robynalana0108 robynalana0108
  • 22-04-2019
  • Mathematics
contestada

For the function f(x,y) = 3x2 + 2x4y2 + 6y3, find the value of fyy(1,1).

Respuesta :

LammettHash
LammettHash LammettHash
  • 25-04-2019

[tex]f_{yy}[/tex] is the second partial derivative with respect to [tex]y[/tex]:

[tex]f(x,y)=3x^2+2x^4y^2+6y^3[/tex]

[tex]\implies f_y(x,y)=4x^4y+18y^2[/tex]

[tex]\implies f_{yy}(x,y)=4x^4+36y[/tex]

Then at the point [tex](x,y)=(1,1)[/tex], we have

[tex]f_{yy}(1,1)=40[/tex]

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