2次の偏微分

■問題

次の関数の第2次偏導関数を求めよ．

$z = \cos^{- 1} 2 x y$ $z = \cos^{- 1} 2 x y$

■答

$\frac{\partial^{2} z}{\partial x^{2}} = - \frac{8 x y^{3}}{(1 - 4 x^{2} y^{2}) \sqrt{1 - 4 x^{2} y^{2}}}$ $\frac{\partial^{2} z}{\partial x^{2}} = - \frac{8 x y^{3}}{(1 - 4 x^{2} y^{2}) \sqrt{1 - 4 x^{2} y^{2}}}$

$\frac{\partial^{2} z}{\partial y^{2}} = - \frac{8 x^{3} y}{(1 - 4 x^{2} y^{2}) \sqrt{1 - 4 x^{2} y^{2}}}$ $\frac{\partial^{2} z}{\partial y^{2}} = - \frac{8 x^{3} y}{(1 - 4 x^{2} y^{2}) \sqrt{1 - 4 x^{2} y^{2}}}$

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

■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

$z = \cos^{- 1} 2 x y$ $z = \cos^{- 1} 2 x y$ 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．

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

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

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

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

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

$z = \cos^{- 1} 2 x y$ $z = \cos^{- 1} 2 x y$ 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．

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

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

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

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

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

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

$= - 2 y \cdot (- \frac{1}{2}) {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \times \frac{\partial}{\partial x} (1 - 4 x^{2} y^{2})$ $= - 2 y \cdot (- \frac{1}{2}) {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \times \frac{\partial}{\partial x} (1 - 4 x^{2} y^{2})$

$= y {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \cdot (- 8 x y^{2})$ $= y {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \cdot (- 8 x y^{2})$

$= - \frac{8 x y^{3}}{(1 - 4 x^{2} y^{2}) \sqrt{1 - 4 x^{2} y^{2}}}$ $= - \frac{8 x y^{3}}{(1 - 4 x^{2} y^{2}) \sqrt{1 - 4 x^{2} y^{2}}}$

● $\frac{\partial^{2} z}{\partial y^{2}}$ $\frac{\partial^{2} z}{\partial y^{2}}$ 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．

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

$= - 2 x \cdot (- \frac{1}{2}) {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \times \frac{\partial}{\partial y} (1 - 4 x^{2} y^{2})$ $= - 2 x \cdot (- \frac{1}{2}) {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \times \frac{\partial}{\partial y} (1 - 4 x^{2} y^{2})$

$= x {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \cdot (- 8 x^{2} y)$ $= x {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \cdot (- 8 x^{2} y)$

$= - \frac{8 x^{3} y}{(1 - 4 x^{2} y^{2}) \sqrt{1 - 4 x^{2} y^{2}}}$ $= - \frac{8 x^{3} y}{(1 - 4 x^{2} y^{2}) \sqrt{1 - 4 x^{2} y^{2}}}$

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

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

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

積の微分の公式を用いて,

$= - 2 {(1 - 4 x^{2} y^{2})}^{- \frac{1}{2}} + (- 2 y) \cdot (- \frac{1}{2}) {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \times \frac{\partial}{\partial y} (1 - 4 x^{2} y^{2})$ $= - 2 {(1 - 4 x^{2} y^{2})}^{- \frac{1}{2}} + (- 2 y) \cdot (- \frac{1}{2}) {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \times \frac{\partial}{\partial y} (1 - 4 x^{2} y^{2})$

$= - 2 {(1 - 4 x^{2} y^{2})}^{- \frac{1}{2}} + y {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \cdot (- 8 x^{2} y)$ $= - 2 {(1 - 4 x^{2} y^{2})}^{- \frac{1}{2}} + y {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \cdot (- 8 x^{2} y)$

$= \frac{- 2 + 8 x^{2} y^{2} - 8 x^{2} y^{2}}{(1 - 4 x^{2} y^{2}) \sqrt{1 - 4 x^{2} y^{2}}}$ $= \frac{- 2 + 8 x^{2} y^{2} - 8 x^{2} y^{2}}{(1 - 4 x^{2} y^{2}) \sqrt{1 - 4 x^{2} y^{2}}}$

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

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

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

$= - 2 {(1 - 4 x^{2} y^{2})}^{- \frac{1}{2}} + (- 2 x) \cdot (- \frac{1}{2}) {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \times \frac{\partial}{\partial x} (1 - 4 x^{2} y^{2})$ $= - 2 {(1 - 4 x^{2} y^{2})}^{- \frac{1}{2}} + (- 2 x) \cdot (- \frac{1}{2}) {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \times \frac{\partial}{\partial x} (1 - 4 x^{2} y^{2})$

$= - 2 {(1 - 4 x^{2} y^{2})}^{- \frac{1}{2}} + x {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \cdot (- 8 x y^{2})$ $= - 2 {(1 - 4 x^{2} y^{2})}^{- \frac{1}{2}} + x {(1 - 4 x^{2} y^{2})}^{- \frac{3}{2}} \cdot (- 8 x y^{2})$

$= \frac{- 2 (1 - 4 x^{2} y^{2}) - 8 x y^{2}}{(1 - 4 x^{2} y^{2}) \sqrt{1 - 4 x^{2} y^{2}}}$ $= \frac{- 2 (1 - 4 x^{2} y^{2}) - 8 x y^{2}}{(1 - 4 x^{2} y^{2}) \sqrt{1 - 4 x^{2} y^{2}}}$

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