Rozwiązanie:
Funkcja:
[tex]$f(x,y,z)=\frac{\sqrt{x^{2}+y^{2}}}{z} [/tex]
Mamy:
[tex]$\frac{\partial }{\partial x} f(x,y,z)=\frac{1}{z} \cdot \frac{\partial f}{\partial x} (\sqrt{x^{2}+y^{2}})=\frac{1}{z} \cdot \frac{1}{2\sqrt{x^{2}+y^{2}} } \cdot 2x=\frac{x}{z\sqrt{x^{2}+y^{2}} } [/tex]
[tex]$\frac{\partial }{\partial z} f(x,y,z)=\sqrt{x^{2}+y^{2}} \cdot \frac{\partial f}{\partial z} \Big(\frac{1}{z}\Big) =-\frac{\sqrt{x^{2}+y^{2}}}{z^{2}} [/tex]
[tex]$\frac{\partial^{2} }{\partial z^{2}} f(x,y,z)=\frac{\partial f}{\partial z} \Big(\frac{\partial f}{\partial z} \Big)=-\sqrt{x^{2}+y^{2}} \cdot \frac{\partial f}{\partial z} \Big(\frac{1}{z^{2}} \Big)=\frac{2\sqrt{x^{2}+y^{2}} }{z^3}} [/tex]
[tex]$\frac{\partial ^{2}}{\partial x \partial z} f(x,y,z)=\frac{\partial f}{\partial x}\Big(\frac{\partial f}{\partial z} \Big) =-\frac{1}{z^{2}} \cdot \frac{\partial f}{\partial x} \Big(\sqrt{x^{2}+y^{2}}\Big)=-\frac{1}{z^{2}} \cdot \frac{2x}{2\sqrt{x^{2}+y^{2}}} =[/tex]
[tex]$=-\frac{x}{z^{2}\sqrt{x^{2}+y^{2}} } [/tex]
