Question

If F(x,y,z)=(X^3+y^2+z)^4 find each of the following:

(1)Fx(x,y,z)=? 2) fy(0,1,1)=? 3) fzz(x,y,z)=?

Answer 1

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


`f_x=\frac\partial{\partial x}(x^3+y^2+z)^4=4(x^3+y^2+z)^3\frac\partial{\partial x}(x^3+y^2+z)=12x^2(x^3+y^2+z)^3`


`f_y=\frac\partial{\partial y}(x^3+y^2+z)^4=4(x^3+y^2+z)^3\frac\partial{\partial y}(x^3+y^2+z)=8y(x^3+y^2+z)^3`

`\implies f_y(0,1,1)=8(2)^3=64`


`f_(zz)=(\partial^2)/(\partial z^2)(x^3+y^2+z)^4=\frac\partial{\partial z}\left(4(x^3+y^2+z)^3\frac\partial{\partial z}(x^3+y^2+z)\right)`

`=4\frac\partial{\partial z}(x^3+y^2+z)^3=12(x^3+y^2+z)^2\frac\partial{\partial z}(x^3+y^2+z)`

`=12(x^3+y^2+z)^2`