2.1 導関数

1.

(a)

$$f'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} = \lim_{h \to 0}\frac{c - c}{h}= 0$$

(b)

$$f'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} = \lim_{h \to 0}\frac{\sqrt{x+h-1} - \sqrt{x-1}}{h}$$

$$= \lim_{h \to 0} \frac{(\sqrt{x+h-1} - \sqrt{x-1})(\sqrt{x+h-1} - \... ...\sqrt{x-1})} = \lim_{h \to 0}\frac{x+h-1 - (x-1)}{h(\sqrt{x+h-1} + \sqrt{x-1})}$$

$$= \lim_{h \to 0}\frac{h}{h(\sqrt{x+h-1} + \sqrt{x-1})} = \frac{1}{2\sqrt{x-1}}$$

(c)

$$f'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} = \lim_{h \to 0}\frac{\frac{1}{(x+h)^{2}} - \frac{1}{x^{2}}}{h}$$

$$= \lim_{h \to 0} \frac{x^{2} - (x+h)^{2}}{hx^{2}(x+h)^{2})} = \lim_{h \to 0}\frac{-2xh - h^{2}}{hx^{2}(x+h)^{2})}$$

$$= \lim_{h \to 0}\frac{h(-2x + h)}{hx^{2}(x+h)^{2})} = \lim_{h \to 0}\frac{-2x + h}{x^{2}(x+h)^{2})} = -\frac{2}{x^{3}}$$

2.

(a)

$$f'(2) = \lim_{h \to 0}\frac{f(2+h) - f(2)}{h} = \lim_{h \to 0}\frac{5(2+h) - (2+h)^{2} - (5\cdot 2 - 2^{2})}{h}$$

$$= \lim_{h \to 0}\frac{10 + 5h - 4 - 4h - h^{2} - 6}{h}$$

$$= \lim_{ h \to 0} \frac{-h^{2} + h}{h} = \lim_{h \to 0}(-h+1) = 1$$

(b)

$$f'(2) = \lim_{h \to 0}\frac{f(2+h) - f(2)}{h} = \lim_{h \to 0}\frac{(3(h+2) - 7)^{2} - (3 \cdot 2 - 7)^{2}}{h}$$

$$= \lim_{h \to 0}\frac{(3h-1)^{2} - 1}{h}$$

$$= \lim_{ h \to 0} \frac{9h^{2} - 6h + 1- 1}{h} = \lim_{h \to 0}\frac{h(9h - 6)}{h} = -6$$

3.

(a) 接線の方程式は，接線が通る点 $(x_{0},y_{0})$と傾き$m$が与えられると， $y - y_{0} = m(x - x_{0})$で求めることができる．$x = a$における接線の傾き$m$は$m = f'(a)$で与えられる．そこで，まず，$f'(x)$を求めると， $f'(x) = 2x - 5$より， $f'(2) = -1$．次に，接線は点 $(a,f(a)) = (2,-3)$を通るので，求める接線の方程式は $y + 3 = -(x-2)$．整理すると $y = -x - 1$となる．

(b) 接線の方程式は，接線が通る点 $(x_{0},y_{0})$と傾き$m$が与えられると， $y - y_{0} = m(x - x_{0})$で求めることができる．$x = a$における接線の傾き$m$は$m = f'(a)$で与えられる．そこで，まず，$f'(x)$を求めると， $f'(x) = -3x^{2}$より， $f'(2) = -12$．次に，接線は点 $(a,f(a)) = (2,-3)$を通るので，求める接線の方程式は $y + 3 = -12(x-2)$．整理すると $y = -12x + 21$となる．

(c) 接線の方程式は，接線が通る点 $(x_{0},y_{0})$と傾き$m$が与えられると， $y - y_{0} = m(x - x_{0})$で求めることができる．$x = a$における接線の傾き$m$は$m = f'(a)$で与えられる．そこで，$f'(2)$を求める．

$$f'(4) = \lim_{h \to 0}\frac{f(4+h) - f(4)}{h}= \lim_{h \to 0}\frac{\sqrt{4+h} - \sqrt{4}}{h}$$

$$= \lim_{h \to 0}\frac{(\sqrt{4+h} - \sqrt{4}(\sqrt{4+h} + \sqrt{4})}{h(\sqrt{4+h} + \sqrt{4})} = \lim_{h \to 0}\frac{4+h-4}{h(\sqrt{4+h} - \sqrt{4})}$$

$$= \lim_{h \to 0}\frac{h}{h(\sqrt{4+h}+\sqrt{4})} = \lim_{h \to 0}\frac{1}{\sqrt{4+h} + \sqrt{4}} = \frac{1}{2\sqrt{4}}$$

より， $${f'(4) = \frac{1}{4}}$$．次に，接線は点 $(a,f(a)) = (4,2)$を通るので，求める接線の方程式は $${y - 2 = \frac{1}{4}(x-4)}$$．整理すると $${y = \frac{1}{4}x + 1}$$となる．

4.

(a)

$$y' = (11x^{5} - 6x^{3} + 8)' = (11x^{5})' - (6x^{3})' + 8' = 55x^{4} - 18x^{2}$$

(b)

$$y' = (-\frac{1}{x^{2}})' = -\frac{-(x^{2})'}{(x^{2})^{2}} = \frac{2x}{x^{4}} = \frac{2}{x^{3}}$$

(c)

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

$$= 2x^{2} - 6x + x^{2} - 1 = 3x^{2} - 6x + 1$$

(d)

$$y' = (\frac{x-1}{x-2})' = \frac{(x-1)'(x-2) - (x-1)(x-2)'}{(x-2)^{2}} = \frac{x-2 - (x-1)}{(x-2)^{2}} = \frac{-1}{(x-2)^{2}}$$

(e)

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

$$= \frac{4x^{2} + 6x - 2x^{2} + 2}{(2x+3)^{2}} = \frac{2x^{2} + 6x +2}{(2x+3)^{2}} = \frac{2(x^{2} + 3x + 1)}{(2x+3)^{2}}$$

(f)

$$y' = (\frac{6 - 1/x}{x - 2})' = \frac{(6 - 1/x)'(x-2) - (6 - 1/x)(x-2)'}{(x-2)^{2}} = \frac{\frac{1}{x^{2}}(x-2) - (6 - \frac{1}{x})}{(x-2)^{2}}$$

$$= \frac{x-2-6x^{2} + x}{x^{2}(x-2)^{2}} = \frac{-6x^{2} + 2x - 2}{x^{2}(x-2)^{2}} = \frac{-2(3x^{2} - x + 1)}{x^{2}(x-2)^{2}}$$

(g)

$$y' = (\frac{1 + x^{4}}{x^{2}})' = \frac{(1+x^{4})'x^{2} - (1+x^{4})(x^{2})'}{(x^{2})^{2}} = \frac{4x^{3}x^{2} - (1+x^{4}\cdot 2x }{x^{4}}$$

$$= \frac{4x^{5} - 2x - 2x^{5}}{x^{4}} = \frac{2x^{5} - 2x}{x^{4}} = \frac{2(x^{4} - 1)}{x^{3}}$$