Частни производни

[Гост]

Не мога и не мога да се справя с това
Z = ln x + y / √xy
някой ще може ли ?

[ammornil]

Може би ще е нещо като това:
[tex]z(x,y)=lnx+\frac{y}{\sqrt{x.y}} \\
DM: \left|x>0 \\ y > 0 \right, \\
z(x,y)=lnx+\frac{\sqrt{y}}{\sqrt{x}}, \\
\frac{\partial z}{\partial x}=\frac{1}{x}+\sqrt{y}.(-\frac{\frac{1}{2.\sqrt{x}}}{|x|})=\frac{1}{x}-\frac{\sqrt
{y}}{2.|x|.\sqrt{x}}; \hspace{120} \frac{\partial z}{\partial y}=\frac{1}{\sqrt{x}}.\frac{1}{2.\sqrt{y}}=\frac{1}{2.\sqrt{x}.\sqrt{y}} \\
\frac{\partial^2 z}{\partial x^2}=-\frac{1}{x^2}-\sqrt{y}.\frac{-2.3.x^2.\frac{1}{2.\sqrt{x^3}}}{4.|x^{3}|}=-\frac{1}{x^2}+\frac{3.\sqrt{y}}{4.|x^3|.\sqrt{x}}; \hspace{20} \frac{\partial^2 z}{\partial y^2}=\frac{1}{2.\sqrt{x}}.(-\frac{1}{2.|y|.\sqrt{y}})=-\frac{1}{4.\sqrt{x}.|y|.\sqrt{y}}[/tex]

[ammornil]

[tex]\frac{\partial z}{\partial x \partial y}=-\frac{1}{2.|x|.\sqrt{x}}.\frac{1}{2.\sqrt{y}}=-\frac{1}{4.|x|.\sqrt{x}.\sqrt{y}}[/tex]

[ammornil]

Грешно записване по-горе. Трябва да е:
[tex]\frac{\partial^2 z}{\partial x \partial y}=-\frac{1}{2.|x|.\sqrt{x}}.\frac{1}{2.\sqrt{y}}=-\frac{1}{4.|x|.\sqrt{x}.\sqrt{y}}[/tex]