練習問題にチャレンジ♪
さっそく練習問題にチャレンジしましょう。
微分の基本計算をマスター
微分の基本計算は,この先ずっと必要となる大切なテクニックです。何度も何度も微分して,自分なりの感覚を身につけましょう。それでは問題にチャレンジです。
次の関数を微分せよ。
↓この問題へのリンクはこちら(右クリックで保存)
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 肩の数 -1}\\ \left(x^{\colBX{bisque}{$n$}}\right)^{\colBX{pink}{$'$}} &= \colBX{bisque}{$n$}\, x^{\colBX{bisque}{$\small n$}-1}\\ & \colMM{red}{肩の数を前に} \end{align*}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{red}{定数を微分}\\ \left(\colBX{palegreen}{$c$}\right)^{\colBX{pink}{$'$}} &= \colBX{palegreen}{0} \colMM{red}{・・・必ず0!} \end{align*}
【解答】
\def\Ak{3} \def\Ai{2} \def\Bk{-4} \def\Bi{1} \def\Ck{+2} \def\Ckb{2} \def\Kotae{6x-4} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} y' &\color{gray}{\ =(\Ak x^{\Ai} \Bk x^{\Bi} \Ck)'}\\ & \colMM{gray}{ \Darr 線形性}\\ &\color{gray}{\ = \Ak (x^{\colBX{bisque}{\Ai}})' \Bk (x^{\Bi})'+(\colBX{palegreen}{$\Ckb$})'}\\ \\ &=\Ak \cdot \colBX{bisque}{\Ai}x^{\colBX{bisque}{\Ai}-1} \Bk \cdot \Bi x^{\Bi -1} + \colBX{palegreen}{$0$}\\ & \colMM{magenta}{ \swarrow x^0=1}\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 肩の数 -1}\\ \left(x^{\colBX{bisque}{$n$}}\right)^{\colBX{pink}{$'$}} &= \colBX{bisque}{$n$}\, x^{\colBX{bisque}{$\small n$}-1}\\ & \colMM{red}{肩の数を前に} \end{align*}
【解答】
\def\Ak{\dfrac{2}{3}} \def\Ai{3} \def\Bk{-\dfrac{5}{2}} \def\Bi{2} \def\Ck{+5} \def\Ci{1} \def\Kotae{2x^2-5x+5} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} y' &\color{gray}{\ = \left(\Ak x^{\Ai} \Bk x^{\Bi} \Ck x^{\Ci} \right)'}\\ & \colMM{gray}{ \Darr 線形性}\\ &\color{gray}{\ = \Ak (x^{\colBX{bisque}{\Ai}})' \Bk (x^{\Bi})' \Ck (x^{\Ci})'}\\ \\ &=\Ak \cdot \colBX{bisque}{\Ai}x^{\colBX{bisque}{\Ai}-1} \Bk \cdot \Bi x^{\Bi-1} \Ck \cdot \Ci x^{\Ci-1}\\ & \colMM{magenta}{ x^0=1\swarrow}\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 肩の数 -1}\\ \left(x^{\colBX{bisque}{$n$}}\right)^{\colBX{pink}{$'$}} &= \colBX{bisque}{$n$}\, x^{\colBX{bisque}{$\small n$}-1}\\ & \colMM{red}{肩の数を前に} \end{align*}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{red}{定数を微分}\\ \left(\colBX{palegreen}{$c$}\right)^{\colBX{pink}{$'$}} &= \colBX{palegreen}{0} \colMM{red}{・・・必ず0!} \end{align*}
【解答】
\def\Ak{4} \def\Ai{2} \def\Bk{+3} \def\Bi{1} \def\Ck{-4} \def\Ckb{-4} \def\Kotae{8x+3} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} y' &\color{gray}{\ =(\Ak x^{\Ai} \Bk x^{\Bi} \Ck)'}\\ & \colMM{gray}{ \Darr 線形性}\\ &\color{gray}{\ = \Ak (x^{\colBX{bisque}{\Ai}})' \Bk (x^{\Bi})'+(\colBX{palegreen}{$\Ckb$})'}\\ \\ &=\Ak \cdot \colBX{bisque}{\Ai}x^{\colBX{bisque}{\Ai}-1} \Bk \cdot \Bi x^{\Bi -1} + \colBX{palegreen}{$0$}\\ & \colMM{magenta}{ \swarrow x^0=1}\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 肩の数 -1}\\ \left(x^{\colBX{bisque}{$n$}}\right)^{\colBX{pink}{$'$}} &= \colBX{bisque}{$n$}\, x^{\colBX{bisque}{$\small n$}-1}\\ & \colMM{red}{肩の数を前に} \end{align*}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{red}{定数を微分}\\ \left(\colBX{palegreen}{$c$}\right)^{\colBX{pink}{$'$}} &= \colBX{palegreen}{0} \colMM{red}{・・・必ず0!} \end{align*}
【解答】
\def\Ak{2} \def\Ai{2} \def\Bk{-5} \def\Bi{1} \def\Ck{+1} \def\Ckb{1} \def\Kotae{4x-5} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} y' &\color{gray}{\ =(\Ak x^{\Ai} \Bk x^{\Bi} \Ck)'}\\ & \colMM{gray}{ \Darr 線形性}\\ &\color{gray}{\ = \Ak (x^{\colBX{bisque}{\Ai}})' \Bk (x^{\Bi})'+(\colBX{palegreen}{$\Ckb$})'}\\ \\ &=\Ak \cdot \colBX{bisque}{\Ai}x^{\colBX{bisque}{\Ai}-1} \Bk \cdot \Bi x^{\Bi -1} + \colBX{palegreen}{$0$}\\ & \colMM{magenta}{ \swarrow x^0=1}\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 肩の数 -1}\\ \left(x^{\colBX{bisque}{$n$}}\right)^{\colBX{pink}{$'$}} &= \colBX{bisque}{$n$}\, x^{\colBX{bisque}{$\small n$}-1}\\ & \colMM{red}{肩の数を前に} \end{align*}
【解答】
\def\Ak{\dfrac{4}{3}} \def\Ai{3} \def\Bk{+\dfrac{3}{4}} \def\Bi{2} \def\Ck{-\dfrac{1}{2}} \def\Ci{1} \def\Kotae{4x^2+\dfrac{3}{2}x-\dfrac{1}{2}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} y' &\color{gray}{\ = \left(\Ak x^{\Ai} \Bk x^{\Bi} \Ck x^{\Ci} \right)'}\\ & \colMM{gray}{ \Darr 線形性}\\ &\color{gray}{\ = \Ak (x^{\colBX{bisque}{\Ai}})' \Bk (x^{\Bi})' \Ck (x^{\Ci})'}\\ \\ &=\Ak \cdot \colBX{bisque}{\Ai}x^{\colBX{bisque}{\Ai}-1} \Bk \cdot \Bi x^{\Bi-1} \Ck \cdot \Ci x^{\Ci-1}\\ & \colMM{magenta}{ x^0=1\swarrow}\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 肩の数 -1}\\ \left(x^{\colBX{bisque}{$n$}}\right)^{\colBX{pink}{$'$}} &= \colBX{bisque}{$n$}\, x^{\colBX{bisque}{$\small n$}-1}\\ & \colMM{red}{肩の数を前に} \end{align*}
【解答】
\def\Ak{-\dfrac{1}{3}} \def\Ai{3} \def\Bk{-\dfrac{3}{2}} \def\Bi{1} \def\Ck{-\dfrac{1}{2}} \def\Ckf{-} \def\Ckb{\dfrac{1}{2}} \def\Kotae{-x^2-\dfrac{3}{2}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} y' &\color{gray}{\ = \left(\Ak x^{\Ai} \Bk x^{\Bi} \Ck \right)'}\\ & \colMM{gray}{ \Darr 線形性}\\ &\color{gray}{\ = \Ak (x^{\colBX{bisque}{\Ai}})' \Bk (x^{\Bi})' \Ckf \left(\colBX{palegreen}{$\Ckb$}\right)'}\\ \\ &=\Ak \cdot \colBX{bisque}{\Ai}x^{\colBX{bisque}{\Ai}-1} \Bk \cdot \Bi x^{\Bi-1} \Ckf \colBX{palegreen}{$0$}\\ & \colMM{magenta}{ x^0=1\swarrow}\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
微分する前に展開を
因数分解されている式は,そのままでは微分できません。まずは展開しましょう。この展開で計算ミスをすると微分も間違えてしまうので気をつけてください。
↓この問題へのリンクはこちら(右クリックで保存)
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 肩の数 -1}\\ \left(x^{\colBX{bisque}{$n$}}\right)^{\colBX{pink}{$'$}} &= \colBX{bisque}{$n$}\, x^{\colBX{bisque}{$\small n$}-1}\\ & \colMM{red}{肩の数を前に} \end{align*}
【解答】
\def\Ak{1} \def\Ai{3} \def\Bk{-1} \def\Bi{2} \def\Ck{-2} \def\Ci{1} \def\Kotae{3x^2-2x-2} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 微分する前に展開!}\\ y &= x(x+1)(x-2)\\ &= x(x^2-x-2)\\ &= x^3-x^2-2x\\ \\ y' &\color{gray}{\ = \left(\Ak x^{\Ai} \Bk x^{\Bi} \Ck x^{\Ci} \right)'}\\ & \colMM{gray}{ \Darr 線形性}\\ &\color{gray}{\ = \Ak (x^{\colBX{bisque}{\Ai}})' \Bk (x^{\Bi})' \Ck (x^{\Ci})'}\\ \\ &=\Ak \cdot \colBX{bisque}{\Ai}x^{\colBX{bisque}{\Ai}-1} \Bk \cdot \Bi x^{\Bi-1} \Ck \cdot \Ci x^{\Ci-1}\\ & \colMM{magenta}{ x^0=1\swarrow}\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 肩の数 -1}\\ \left(x^{\colBX{bisque}{$n$}}\right)^{\colBX{pink}{$'$}} &= \colBX{bisque}{$n$}\, x^{\colBX{bisque}{$\small n$}-1}\\ & \colMM{red}{肩の数を前に} \end{align*}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{red}{定数を微分}\\ \left(\colBX{palegreen}{$c$}\right)^{\colBX{pink}{$'$}} &= \colBX{palegreen}{0} \colMM{red}{・・・必ず0!} \end{align*}
【解答】
\def\Ak{1} \def\Ai{2} \def\Bk{+5} \def\Bi{1} \def\Ck{+6} \def\Ckb{6} \def\Kotae{2x+5} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 微分する前に展開!}\\ y &= (x+2)(x+3)\\ &= x^2+5x+6\\ \\ y' &\color{gray}{\ =(\Ak x^{\Ai} \Bk x^{\Bi} \Ck)'}\\ & \colMM{gray}{ \Darr 線形性}\\ &\color{gray}{\ = \Ak (x^{\colBX{bisque}{\Ai}})' \Bk (x^{\Bi})'+(\colBX{palegreen}{$\Ckb$})'}\\ \\ &=\Ak \cdot \colBX{bisque}{\Ai}x^{\colBX{bisque}{\Ai}-1} \Bk \cdot \Bi x^{\Bi -1} + \colBX{palegreen}{$0$}\\ & \colMM{magenta}{ \swarrow x^0=1}\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 肩の数 -1}\\ \left(x^{\colBX{bisque}{$n$}}\right)^{\colBX{pink}{$'$}} &= \colBX{bisque}{$n$}\, x^{\colBX{bisque}{$\small n$}-1}\\ & \colMM{red}{肩の数を前に} \end{align*}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{red}{定数を微分}\\ \left(\colBX{palegreen}{$c$}\right)^{\colBX{pink}{$'$}} &= \colBX{palegreen}{0} \colMM{red}{・・・必ず0!} \end{align*}
【解答】
\def\Ak{3} \def\Ai{2} \def\Bk{-12} \def\Bi{1} \def\Ck{+12} \def\Ckb{12} \def\Kotae{6x-12} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 微分する前に展開!}\\ y &= 3(x-2)^2\\ &= 3(x^2-4x+4)\\ &= 3x^2-12x+12\\ \\ y' &\color{gray}{\ =(\Ak x^{\Ai} \Bk x^{\Bi} \Ck)'}\\ & \colMM{gray}{ \Darr 線形性}\\ &\color{gray}{\ = \Ak (x^{\colBX{bisque}{\Ai}})' \Bk (x^{\Bi})'+(\colBX{palegreen}{$\Ckb$})'}\\ \\ &=\Ak \cdot \colBX{bisque}{\Ai}x^{\colBX{bisque}{\Ai}-1} \Bk \cdot \Bi x^{\Bi -1} + \colBX{palegreen}{$0$}\\ & \colMM{magenta}{ \swarrow x^0=1}\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 肩の数 -1}\\ \left(x^{\colBX{bisque}{$n$}}\right)^{\colBX{pink}{$'$}} &= \colBX{bisque}{$n$}\, x^{\colBX{bisque}{$\small n$}-1}\\ & \colMM{red}{肩の数を前に} \end{align*}
【解答】
\def\Ak{1} \def\Ai{3} \def\Bk{0} \def\Bi{0} \def\Ck{-4} \def\Ci{1} \def\Kotae{3x^2-4} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 微分する前に展開!}\\ y &= x(x+2)(x-2)\\ &= x(x^2-4)\\ &= x^3-4x\\ \\ y' &\color{gray}{\ = \left(\Ak x^{\Ai} \Ck x^{\Ci} \right)'}\\ & \colMM{gray}{ \Darr 線形性}\\ &\color{gray}{\ = \Ak (x^{\colBX{bisque}{\Ai}})' \Ck (x^{\Ci})'}\\ \\ &=\Ak \cdot \colBX{bisque}{\Ai}x^{\colBX{bisque}{\Ai}-1} \Ck \cdot \Ci x^{\Ci-1}\\ & \colMM{magenta}{ x^0=1\swarrow}\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 肩の数 -1}\\ \left(x^{\colBX{bisque}{$n$}}\right)^{\colBX{pink}{$'$}} &= \colBX{bisque}{$n$}\, x^{\colBX{bisque}{$\small n$}-1}\\ & \colMM{red}{肩の数を前に} \end{align*}
【解答】
\def\Ak{2} \def\Ai{3} \def\Bk{-4} \def\Bi{2} \def\Ck{-6} \def\Ci{1} \def\Kotae{6x^2-8x-6} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 微分する前に展開!}\\ y &= 2x(x+1)(x-3)\\ &= 2x(x^2-2x-3)\\ &= 2x^3-4x^2-6x\\ \\ y' &\color{gray}{\ = \left(\Ak x^{\Ai} \Bk x^{\Bi} \Ck x^{\Ci} \right)'}\\ & \colMM{gray}{ \Darr 線形性}\\ &\color{gray}{\ = \Ak (x^{\colBX{bisque}{\Ai}})' \Bk (x^{\Bi})' \Ck (x^{\Ci})'}\\ \\ &=\Ak \cdot \colBX{bisque}{\Ai}x^{\colBX{bisque}{\Ai}-1} \Bk \cdot \Bi x^{\Bi-1} \Ck \cdot \Ci x^{\Ci-1}\\ & \colMM{magenta}{ x^0=1\swarrow}\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
微分する文字が変わっても同じ
ここを超えられると数学が楽しくなってきます。文字が x から別の文字に変わっても,微分は変わりません。この感覚を感じ取ってください。
次の関数を[ ]内に示された変数で微分せよ。ただし,右辺において変数以外の文字は定数である。
↓この問題へのリンクはこちら(右クリックで保存)
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 肩の数 -1}\\ \left(x^{\colBX{bisque}{$n$}}\right)^{\colBX{pink}{$'$}} &= \colBX{bisque}{$n$}\, x^{\colBX{bisque}{$\small n$}-1}\\ & \colMM{red}{肩の数を前に} \end{align*}
【解答】
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \begin{align*} & \colMM{orange}{ t\ の係数}\\ s &= \colBX{bisque}{$-\dfrac12g$}\,t^{\colBX{palegreen}{$\scriptsize2$}}\\ \colMM{red}{sをtで微分 \Darr} & \ \colMM{green}{指数}\colMM{red}{-1}\\ \colBX{mistyrose}{$\dfrac{ds}{dt}$} &= \colBX{bisque}{$-\dfrac12g$}\cdot\colBX{palegreen}{$2$}\,t^{\colBX{palegreen}{$\scriptsize 2$}-1}\\ & \colMM{orange}{係数}\colMM{red}{\times}\colMM{green}{指数}\\ &=-g\,t^1\\ \\ &= -g\,t \end{align*}
↓この問題へのリンクはこちら(右クリックで保存)
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 肩の数 -1}\\ \left(x^{\colBX{bisque}{$n$}}\right)^{\colBX{pink}{$'$}} &= \colBX{bisque}{$n$}\, x^{\colBX{bisque}{$\small n$}-1}\\ & \colMM{red}{肩の数を前に} \end{align*}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{red}{定数を微分}\\ \left(\colBX{palegreen}{$c$}\right)^{\colBX{pink}{$'$}} &= \colBX{palegreen}{0} \colMM{red}{・・・必ず0!} \end{align*}
【解答】文字が変わっているだけ!怖くないよ。
\def\Ak{3} \def\Ai{2} \def\Bk{-4} \def\Bi{1} \def\Ck{+2} \def\Ckb{2} \def\Kotae{6t-4} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{red}{sをtで微分 \Darr} & \\ \colBX{mistyrose}{$\dfrac{ds}{dt}$} &\color{gray}{\ =(\Ak t^{\Ai} \Bk t^{\Bi} \Ck)'}\\ & \colMM{gray}{ \Darr 線形性}\\ &\color{gray}{\ = \Ak (t^{\colBX{bisque}{\Ai}})' \Bk (t^{\Bi})'+(\colBX{palegreen}{$\Ckb$})'}\\ \\ &=\Ak \cdot \colBX{bisque}{\Ai}t^{\colBX{bisque}{\Ai}-1} \Bk \cdot \Bi t^{\Bi -1} + \colBX{palegreen}{$0$}\\ & \colMM{magenta}{ \swarrow t^0=1}\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ 肩の数 -1}\\ \left(x^{\colBX{bisque}{$n$}}\right)^{\colBX{pink}{$'$}} &= \colBX{bisque}{$n$}\, x^{\colBX{bisque}{$\small n$}-1}\\ & \colMM{red}{肩の数を前に} \end{align*}
【解答】文字が変わっているだけ!怖くないよ。
\def\Ak{a} \def\Ai{3} \def\Bk{+b} \def\Bi{2} \def\Kotae{3at^2+2bt} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{red}{f(t)をtで微分 \Darr} & \\ \colBX{mistyrose}{$\dfrac{d}{dt}f(t)$} &\color{gray}{\ =(\Ak t^{\Ai} \Bk t^{\Bi})'}\\ & \colMM{gray}{ \Darr 線形性}\\ &\color{gray}{\ = \Ak (t^{\colBX{bisque}{\Ai}})' \Bk (t^{\Bi})'}\\ \\ &=\Ak \cdot \colBX{bisque}{\Ai}t^{\colBX{bisque}{\Ai}-1} \Bk \cdot \Bi t^{\Bi -1}\\ \\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
導関数を利用して微分係数を求めよう
導関数を求めることを「微分する」といいます。つまり微分すれば導関数を求めることができるわけです。ところで導関数って何を導く関数でしたっけ。そう 微分係数 です。ということで、ここでは導関数を利用して微分係数を求める問題を練習しましょう。
関数 f(x)=-x^2+4x+1 について,次の x の値における微分係数を求めよ。
↓この問題へのリンクはこちら(右クリックで保存)
\def\atX{-1} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & 関数\ f(x)\ の \\ & \ x=\colBX{mistyrose}{$\atX$}\ における \colFR{red}{\color{red}\bf微分係数} は\\ \\ & f'(\colBX{mistyrose}{$\atX$})& \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
【解答】
\def\atX{-1} \def\left{2} \def\Kotae{6} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{orange}{微分して}\\ \colBX{bisque}{$f'(x)$} &= -2x+4\\ \\ よって,求める & 微分係数は\\ \\ f'(\colBX{mistyrose}{$\atX$}) &= -2 (\colBX{mistyrose}{$\atX$})+4\\ &= \left +4\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
\def\atX{2} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & 関数\ f(x)\ の \\ & \ x=\colBX{mistyrose}{$\atX$}\ における \colFR{red}{\color{red}\bf微分係数} は\\ \\ & f'(\colBX{mistyrose}{$\atX$})& \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
【解答】
\def\atX{2} \def\left{-4} \def\Kotae{0} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{orange}{微分して}\\ \colBX{bisque}{$f'(x)$} &= -2x+4\\ \\ よって,求める & 微分係数は\\ \\ f'(\colBX{mistyrose}{$\atX$}) &= -2 \cdot \colBX{mistyrose}{$\atX$}+4\\ &= \left +4\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
\def\atX{0} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & 関数\ f(x)\ の \\ & \ x=\colBX{mistyrose}{$\atX$}\ における \colFR{red}{\color{red}\bf微分係数} は\\ \\ & f'(\colBX{mistyrose}{$\atX$})& \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
【解答】
\def\atX{0} \def\left{0} \def\Kotae{4} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{orange}{微分して}\\ \colBX{bisque}{$f'(x)$} &= -2x+4\\ \\ よって,求める & 微分係数は\\ \\ f'(\colBX{mistyrose}{$\atX$}) &= -2 \cdot \colBX{mistyrose}{$\atX$}+4\\ &= \left +4\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
\def\atX{-2} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & 関数\ f(x)\ の \\ & \ x=\colBX{mistyrose}{$\atX$}\ における \colFR{red}{\color{red}\bf微分係数} は\\ \\ & f'(\colBX{mistyrose}{$\atX$})& \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
【解答】
\def\atX{-2} \def\left{4} \def\Kotae{8} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{orange}{微分して}\\ \colBX{bisque}{$f'(x)$} &= -2x+4\\ \\ よって,求める & 微分係数は\\ \\ f'(\colBX{mistyrose}{$\atX$}) &= -2 (\colBX{mistyrose}{$\atX$})+4\\ &= \left +4\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
関数 f(x)=x^3-3x^2+3 について,次の x の値における微分係数を求めよ。
↓この問題へのリンクはこちら(右クリックで保存)
\def\atX{2} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & 関数\ f(x)\ の \\ & \ x=\colBX{mistyrose}{$\atX$}\ における \colFR{red}{\color{red}\bf微分係数} は\\ \\ & f'(\colBX{mistyrose}{$\atX$})& \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
【解答】
\def\atX{2} \def\left{12} \def\right{-12} \def\Kotae{0} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{orange}{微分して}\\ \colBX{bisque}{$f'(x)$} &= 3x^2-6x\\ \\ よって,求める & 微分係数は\\ \\ f'(\colBX{mistyrose}{$\atX$}) &= 3 \cdot \colBX{mistyrose}{$\atX$}^2 -6 \cdot \colBX{mistyrose}{$\atX$} \\ &= \left \right\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
\def\atX{0} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & 関数\ f(x)\ の \\ & \ x=\colBX{mistyrose}{$\atX$}\ における \colFR{red}{\color{red}\bf微分係数} は\\ \\ & f'(\colBX{mistyrose}{$\atX$})& \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
【解答】
\def\atX{0} \def\left{0} \def\right{-0} \def\Kotae{0} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{orange}{微分して}\\ \colBX{bisque}{$f'(x)$} &= 3x^2-6x\\ \\ よって,求める & 微分係数は\\ \\ f'(\colBX{mistyrose}{$\atX$}) &= 3 \cdot \colBX{mistyrose}{$\atX$}^2 -6 \cdot \colBX{mistyrose}{$\atX$} \\ &= \left \right\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
\def\atX{-2} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & 関数\ f(x)\ の \\ & \ x=\colBX{mistyrose}{$\atX$}\ における \colFR{red}{\color{red}\bf微分係数} は\\ \\ & f'(\colBX{mistyrose}{$\atX$})& \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
【解答】
\def\atX{-2} \def\left{12} \def\right{+12} \def\Kotae{24} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{orange}{微分して}\\ \colBX{bisque}{$f'(x)$} &= 3x^2-6x\\ \\ よって,求める & 微分係数は\\ \\ f'(\colBX{mistyrose}{$\atX$}) &= 3 (\colBX{mistyrose}{$\atX$})^2 -6 (\colBX{mistyrose}{$\atX$}) \\ &= \left \right\\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan