与えられた関数 $z$ について、$x$ と $y$ に関する偏導関数 $\frac{\partial z}{\partial x}$ と $\frac{\partial z}{\partial y}$ をそれぞれ求めます。

はい、承知いたしました。与えられた6つの関数について、偏微分を求めます。

1. 問題の内容

与えられた関数 $z$ について、$x$ と $y$ に関する偏導関数 $\frac{\partial z}{\partial x}$ と $\frac{\partial z}{\partial y}$ をそれぞれ求めます。

2. 解き方の手順

(1) $z = x^3 + y^3 - 3axy$

* $\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(x^3 + y^3 - 3axy) = 3x^2 - 3ay$

* $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(x^3 + y^3 - 3axy) = 3y^2 - 3ax$

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

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

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

(3) $z = e^{ax} \cos{by}$

* $\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(e^{ax} \cos{by}) = a e^{ax} \cos{by}$

* $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(e^{ax} \cos{by}) = -b e^{ax} \sin{by}$

(4) $z = \log{(x^2 + y^2)}$

* $\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(\log{(x^2 + y^2)}) = \frac{1}{x^2 + y^2} \cdot 2x = \frac{2x}{x^2 + y^2}$

* $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(\log{(x^2 + y^2)}) = \frac{1}{x^2 + y^2} \cdot 2y = \frac{2y}{x^2 + y^2}$

(5) $z = x^y$

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

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

(6) $z = \sin^{-1}{\left(\frac{x}{y}\right)}$

* $\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}\left(\sin^{-1}{\left(\frac{x}{y}\right)}\right) = \frac{1}{\sqrt{1 - \left(\frac{x}{y}\right)^2}} \cdot \frac{1}{y} = \frac{1}{y \sqrt{1 - \frac{x^2}{y^2}}} = \frac{1}{\sqrt{y^2 - x^2}}$

* $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}\left(\sin^{-1}{\left(\frac{x}{y}\right)}\right) = \frac{1}{\sqrt{1 - \left(\frac{x}{y}\right)^2}} \cdot \left(-\frac{x}{y^2}\right) = -\frac{x}{y^2 \sqrt{1 - \frac{x^2}{y^2}}} = -\frac{x}{y\sqrt{y^2 - x^2}}$

3. 最終的な答え

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

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

(3) $\frac{\partial z}{\partial x} = ae^{ax}\cos{by}$, $\frac{\partial z}{\partial y} = -be^{ax}\sin{by}$

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

(5) $\frac{\partial z}{\partial x} = yx^{y-1}$, $\frac{\partial z}{\partial y} = x^y \log{x}$

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