$z = y^x$ について、以下の偏導関数を求める問題です。 1. $z_x$, $z_y$ を計算する。解析学偏微分偏導関数指数関数2025/6/261. 問題の内容z=yxz = y^xz=yx について、以下の偏導関数を求める問題です。1. $z_x$, $z_y$ を計算する。2. 2階偏導関数($z_{xx}$, $z_{xy}$, $z_{yx}$, $z_{yy}$)を計算する。2. 解き方の手順1. $z_x$ と $z_y$ を計算します。z=yxz = y^xz=yx を xxx で偏微分すると、 zx=∂z∂x=yxlogyz_x = \frac{\partial z}{\partial x} = y^x \log yzx=∂x∂z=yxlogy z=yxz = y^xz=yx を yyy で偏微分すると、 zy=∂z∂y=xyx−1z_y = \frac{\partial z}{\partial y} = x y^{x-1}zy=∂y∂z=xyx−12. 2階偏導関数を計算します。zxx=∂2z∂x2=∂∂x(yxlogy)=yx(logy)2z_{xx} = \frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x} (y^x \log y) = y^x (\log y)^2zxx=∂x2∂2z=∂x∂(yxlogy)=yx(logy)2 zxy=∂2z∂x∂y=∂∂y(yxlogy)=∂∂y(yx)logy+yx∂∂y(logy)=xyx−1logy+yx1y=xyx−1logy+yx−1z_{xy} = \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y} (y^x \log y) = \frac{\partial}{\partial y} (y^x) \log y + y^x \frac{\partial}{\partial y} (\log y) = xy^{x-1} \log y + y^x \frac{1}{y} = xy^{x-1} \log y + y^{x-1}zxy=∂x∂y∂2z=∂y∂(yxlogy)=∂y∂(yx)logy+yx∂y∂(logy)=xyx−1logy+yxy1=xyx−1logy+yx−1 zyx=∂2z∂y∂x=∂∂x(xyx−1)=∂∂x(x)yx−1+x∂∂x(yx−1)=yx−1+xyx−1logyz_{yx} = \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial x} (x y^{x-1}) = \frac{\partial}{\partial x} (x) y^{x-1} + x \frac{\partial}{\partial x} (y^{x-1}) = y^{x-1} + x y^{x-1} \log yzyx=∂y∂x∂2z=∂x∂(xyx−1)=∂x∂(x)yx−1+x∂x∂(yx−1)=yx−1+xyx−1logy zyy=∂2z∂y2=∂∂y(xyx−1)=x(x−1)yx−2z_{yy} = \frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y} (xy^{x-1}) = x(x-1)y^{x-2}zyy=∂y2∂2z=∂y∂(xyx−1)=x(x−1)yx−23. 最終的な答え1. $z_x = y^x \log y$zy=xyx−1z_y = x y^{x-1}zy=xyx−12. $z_{xx} = y^x (\log y)^2$zxy=xyx−1logy+yx−1=yx−1(xlogy+1)z_{xy} = x y^{x-1} \log y + y^{x-1} = y^{x-1}(x \log y + 1)zxy=xyx−1logy+yx−1=yx−1(xlogy+1) zyx=yx−1+xyx−1logy=yx−1(1+xlogy)z_{yx} = y^{x-1} + x y^{x-1} \log y = y^{x-1}(1 + x \log y)zyx=yx−1+xyx−1logy=yx−1(1+xlogy) zyy=x(x−1)yx−2z_{yy} = x(x-1)y^{x-2}zyy=x(x−1)yx−2