The derivative of Multivariate function For Multivariate function
$$f(x,y)=x^3+y^3$$
How to express $$f''(x,y)$$
 A: Assuming you are using the Hessian for your derivative, which is the second partials, it would be given by:
$$f''(x,y) = \pmatrix{f_{xx} & f_{xy}
\\ f_{yx} & f_{yy}}
$$
Using:
$$f(x,y)=x^3+y^3$$
We find:
$f_{xx} = 6x$, $f_{xy} = 0$, $f_{yx} = 0$, $f_{yy} = 6y$, hence:
$$f''(x,y) = \pmatrix{6x & 0
\\ 0 & 6y}
$$
Regards
A: Total derivative of $f(x,y)$
$$df(x,y)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$$
$$d^2f(x,y)=d\big(df(x,y)\big)$$
$$d^2f(x,y)=d\bigg(\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy\bigg)$$
$$d^2f(x,y)=\frac{\partial^2 f}{\partial x^2}dx^2+\frac{\partial^2 f}{\partial x\partial y}dxdy+\frac{\partial^2 f}{\partial y\partial x}dydx+\frac{\partial^2 f}{\partial y^2}dy^2$$
In your particular case where $f(x,y)=x^3+y^3$$
$$\frac{\partial^2 f}{\partial x^2}=6x$$
$$\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial (3x^2)}{\partial y}=0$$
$$\frac{\partial^2 f}{\partial y\partial x}=\frac{\partial (3y^2)}{\partial x}=0$$
$$\frac{\partial^2 f}{\partial y^2}=6y$$
$$\Rightarrow d^2f(x,y)=6x\ dx^2+6y\ dy^2$$
