気になるところをタップして確認しましょう。
関数 f(x),g(x) がともに微分可能であるとき
- \displaystyle\left\{ \dfrac{1}{g(x)} \right\}' = -\dfrac{g'(x)}{\left\{ g(x) \right\}^2}
- \displaystyle\left\{ \dfrac{f(x)}{g(x)} \right\}' = \dfrac{f'(x)g(x)-f(x)g'(x)}{\left\{g(x)\right\}^2}
\newcommand\cmr[1]{\colorbox{mistyrose}{$#1$}} \newcommand\clc[1]{\colorbox{lightcyan}{$#1$}}\small \left\{\dfrac{\cmr{分子}}{\clc{分母}}\right\}' = \dfrac{\cmr{分子}' \cdot \clc{分母} - \cmr{分子} \cdot \clc{分母}'}{\clc{分母}^2}
何度も解いて体で覚えましょう!
次の関数を微分せよ。
【解答】
\def\bunsi{1} \def\bunbo{3x+2} \newcommand\cmr[1]{\colorbox{mistyrose}{$#1$}} \newcommand\clc[1]{\colorbox{lightcyan}{$#1$}} \newcommand\hosoku[1]{\textcolor{red}{\scriptsize #1}} \begin{align*} y &= \dfrac{\cmr{\bunsi}}{\clc{\bunbo}}\\ & \hosoku{\Downarrow 分子'・分母 - 分子・分母' \Downarrow}\\ y' &= \dfrac{\cmr{\bunsi}' \cdot \clc{(\bunbo)} - \cmr{\bunsi} \cdot \clc{(\bunbo)'}}{\clc{(\bunbo)^2}}\\ & \hosoku{\Uparrow 分母^2 \Uparrow}\\ &= \dfrac{0 \cdot (\bunbo) - 1 \cdot 3}{(\bunbo)^2}\\ \\ &= \dfrac{-3}{(\bunbo)^2} \end{align*}
【解答】
\def\bunsi{3x} \def\bunbo{x^2-1} \newcommand\cmr[1]{\colorbox{mistyrose}{$#1$}} \newcommand\clc[1]{\colorbox{lightcyan}{$#1$}} \newcommand\hosoku[1]{\textcolor{red}{\scriptsize #1}} \begin{align*} y &= \dfrac{\cmr{\bunsi}}{\clc{\bunbo}}\\ & \hosoku{\Downarrow 分子'・分母 - 分子・分母' \Downarrow}\\ y' &= \dfrac{\cmr{(\bunsi)}' \cdot \clc{(\bunbo)} - \cmr{\bunsi} \cdot \clc{(\bunbo)'}}{\clc{(\bunbo)^2}}\\ & \hosoku{\Uparrow 分母^2 \Uparrow}\\ &= \dfrac{3 \cdot (\bunbo) - 3x \cdot 2x}{(\bunbo)^2}\\ \\ &= \dfrac{3x^2-3-6x^2}{(\bunbo)^2}\\ \\ &= \dfrac{-3x^2-3}{(\bunbo)^2} = -\dfrac{3x^2+3}{(\bunbo)^2}\\ \end{align*}
【解答】
\def\bunsi{1} \def\bunbo{2x-3} \newcommand\cmr[1]{\colorbox{mistyrose}{$#1$}} \newcommand\clc[1]{\colorbox{lightcyan}{$#1$}} \newcommand\hosoku[1]{\textcolor{red}{\scriptsize #1}} \begin{align*} y &= \dfrac{\cmr{\bunsi}}{\clc{\bunbo}}\\ & \hosoku{\Downarrow 分子'・分母 - 分子・分母' \Downarrow}\\ y' &= \dfrac{\cmr{(\bunsi)}' \cdot \clc{(\bunbo)} - \cmr{\bunsi} \cdot \clc{(\bunbo)'}}{\clc{(\bunbo)^2}}\\ & \hosoku{\Uparrow 分母^2 \Uparrow}\\ &= \dfrac{0 \cdot (\bunbo) - 1 \cdot 2}{(\bunbo)^2}\\ \\ &= \dfrac{-2}{(\bunbo)^2} \end{align*}
【解答】
\def\bunsi{x} \def\bunbo{x^2-2} \newcommand\cmr[1]{\colorbox{mistyrose}{$#1$}} \newcommand\clc[1]{\colorbox{lightcyan}{$#1$}} \newcommand\hosoku[1]{\textcolor{red}{\scriptsize #1}} \begin{align*} y &= \dfrac{\cmr{\bunsi}}{\clc{\bunbo}}\\ & \hosoku{\Downarrow 分子'・分母 - 分子・分母' \Downarrow}\\ y' &= \dfrac{\cmr{(\bunsi)}' \cdot \clc{(\bunbo)} - \cmr{\bunsi} \cdot \clc{(\bunbo)'}}{\clc{(\bunbo)^2}}\\ & \hosoku{\Uparrow 分母^2 \Uparrow}\\ &= \dfrac{1 \cdot (\bunbo) - x \cdot 2x}{(\bunbo)^2}\\ \\ &= \dfrac{x^2-2-2x^2}{(\bunbo)^2}\\ \\ &= \dfrac{-x^2-2}{(\bunbo)^2} = -\dfrac{x^2+2}{(\bunbo)^2}\\ \end{align*}
【解答】
\def\bunsi{2x-1} \def\bunbo{x^2+1} \newcommand\cmr[1]{\colorbox{mistyrose}{$#1$}} \newcommand\clc[1]{\colorbox{lightcyan}{$#1$}} \newcommand\hosoku[1]{\textcolor{red}{\scriptsize #1}} \begin{align*} y &= \dfrac{\cmr{\bunsi}}{\clc{\bunbo}}\\ & \hosoku{\Downarrow 分子'・分母 - 分子・分母' \Downarrow}\\ y' &= \dfrac{\cmr{(\bunsi)}' \cdot \clc{(\bunbo)} - \cmr{(\bunsi)} \clc{(\bunbo)'}}{\clc{(\bunbo)^2}}\\ & \hosoku{\Uparrow 分母^2 \Uparrow}\\ &= \dfrac{2 \cdot (\bunbo) - (\bunsi) \cdot 2x}{(\bunbo)^2}\\ \\ &= \dfrac{2x^2+2-4x^2+2x}{(\bunbo)^2}\\ \\ &= \dfrac{-2x^2+2x+2}{(\bunbo)^2} \end{align*}
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- 20210913…初版公開。問題数5。