2次の偏微分

■問題

次の関数の第2次偏導関数を求めよ．

$z = \sin^{- 1} x y$ $z = \sin^{- 1} x y$

■答

$\frac{\partial^{2} z}{\partial x^{2}} = \frac{x y^{3}}{(1 - x {}^{2}y^{2}) \sqrt{1 - x {}^{2}y^{2}}}$ $\frac{\partial^{2} z}{\partial x^{2}} = \frac{x y^{3}}{(1 - x {}^{2}y^{2}) \sqrt{1 - x {}^{2}y^{2}}}$

$\frac{\partial^{2} z}{\partial y^{2}} = \frac{x^{3} y}{(1 - x {}^{2}y^{2}) \sqrt{1 - x {}^{2}y^{2}}}$ $\frac{\partial^{2} z}{\partial y^{2}} = \frac{x^{3} y}{(1 - x {}^{2}y^{2}) \sqrt{1 - x {}^{2}y^{2}}}$

$\frac{\partial^{2} z}{\partial y \partial x} = \frac{\partial^{2} z}{\partial x \partial y}$ $\frac{\partial^{2} z}{\partial y \partial x} = \frac{\partial^{2} z}{\partial x \partial y}$ $= \frac{1}{(1 - x {}^{2}y^{2}) \sqrt{1 - x {}^{2}y^{2}}}$ $= \frac{1}{(1 - x {}^{2}y^{2}) \sqrt{1 - x {}^{2}y^{2}}}$

■ヒント

2次偏導関数 $\frac{\partial^{2} z}{\partial x^{2}}$ $\frac{\partial^{2} z}{\partial x^{2}}$ ， $\frac{\partial^{2} z}{\partial y^{2}}$ $\frac{\partial^{2} z}{\partial y^{2}}$ ， $\frac{\partial^{2} z}{\partial y \partial x}$ $\frac{\partial^{2} z}{\partial y \partial x}$ ， $\frac{\partial^{2} z}{\partial x \partial y}$ $\frac{\partial^{2} z}{\partial x \partial y}$ の4つを求める．

$\frac{\partial z}{\partial x}$ $\frac{\partial z}{\partial x}$ ， $\frac{\partial z}{\partial y}$ $\frac{\partial z}{\partial y}$ を計算してから，それぞれを更に $x$ $x$ ， $y$ $y$ で偏微分する．

■解説

● $\frac{\partial z}{\partial x}$ $\frac{\partial z}{\partial x}$ の計算

$z = \sin^{- 1} x y$ $z = \sin^{- 1} x y$ を偏導関数の定義より， $y$ $y$ を定数とみなして $x$ $x$ で微分する．

$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} (\sin^{- 1} x y)$ $\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} (\sin^{- 1} x y)$

$= \frac{1}{\sqrt{1 - x {}^{2}y^{2}}} \times \frac{\partial}{\partial x} (x y)$ $= \frac{1}{\sqrt{1 - x {}^{2}y^{2}}} \times \frac{\partial}{\partial x} (x y)$

$= \frac{y}{\sqrt{1 - x {}^{2}y^{2}}}$ $= \frac{y}{\sqrt{1 - x {}^{2}y^{2}}}$

$= y {(1 - x^{2} y^{2})}^{- \frac{1}{2}}$ $= y {(1 - x^{2} y^{2})}^{- \frac{1}{2}}$ 　･･････(1)

● $\frac{\partial z}{\partial y}$ $\frac{\partial z}{\partial y}$ の計算

$z = \sin^{- 1} x y$ $z = \sin^{- 1} x y$ を偏導関数の定義より， $x$ $x$ を定数とみなして $y$ $y$ で微分する．

$\frac{\partial z}{\partial y} = \frac{\partial}{\partial y} (\cos x^{2} y^{2})$ $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y} (\cos x^{2} y^{2})$

$= \frac{1}{\sqrt{1 - {(x y)}^{2}}} \times \frac{\partial}{\partial x} (x y)$ $= \frac{1}{\sqrt{1 - {(x y)}^{2}}} \times \frac{\partial}{\partial x} (x y)$

$= \frac{x}{\sqrt{1 - x^{2} y^{2}}}$ $= \frac{x}{\sqrt{1 - x^{2} y^{2}}}$

$= x {(1 - x^{2} y^{2})}^{- \frac{1}{2}}$ $= x {(1 - x^{2} y^{2})}^{- \frac{1}{2}}$ 　･･････(2)

● $\frac{\partial^{2} z}{\partial x^{2}}$ $\frac{\partial^{2} z}{\partial x^{2}}$ の計算

(1)を更に，偏導関数の定義より， $y$ $y$ を定数とみなして $x$ $x$ で微分する．

$\frac{\partial^{2} z}{\partial x^{2}}$ $\frac{\partial^{2} z}{\partial x^{2}}$ $= \frac{\partial}{\partial x} (\frac{\partial z}{\partial x})$ $= \frac{\partial}{\partial x} (\frac{\partial z}{\partial x})$

$= \frac{\partial}{\partial x} {y {(1 - x^{2} y^{2})}^{- \frac{1}{2}}}$ $= \frac{\partial}{\partial x} {y {(1 - x^{2} y^{2})}^{- \frac{1}{2}}}$

$= y \cdot (- \frac{1}{2}) \cdot {(1 - x^{2} y^{2})}^{- \frac{3}{2}} \cdot \frac{\partial}{\partial x} (1 - x^{2} y^{2})$ $= y \cdot (- \frac{1}{2}) \cdot {(1 - x^{2} y^{2})}^{- \frac{3}{2}} \cdot \frac{\partial}{\partial x} (1 - x^{2} y^{2})$

$= - \frac{1}{2} y {(1 - x^{2} y^{2})}^{- \frac{3}{2}} \cdot (- 2 x y^{2})$ $= - \frac{1}{2} y {(1 - x^{2} y^{2})}^{- \frac{3}{2}} \cdot (- 2 x y^{2})$

$= \frac{x y^{3}}{(1 - x^{2} y^{2}) \sqrt{1 - x^{2} y^{2}}}$ $= \frac{x y^{3}}{(1 - x^{2} y^{2}) \sqrt{1 - x^{2} y^{2}}}$

● $\frac{\partial^{2} z}{\partial y^{2}}$ $\frac{\partial^{2} z}{\partial y^{2}}$ の計算

(2)を更に，偏導関数の定義より， $x$ $x$ を定数とみなして $y$ $y$ で微分する．

$\frac{\partial^{2} z}{\partial y^{2}}$ $\frac{\partial^{2} z}{\partial y^{2}}$ $= \frac{\partial}{\partial y} (\frac{\partial z}{\partial y})$ $= \frac{\partial}{\partial y} (\frac{\partial z}{\partial y})$

$= x \cdot (- \frac{1}{2}) {(1 - x^{2} y^{2})}^{- \frac{3}{2}} \times \frac{\partial}{\partial y} (1 - x^{2} y^{2})$ $= x \cdot (- \frac{1}{2}) {(1 - x^{2} y^{2})}^{- \frac{3}{2}} \times \frac{\partial}{\partial y} (1 - x^{2} y^{2})$

$= - \frac{1}{2} x {(1 - x^{2} y^{2})}^{- \frac{3}{2}} \cdot (- 2 x^{2} y)$ $= - \frac{1}{2} x {(1 - x^{2} y^{2})}^{- \frac{3}{2}} \cdot (- 2 x^{2} y)$

$= \frac{x^{3} y}{(1 - x^{2} y^{2}) \sqrt{1 - x^{2} y^{2}}}$ $= \frac{x^{3} y}{(1 - x^{2} y^{2}) \sqrt{1 - x^{2} y^{2}}}$

● $\frac{\partial^{2} z}{\partial y \partial x}$ $\frac{\partial^{2} z}{\partial y \partial x}$ の計算

(1)を更に，偏導関数の定義より， $x$ $x$ を定数とみなして $y$ $y$ で微分する．

$\frac{\partial^{2} z}{\partial y \partial x}$ $\frac{\partial^{2} z}{\partial y \partial x}$ $= \frac{\partial}{\partial y} (\frac{\partial z}{\partial x})$ $= \frac{\partial}{\partial y} (\frac{\partial z}{\partial x})$

$= \frac{\partial}{\partial y} {y {(1 - x^{2} y^{2})}^{- \frac{1}{2}}}$ $= \frac{\partial}{\partial y} {y {(1 - x^{2} y^{2})}^{- \frac{1}{2}}}$

積の微分の公式を用いて

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

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

$= \frac{1 - x^{2} y^{2} + x^{2} y^{2}}{(1 - x^{2} y^{2}) \sqrt{1 - x^{2} y^{2}}}$ $= \frac{1 - x^{2} y^{2} + x^{2} y^{2}}{(1 - x^{2} y^{2}) \sqrt{1 - x^{2} y^{2}}}$

$= \frac{1}{(1 - x^{2} y^{2}) \sqrt{1 - x^{2} y^{2}}}$ $= \frac{1}{(1 - x^{2} y^{2}) \sqrt{1 - x^{2} y^{2}}}$ 　･･････(3)

● $\frac{\partial^{2} z}{\partial x \partial y}$ $\frac{\partial^{2} z}{\partial x \partial y}$ の計算

(2)を更に，偏導関数の定義より， $y$ $y$ を定数とみなして $x$ $x$ で微分する．

$= \frac{\partial}{\partial x} {x {(1 - x^{2} y^{2})}^{- \frac{1}{2}}}$ $= \frac{\partial}{\partial x} {x {(1 - x^{2} y^{2})}^{- \frac{1}{2}}}$

$= {(1 - x^{2} y^{2})}^{- \frac{1}{2}} + x \cdot (- \frac{1}{2}) \cdot {(1 - x^{2} y^{2})}^{- \frac{3}{2}} \times \frac{\partial}{\partial x} (1 - x^{2} y^{2})$ $= {(1 - x^{2} y^{2})}^{- \frac{1}{2}} + x \cdot (- \frac{1}{2}) \cdot {(1 - x^{2} y^{2})}^{- \frac{3}{2}} \times \frac{\partial}{\partial x} (1 - x^{2} y^{2})$

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

$= \frac{1 - x^{2} y^{2} + x^{2} y^{2}}{(1 - x^{2} y^{2}) \sqrt{1 - x^{2} y^{2}}}$ $= \frac{1 - x^{2} y^{2} + x^{2} y^{2}}{(1 - x^{2} y^{2}) \sqrt{1 - x^{2} y^{2}}}$

$= \frac{1}{(1 - x^{2} y^{2}) \sqrt{1 - x^{2} y^{2}}}$ $= \frac{1}{(1 - x^{2} y^{2}) \sqrt{1 - x^{2} y^{2}}}$ 　･･････(4)

(3)，(4)より

$\frac{\partial^{2} z}{\partial y \partial x} = \frac{\partial^{2} z}{\partial x \partial y}$ $\frac{\partial^{2} z}{\partial y \partial x} = \frac{\partial^{2} z}{\partial x \partial y}$

となり，2次偏導関数は偏微分する順序には無関係であることが確かめられた．

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最終更新日： 2023年8月31日

2次の偏微分

■問題

■答

■ヒント

■解説

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