2.1 導関数

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

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

$\displaystyle f'(x)$	$\displaystyle =$	$\displaystyle \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} = \lim_{h \to 0}\frac{\frac{1}{(x+h)^{2}} - \frac{1}{x^{2}}}{h}$
	$\displaystyle =$	$\displaystyle \lim_{h \to 0} \frac{x^{2} - (x+h)^{2}}{hx^{2}(x+h)^{2})} = \lim_{h \to 0}\frac{-2xh - h^{2}}{hx^{2}(x+h)^{2})}$
	$\displaystyle =$	$\displaystyle \lim_{h \to 0}\frac{h(-2x + h)}{hx^{2}(x+h)^{2})} = \lim_{h \to 0}\frac{-2x + h}{x^{2}(x+h)^{2})} = -\frac{2}{x^{3}}$

$\displaystyle f'(2)$	$\displaystyle =$	$\displaystyle \lim_{h \to 0}\frac{f(2+h) - f(2)}{h} = \lim_{h \to 0}\frac{5(2+h) - (2+h)^{2} - (5\cdot 2 - 2^{2})}{h}$
	$\displaystyle =$	$\displaystyle \lim_{h \to 0}\frac{10 + 5h - 4 - 4h - h^{2} - 6}{h}$
	$\displaystyle =$	$\displaystyle \lim_{ h \to 0} \frac{-h^{2} + h}{h} = \lim_{h \to 0}(-h+1) = 1$

$\displaystyle f'(2)$	$\displaystyle =$	$\displaystyle \lim_{h \to 0}\frac{f(2+h) - f(2)}{h} = \lim_{h \to 0}\frac{(3(h+2) - 7)^{2} - (3 \cdot 2 - 7)^{2}}{h}$
	$\displaystyle =$	$\displaystyle \lim_{h \to 0}\frac{(3h-1)^{2} - 1}{h}$
	$\displaystyle =$	$\displaystyle \lim_{ h \to 0} \frac{9h^{2} - 6h + 1- 1}{h} = \lim_{h \to 0}\frac{h(9h - 6)}{h} = -6$

(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

は

で与えられる．そこで，まず， $f'(x)$

を求めると，

より，

．次に，接線は点

を通るので，求める接線の方程式は $y + 3 = -(x-2)$

．整理すると

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

は

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

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

となる．

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

は

で与えられる．そこで， $f'(2)$

を求める．

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

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

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

$\displaystyle y'$	$\displaystyle =$	$\displaystyle ((x^{2}-1)(x-3))' = (x^{2}-1)'(x-3) + (x^{2}-1)(x-3)' = 2x(x-3) + x^{2} - 1$
	$\displaystyle =$	$\displaystyle 2x^{2} - 6x + x^{2} - 1 = 3x^{2} - 6x + 1$

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

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

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

$\displaystyle y'$	$\displaystyle =$	$\displaystyle (\frac{1 + x^{4}}{x^{2}})' = \frac{(1+x^{4})'x^{2} - (1+x^{4})(x^{2})'}{(x^{2})^{2}} = \frac{4x^{3}x^{2} - (1+x^{4}\cdot 2x }{x^{4}}$
	$\displaystyle =$	$\displaystyle \frac{4x^{5} - 2x - 2x^{5}}{x^{4}} = \frac{2x^{5} - 2x}{x^{4}} = \frac{2(x^{4} - 1)}{x^{3}}$