2.24 Знайти чсатинні похідні першого порядку функцій
3.24 опис завдання на фотографії
[m]\frac{ ∂z }{ ∂x }=(cos\sqrt{\frac{x}{y^5}})`_{x}=-sin\sqrt{\frac{x}{y^5}}\cdot (\sqrt{\frac{x}{y^5}})`_{x}=-sin\sqrt{\frac{x}{y^5}}\cdot\cdot\frac{1}{2\sqrt{\frac{x}{y^5}}}\cdot (\frac{x}{y^5})`_{x}=-sin\sqrt{\frac{x}{y^5}}\cdot \frac{1}{2\sqrt{\frac{x}{y^5}}}\cdot (\frac{1}{y^5})\cdot (x)`_{x}=-sin\sqrt{\frac{x}{y^5}}\cdot \frac{1}{2\sqrt{\frac{x}{y^5}}}\cdot (\frac{1}{y^5})[/m]
[m]\frac{ ∂z }{ ∂y }=(cos\sqrt{\frac{x}{y^5}})`_{y}=-sin\sqrt{\frac{x}{y^5}}\cdot (\sqrt{\frac{x}{y^5}})`_{y}-sin\sqrt{\frac{x}{y^5}}\cdot\frac{1}{2\sqrt{\frac{x}{y^5}}}\cdot (\frac{x}{y^5})`_{y}=-sin\sqrt{\frac{x}{y^5}}\cdot\frac{1}{2\sqrt{\frac{x}{y^5}}}\cdot x \cdot (-5y^{-6})[/m]
2)
[m]\frac{ ∂z }{ ∂x }=(sin(x^3-y))`_{x}=cos(x^3-y)\cdot (x^3-y)`_{x}=cos(x^3-y)\cdot (3x^2)=3x^2\cdot cos(x^3-y)[/m]
[m]\frac{ ∂z }{ ∂y }=(sin(x^3-y)`_{y}=cos(x^3-y)\cdot (x^3-y)`_{y}cos(x^3-y)\cdot (-1)=-cos(x^3-y)[/m]
[m]\frac{ ∂^2z }{ ∂x ∂y }=(\frac{ ∂z }{ ∂x })`_{y}=()`_{y}=(3x^2\cdot cos(x^3-y))`_{y}=3x^2\cdot (-sin(x^3-y)\cdot (x^3-y)`_{y}=3x^2\cdot (-sin(x^3-y)\cdot (-1)=3x^2\cdot cos(x^3-y)[/m]
[m]\frac{ ∂^2z }{ ∂y ∂x }=(\frac{ ∂z }{ ∂y })`_{x}=(-cos(x^3-y))`_{x}=-(-sin(x^3-y))\cdot (x^3-y)`_{y}=sin(x^3-y)\cdot (3x^2)=3x^2\cdot cos(x^3-y)[/m]
[m]\frac{ ∂^2z }{ ∂x ∂y}=\frac{ ∂^2z }{ ∂y ∂x }[/m]