Respuesta :

LammettHash
[tex]f(x,y,z)=(x^3+y^2+z)^4[/tex]

[tex]f_x=\dfrac\partial{\partial x}(x^3+y^2+z)^4=4(x^3+y^2+z)^3\dfrac\partial{\partial x}(x^3+y^2+z)=12x^2(x^3+y^2+z)^3[/tex]

[tex]f_y=\dfrac\partial{\partial y}(x^3+y^2+z)^4=4(x^3+y^2+z)^3\dfrac\partial{\partial y}(x^3+y^2+z)=8y(x^3+y^2+z)^3[/tex]
[tex]\implies f_y(0,1,1)=8(2)^3=64[/tex]

[tex]f_{zz}=\dfrac{\partial^2}{\partial z^2}(x^3+y^2+z)^4=\dfrac\partial{\partial z}\left(4(x^3+y^2+z)^3\dfrac\partial{\partial z}(x^3+y^2+z)\right)[/tex]
[tex]=4\dfrac\partial{\partial z}(x^3+y^2+z)^3=12(x^3+y^2+z)^2\dfrac\partial{\partial z}(x^3+y^2+z)[/tex]
[tex]=12(x^3+y^2+z)^2[/tex]

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