【高校数学Ⅲ】3-1-1 微分係数と導関数｜問題集
1.次の微分係数を求めなさい。
(1)関数\(\displaystyle f(x)=\frac{2}{x}\)の\(x=1\)における微分係数
\(\displaystyle f'(1)=\lim_{h\to0}\frac{\frac{2}{1+h}-\frac{2}{1}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{2-2(1+h)}{1+h}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{-2h}{1+h}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{-2}{1+h}\)
\(=-2\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{2-2(1+h)}{1+h}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{-2h}{1+h}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{-2}{1+h}\)
\(=-2\)
(2)関数\(\displaystyle f(x)=\frac{3}{x+1}\)の\(x=2\)における微分係数
\(\displaystyle f'(2)=\lim_{h\to0}\frac{\frac{3}{2+h+1}-\frac{3}{2+1}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{3-(3+h)}{3+h}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{-h}{3+h}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{-1}{3+h}\)
\(\displaystyle =-\frac{1}{3}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{3-(3+h)}{3+h}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{-h}{3+h}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{-1}{3+h}\)
\(\displaystyle =-\frac{1}{3}\)
(3)関数\(f(x)=\sqrt{x}\)の\(x=2\)における微分係数
\(\displaystyle f'(2)=\lim_{h\to0}\frac{\sqrt{2+h}-\sqrt{2}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{(2+h)-2}{h(\sqrt{2+h}+\sqrt{2})}\)
\(\displaystyle =\lim_{h\to0}\frac{1}{\sqrt{2+h}+\sqrt{2}}\)
\(\displaystyle =\frac{1}{2\sqrt{2}}\)
\(\displaystyle =\lim_{h\to0}\frac{(2+h)-2}{h(\sqrt{2+h}+\sqrt{2})}\)
\(\displaystyle =\lim_{h\to0}\frac{1}{\sqrt{2+h}+\sqrt{2}}\)
\(\displaystyle =\frac{1}{2\sqrt{2}}\)
2.関数\(f(x)=\left\{\begin{array}{l}2\sqrt{x+1} (x\geqq0) \\ \displaystyle \frac{1}{2}x^2+x+2 (x<0)\end{array}\right.\)
について、次の問いに答えなさい。
(1)\(x=0\)において連続か答えなさい。
\(\displaystyle \lim_{x\to+0}2\sqrt{x+1}=2\)
\(\displaystyle \lim_{x\to-0}\left(\frac{1}{2}x^2+x+2\right)=2\)
\(\displaystyle \lim_{x\to+0}f(x)=\lim_{x\to-0}f(x)\)より、\(f(x)\)は\(x=0\)で連続である。
\(\displaystyle \lim_{x\to-0}\left(\frac{1}{2}x^2+x+2\right)=2\)
\(\displaystyle \lim_{x\to+0}f(x)=\lim_{x\to-0}f(x)\)より、\(f(x)\)は\(x=0\)で連続である。
(2)\(x=0\)において微分可能か答えなさい。
\(\displaystyle \lim_{x\to+0}\frac{f(0+h)-f(0)}{h}=\lim_{x\to+0}\frac{2}{\sqrt{h+1}+1}=1\)
\(\displaystyle \lim_{x\to-0}\frac{f(0+h)-f(0)}{h}=\lim_{x\to-0}\left(\frac{1}{2}h+1\right)=1\)
\(\displaystyle \lim_{x\to+0}\frac{f(0+h)-f(0)}{h}=\lim_{x\to-0}\frac{f(0+h)-f(0)}{h}\)より、\(f(x)\)は\(x=0\)で微分可能である。
\(\displaystyle \lim_{x\to-0}\frac{f(0+h)-f(0)}{h}=\lim_{x\to-0}\left(\frac{1}{2}h+1\right)=1\)
\(\displaystyle \lim_{x\to+0}\frac{f(0+h)-f(0)}{h}=\lim_{x\to-0}\frac{f(0+h)-f(0)}{h}\)より、\(f(x)\)は\(x=0\)で微分可能である。
3.関数\(f(x)=|x^2-1|\)について、次の問いに答えなさい。
(1)\(x=1\)において連続か答えなさい。
\(\displaystyle \lim_{x\to1+0}(x^2-1)=0\)
\(\displaystyle \lim_{x\to1-0}\{-(x^2-1)\}=0\)
\(\displaystyle \lim_{x\to1+0}|x^2-1|=\lim_{x\to1-0}|x^2-1|\)より、\(f(x)\)は\(x=1\)で連続である。
\(\displaystyle \lim_{x\to1-0}\{-(x^2-1)\}=0\)
\(\displaystyle \lim_{x\to1+0}|x^2-1|=\lim_{x\to1-0}|x^2-1|\)より、\(f(x)\)は\(x=1\)で連続である。
(2)\(x=1\)において微分可能か答えなさい。
\(\displaystyle \lim_{x\to+0}\frac{f(1+h)-f(1)}{h}\)
\(\displaystyle \ \ \ =\lim_{x\to+0}\frac{|(1+h)^2-1|-0}{h}=2\)
\(\displaystyle \lim_{x\to-0}\frac{f(1+h)-f(1)}{h}\)
\(\displaystyle \ \ \ =\lim_{x\to-0}\frac{|-(1+h)^2+1|-0}{h}=-2\)
\(\displaystyle \lim_{x\to+0}\frac{f(1+h)-f(1)}{h}\neq\lim_{x\to-0}\frac{f(1+h)-f(1)}{h}\)より、\(f(x)\)は\(x=1\)で微分可能ではない。
\(\displaystyle \ \ \ =\lim_{x\to+0}\frac{|(1+h)^2-1|-0}{h}=2\)
\(\displaystyle \lim_{x\to-0}\frac{f(1+h)-f(1)}{h}\)
\(\displaystyle \ \ \ =\lim_{x\to-0}\frac{|-(1+h)^2+1|-0}{h}=-2\)
\(\displaystyle \lim_{x\to+0}\frac{f(1+h)-f(1)}{h}\neq\lim_{x\to-0}\frac{f(1+h)-f(1)}{h}\)より、\(f(x)\)は\(x=1\)で微分可能ではない。
4.次の関数を導関数の定義に従って微分しなさい。
(1)\(\displaystyle y=\frac{1}{2x}\)
\(\displaystyle y'=\lim_{h\to0}\frac{\frac{1}{2(x+h)}-\frac{1}{2x}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{x-(x+h)}{2x(x+h)}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{-h}{2x(x+h)}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{-1}{2x(x+h)}\)
\(\displaystyle =-\frac{1}{2x^2}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{x-(x+h)}{2x(x+h)}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{-h}{2x(x+h)}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{-1}{2x(x+h)}\)
\(\displaystyle =-\frac{1}{2x^2}\)
(2)\(\displaystyle y=\sqrt{x}\)
\(\displaystyle y'=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}\)
\(\displaystyle =\lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt{x}}\)
\(\displaystyle =\frac{1}{2\sqrt{x}}\)
\(\displaystyle =\lim_{h\to0}\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}\)
\(\displaystyle =\lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt{x}}\)
\(\displaystyle =\frac{1}{2\sqrt{x}}\)
(3)\(\displaystyle y=\frac{3}{x+1}\)
\(\displaystyle y'=\lim_{h\to0}\frac{\frac{3}{x+h+1}-\frac{3}{x+1}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{3(x+1)-3(x+h+1)}{(x+h+1)(x+1)}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{-3h}{(x+h+1)(x+1)}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{-3}{(x+h+1)(x+1)}\)
\(\displaystyle =-\frac{3}{(x+1)^2}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{3(x+1)-3(x+h+1)}{(x+h+1)(x+1)}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{-3h}{(x+h+1)(x+1)}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{-3}{(x+h+1)(x+1)}\)
\(\displaystyle =-\frac{3}{(x+1)^2}\)
(4)\(\displaystyle y=2\sqrt{x+1}\)
\(\displaystyle y'=\lim_{h\to0}\frac{2\sqrt{x+h+1}-2\sqrt{x+1}}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{2(x+h+1-x-1)}{h(\sqrt{x+h+1}+\sqrt{x+1})}\)
\(\displaystyle =\lim_{h\to0}\frac{2}{\sqrt{x+h+1}+\sqrt{x+1}}\)
\(\displaystyle =\frac{2}{2\sqrt{x+1}}\)
\(\displaystyle =\frac{1}{\sqrt{x+1}}\)
\(\displaystyle =\lim_{h\to0}\frac{2(x+h+1-x-1)}{h(\sqrt{x+h+1}+\sqrt{x+1})}\)
\(\displaystyle =\lim_{h\to0}\frac{2}{\sqrt{x+h+1}+\sqrt{x+1}}\)
\(\displaystyle =\frac{2}{2\sqrt{x+1}}\)
\(\displaystyle =\frac{1}{\sqrt{x+1}}\)
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