Question

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

Answer 1

`f_(yy)` is the second partial derivative with respect to
`y`:

`f(x,y)=3x^2+2x^4y^2+6y^3`

`\implies f_y(x,y)=4x^4y+18y^2`

`\implies f_(yy)(x,y)=4x^4+36y`

Then at the point
`(x,y)=(1,1)`, we have

`f_(yy)(1,1)=40`