0}{\lim }\left(2x+h\right)\\ =2x\end{array}$
6. 
   $\begin{array}{l}\underset{h\to 0}{\lim }\frac{2{\left(x+h\right)}^2+\left(x+h\right)-\left(2x^2+x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{4xh+2h^2+h}{h}\\ =\underset{h\to 0}{\lim }\left(4x+2h+1\right)\\ =4x+1\end{array}$
7. 
   $\begin{array}{l}\underset{h\to 0}{\lim }\frac{2{\left(x+h\right)}^2-3\left(x+h\right)-\left(2x^2-3x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{4xh+2h^2-3h}{h}\\ =\underset{h\to 0}{\lim }\left(4x+2h-3\right)\\ =4x-3\end{array}$
8. 
   $\begin{array}{l}\underset{h\to 0}{\lim }\frac{\frac{1}{2}{\left(x+h\right)}^3+2\left(x+h\right)-\left(\frac{1}{2}{x}^3+2x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{3}{2}{x}^{2}h+\frac{3}{2}x{h}^2+\frac{1}{2}{h}^3+2h}{h}\\ =\underset{h\to 0}{\lim }\left(\frac{3}{2}{x}^2+\frac{3}{2}xh+\frac{1}{2}{h}^2+2\right)\\ =\frac{3}{2}{x}^2+2\end{array}$
9. 
   $\begin{array}{l}\underset{h\to 0}{\lim }\frac{\frac{1}{x+h+1}-\frac{1}{x+1}}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{x+1-x-h-1}{\left(x+h+1\right)\left(x+1\right)}}{h}\\ =\underset{h\to 0}{\lim }\frac{-1}{\left(x+h+1\right)\left(x+1\right)}\\ =-\frac{1}{{\left(x+1\right)}^2}\end{array}$
10. 
    $\begin{array}{l}\underset{h\to 0}{\lim }\frac{\frac{2}{x+h+1}-\frac{2}{x+1}}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{2x+2-2x-2h-2}{\left(x+h+1\right)\left(x+1\right)}}{h}\\ =\underset{h\to 0}{\lim }\frac{-2}{\left(x+h+1\right)\left(x+1\right)}\\ =-\frac{2}{{\left(x+1\right)}^2}\end{array}$

練習問題11.

1. 
   $\begin{array}{l}y-5=4\left(x-2\right)\\ y=4x-3\end{array}$
2. 
   $\begin{array}{l}y-8=12\left(x-2\right)\\ y=12x-16\end{array}$
3. 
   $\begin{array}{l}y-16=24\left(x-2\right)\\ y=24x-32\end{array}$
4. 
   $\begin{array}{l}y-12=12\left(x-2\right)\\ y=12x-12\end{array}$
5. 
   $\begin{array}{l}y+1=4\left(x-2\right)\\ y=4x-9\end{array}$
6. 
   $\begin{array}{l}y-10=9\left(x-2\right)\\ y=9x-8\end{array}$
7. 
   $\begin{array}{l}y-2=5\left(x-2\right)\\ y=5x-8\end{array}$
8. 
   $\begin{array}{l}y-8=8\left(x-2\right)\\ y=8x-8\end{array}$
9. 
   $\begin{array}{l}y-\frac{1}{3}=-\frac{1}{9}\left(x-2\right)\\ y=-\frac{1}{9}x+\frac{5}{9}\end{array}$
10. 
    $\begin{array}{l}y-\frac{2}{3}=-\frac{2}{9}\left(x-2\right)\\ y=-\frac{2}{9}x+\frac{10}{9}\end{array}$