微積2.1.1b

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問題2.1.1b

　次の関数の導関数を求めよ。

（１０）$e^{{\tan }^{-1}x}$

（１１）$x\sqrt{a^2-x^2}+a^2{\sin }^{-1}{\displaystyle \frac{x}{a}}$

（１２）${\sin }^{-1}{\displaystyle \frac{x}{\sqrt{1+x^2}}}$

（１３）$2{\cos }^{-1}\sqrt{{\displaystyle \frac{x+1}{2}}}$

（１４）$\sqrt{{\displaystyle \frac{(x-1)(x-2)}{(x-3)(x-4)}}}$

（１５）$\sqrt[3]{{\displaystyle \frac{x^2+1}{(x-1{)}^2}}}$

 

《ポイント》

適宜、対数微分法を利用しましょう。

 

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《解答例》

（１０）

$y=e^{{\tan }^{-1}}x$ の両辺に自然対数を取ると、$\log y={\tan }^{-1}x$となる。

よって ${\displaystyle \frac{dy}{dx}}=y\cdot {\displaystyle \frac{1}{1+x^2}}$ である。

$\therefore {\displaystyle \frac{d}{dx}}{e}^{{\tan }^{-1}x}=e^{{\tan }^{-1}}{\displaystyle \frac{x}{1+x^2}}\cdots \cdots \text{(答)}$

$\begin{array}{rl}& {\displaystyle \frac{d}{dx}}(x^2+1{)}^5(x^3-2{)}^3\\ & =(x^3-2{)}^2(x^2+1{)}^4\{9x^2(x^2+1)+10x(x^3-2)\}\\ & =x(x^3-2{)}^2(x^2+1{)}^4(19x^3+9x-20)\cdots \cdots \text{(答)}\end{array}$

 

（１１）

${\displaystyle \frac{d}{dx}}x\sqrt{a^2-x^2}=-{\displaystyle \frac{x^2}{\sqrt{a^2-x^2}}}+\sqrt{a^2-x^2}\cdots \cdots$①

${\displaystyle \frac{d}{dx}}{a}^2{\sin }^{-1}{\displaystyle \frac{x}{a}}={\displaystyle \frac{a}{\sqrt{1-{\left({\displaystyle \frac{x}{a}}\right)}^2}}}={\displaystyle \frac{a^2}{\sqrt{a^2-x^2}}}\cdots \cdots$②

①、②の辺々を加えると

$\begin{array}{rl}& {\displaystyle \frac{d}{dx}}(x\sqrt{a^2-x^2}+a^2{\sin }^{-1}{\displaystyle \frac{x}{a}})\\ & ={\displaystyle \frac{a^2-x^2}{\sqrt{a^2-x^2}}}+\sqrt{a^2-x^2}\\ & =2\sqrt{a^2-x^2}\cdots \cdots \text{(答)}\end{array}$

 

（１２）

$\begin{array}{rl}& {\displaystyle \frac{d}{dx}}({\sin }^{-1}{\displaystyle \frac{x}{\sqrt{1+x^2}}})\\ & ={\left({\displaystyle \frac{x}{\sqrt{1+x^2}}}\right)}^{´}\cdot {\displaystyle \frac{1}{\sqrt{1-{\left({\displaystyle \frac{x}{\sqrt{1+x^2}}}\right)}^2}}}\\ & ={\displaystyle \frac{(1+x^2)-x^2}{(1+x^2)\sqrt{1+x^2}}}\sqrt{1+x^2}\\ & ={\displaystyle \frac{1}{1+x^2}}\cdots \cdots \text{(答)}\end{array}$

 

（１３）

$\begin{array}{rl}& {\displaystyle \frac{d}{dx}}(2{\cos }^{-1}\sqrt{{\displaystyle \frac{x+1}{2}}})\\ & =2{(\sqrt{{\displaystyle \frac{x+1}{2}}})}^{´}\cdot {\displaystyle \frac{-1}{\sqrt{1-{(\sqrt{{\displaystyle \frac{x+1}{2}}})}^2}}}\\ & =-2\cdot {\displaystyle \frac{1}{4\sqrt{{\displaystyle \frac{x+1}{2}}}}}\cdot {\displaystyle \frac{1}{\sqrt{{\displaystyle \frac{1-x}{2}}}}}\\ & =-{\displaystyle \frac{1}{\sqrt{1-x^2}}}\cdots \cdots \text{(答)}\end{array}$

 

（１４）

$y=\sqrt{{\displaystyle \frac{(x-1)(x-2)}{(x-3)(x-4)}}}$ の両辺に自然対数を取ると、

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

$\begin{array}{rl}& \therefore {\displaystyle \frac{dy}{dx}}\\ & =y\cdot {\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{x-1}}+{\displaystyle \frac{1}{x-2}}-{\displaystyle \frac{1}{x-3}}-{\displaystyle \frac{1}{x-4}})\\ & ={\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{x-1}}+{\displaystyle \frac{1}{x-2}}-{\displaystyle \frac{1}{x-3}}-{\displaystyle \frac{1}{x-4}})\sqrt{{\displaystyle \frac{(x-1)(x-2)}{(x-3)(x-4)}}}\cdots \cdots \text{(答)}\end{array}$

※$-{\displaystyle \frac{2x^2-10x+11}{x-1)(x-2)(x-3)(x-4)}}\sqrt{{\displaystyle \frac{(x-1)(x-2)}{(x-3)(x-4)}}}$ でも可

 

（１５）

$y=\sqrt[3]{{\displaystyle \frac{x^2+1}{(x-1{)}^2}}}$ の両辺に自然対数を取ると、

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

$\begin{array}{rl}& \therefore {\displaystyle \frac{dy}{dx}}\\ & =y\cdot {\displaystyle \frac{1}{3}}({\displaystyle \frac{2x}{x^2+1}}-{\displaystyle \frac{2}{x-1}})\\ & =\sqrt[3]{{\displaystyle \frac{x^2+1}{(x-1{)}^2}}}({\displaystyle \frac{2x}{x^2+1}}-{\displaystyle \frac{2}{x-1}})\cdots \cdots \text{(答)}\end{array}$

※$-\left({\displaystyle \frac{2(x+1)}{3(x^2+1)(x-1)}}\right)\sqrt[3]{{\displaystyle \frac{x^2+1}{(x-1{)}^2}}}$でも可

 

 

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復習例題2.1.1b

　次の関数の導関数を求めよ。

（１）${\displaystyle \frac{1}{{\sin }^{\frac{3}{2}}(x)}}$

（２）$2\arcsin \sqrt{{\displaystyle \frac{x+1}{2}}}$

（３）$\sqrt{{\displaystyle \frac{\arcsin x}{x^2+1}}}$

＞＞解答・解説

 

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