The derivative of Multivariate function

Question

For Multivariate function $$f(x,y)=x^3+y^3$$ How to express $$f''(x,y)$$

Answer 1

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

Answer 2

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$$

$$\frac{\partial^2 f}{\partial x^2}=6x$$ $$\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial (3x^2)}{\partial y}=0$$ $$\frac{\partial^2 f}{\partial y\partial x}=\frac{\partial (3y^2)}{\partial x}=0$$ $$\frac{\partial^2 f}{\partial y^2}=6y$$ $$\Rightarrow d^2f(x,y)=6x\ dx^2+6y\ dy^2$$