## 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

1. $\begin{array}{l}f\text{'}\left(x\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2{\left(x+h\right)}^{2}+3\left(x+h\right)-\left(2{x}^{2}+3x\right)}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{4hx+2{h}^{2}+3h}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\left(4x+2h+3\right)\\ =\underset{h\to 0}{\mathrm{lim}}4x+\underset{h\to 0}{\mathrm{lim}}2h+\underset{h\to 0}{\mathrm{lim}}3\\ =4x+3\end{array}$

2. $\begin{array}{l}f\text{'}\left(x\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\frac{1}{2\left(x+h\right)+1}-\frac{1}{2x+1}}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{-2h}{h\left(2x+2h+1\right)\left(2x+1\right)}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{-2}{\left(2x+2h+1\right)\left(2x+1\right)}\\ =\frac{\underset{h\to 0}{\mathrm{lim}}\left(-2\right)}{\underset{h\to 0}{\mathrm{lim}}\left(2x+2h+1\right)·\underset{h\to 0}{\mathrm{lim}}\left(2x+1\right)}\\ =\frac{-2}{{\left(2x+1\right)}^{2}}\end{array}$

3. $\begin{array}{l}f\text{'}\left(x\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\frac{x+h}{x+h+1}-\frac{x}{x+1}}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{\left(x+h\right)\left(x+1\right)-x\left(x+h+1\right)}{h\left(x+h+1\right)\left(x+1\right)}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{h}{h\left(x+h+1\right)\left(x+1\right)}\\ =\frac{1}{\underset{h\to 0}{\mathrm{lim}}\left(x+h+1\right)·\underset{h\to 0}{\mathrm{lim}}\left(x+1\right)}\\ =\frac{1}{{\left(x+1\right)}^{2}}\end{array}$

4. $\begin{array}{l}f\text{'}\left(x\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\left(x+h\right)\left(x+h+1\right)-x\left(x+1\right)}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{2hx+h}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\left(2x+1\right)\\ =2x+1\end{array}$

5. $\begin{array}{l}f\text{'}\left(x\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\frac{x+h}{2\left(x+h\right)-1}-\frac{x}{2x-1}}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{\left(x+h\right)\left(2x-1\right)-x\left(2x+2h-1\right)}{h\left(2x+2h-1\right)\left(2x-1\right)}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{-h}{h\left(2x+2h-1\right)\left(2x-1\right)}\\ =\frac{\underset{h\to 0}{\mathrm{lim}}\left(-1\right)}{\underset{h\to 0}{\mathrm{lim}}\left(2x+2h-1\right)·\underset{h\to 0}{\mathrm{lim}}\left(2x-1\right)}\\ =\frac{-1}{{\left(2x-1\right)}^{2}}\end{array}$

6. $\begin{array}{l}f\text{'}\left(x\right)=\underset{h\to 0}{\mathrm{lim}}\frac{3{\left(x+h\right)}^{3}-3{x}^{3}}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{3\left(3h{x}^{2}+3{h}^{2}x+{h}^{3}\right)}{h}\\ =\underset{h\to 0}{\mathrm{lim}}3·\left(\underset{h\to 0}{\mathrm{lim}}3{x}^{2}+\underset{h\to 0}{\mathrm{lim}}3hx+\underset{h\to 0}{\mathrm{lim}}{h}^{2}\right)\\ =3\left(3{x}^{2}\right)\\ =9{x}^{2}\end{array}$

7. $\begin{array}{l}f\text{'}\left(x\right)=\underset{h\to 0}{\mathrm{lim}}\frac{{\left(x+h\right)}^{4}-{x}^{4}}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{4h{x}^{3}+6{h}^{2}{x}^{2}+4{h}^{3}x+{h}^{4}}{h}\\ =\underset{h\to 0}{\mathrm{lim}}4{x}^{3}+\underset{h\to 0}{\mathrm{lim}}6h{x}^{2}+\underset{h\to 0}{\mathrm{lim}}{h}^{2}x+\underset{h\to 0}{\mathrm{lim}}{h}^{3}\\ =4{x}^{3}\end{array}$

8. $\begin{array}{l}f\text{'}\left(x\right)=\underset{h\to 0}{\mathrm{lim}}\frac{{\left(x+h\right)}^{5}-{x}^{5}}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{5h{x}^{4}+10{h}^{2}{x}^{3}+10{h}^{3}{x}^{2}+5{h}^{4}x+{h}^{5}}{h}\\ =\underset{h\to 0}{\mathrm{lim}}5{x}^{4}+\underset{h\to 0}{\mathrm{lim}}10h{x}^{3}+\underset{h\to 0}{\mathrm{lim}}10{h}^{2}x+\underset{h\to 0}{\mathrm{lim}}5{h}^{3}x+\underset{h\to 0}{\mathrm{lim}}{h}^{4}\\ =5{x}^{4}\end{array}$

9. $\begin{array}{l}f\text{'}\left(x\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2{\left(x+h\right)}^{3}-2{x}^{3}}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{2\left(3h{x}^{2}+3{h}^{2}x+{h}^{3}\right)}{h}\\ =\underset{h\to 0}{\mathrm{lim}}2·\left(\underset{h\to 0}{\mathrm{lim}}3{x}^{2}+\underset{h\to 0}{\mathrm{lim}}3hx+\underset{h\to 0}{\mathrm{lim}}{h}^{2}\right)\\ =6{x}^{2}\end{array}$

10. $\begin{array}{l}f\text{'}\left(x\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\left(\frac{1}{2}{\left(x+h\right)}^{3}+\left(x+h\right)\right)-\left(\frac{1}{2}{x}^{3}+x\right)}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\frac{1}{2}\left(3h{x}^{2}+3{h}^{2}x+{h}^{3}\right)+h}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{1}{2}·\left(\underset{h\to 0}{\mathrm{lim}}3{x}^{2}+\underset{h\to 0}{\mathrm{lim}}3hx+\underset{h\to 0}{\mathrm{lim}}{h}^{2}\right)+\underset{h\to 0}{\mathrm{lim}}1\\ =\frac{3}{2}{x}^{2}+1\end{array}$