完成度２０％
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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