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

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One way to do that would be to compute the Hessian matrix: en.wikipedia.org/wiki/Hessian_matrix –  1015 Feb 17 '13 at 4:07
    
What's the definition of the derivative that you're using? Do you want the Hessian matrix or an element of $L(\mathbb{R}^2,L(\mathbb{R}^2,\mathbb{R}))$? –  wj32 Feb 17 '13 at 4:07
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Please expound in what sense $''$ means "derivative". –  JohnD Feb 17 '13 at 4:25

2 Answers 2

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

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This is a function from $R^2$ to $R$ so how can the derivative be a $2\times 2$ matrix. It must be a $1\times 2$ matrix? Consider this as a question and try explaining me the reason. –  Abhra Abir Kundu Feb 17 '13 at 4:35
    
@AbhraAbirKundu: see the link I included for second derivative of a function $f(x, y)$. Regards –  Amzoti Feb 17 '13 at 4:38
    
Thanks buddy..... I was thinking abt the first derivative. I guess the first derivative is a $1\times 2$ matrix. What do you say????? –  Abhra Abir Kundu Feb 17 '13 at 4:40
    
@AbhraAbirKundu: The first derivatives would be $f_x = 3x^{2}$ and $f_y = 3y^{2}$ and certainly can be represented however you need it (as a vector if you are looking for extremum can work). Regards –  Amzoti Feb 17 '13 at 4:42
    
We need an upvote here! +1 –  amWhy May 2 '13 at 1:01

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$$

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

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