91Ó°ÊÓ

In this question, it is provided that $w=x^3-y^3-2xy+6$ and

Now, find $\frac{{∂}^{2}w}{∂x^2}\mathrm{and}\frac{{∂}^2\mathrm{w}}{∂\mathrm{y}}=0$

The procedure of calculating the partial derivative of a function is called partial differentiation. This method is used to get the partial derivative of a function with respect to one variable while keeping the other constant.

Putt $\frac{∂w}{∂x}=0$and $\frac{∂w}{∂y}=0$,

Consider,

role="math" localid="1659089100795" $$\begin{gathered} 3x^2-2y=0...\left(1\right) \\ -3y^2-2x=0...\left(2\right) \end{gathered}$$

Solve equation 1 and 2, (0,0)and$\left(\frac{-2}{3},\frac{2}{3}\right)$

Now,

$w=x^3-y^3-2xy+6$

Implies that,

$\frac{∂w}{∂x}=3x^2-2y$

Since,

$\frac{{∂}^{2}w}{∂x^2}=6x$

Therefore, At (0,0), $\frac{{∂}^{2}w}{∂x^2}=0$

And at role="math" localid="1659089309920" $\left(\frac{-2}{3},\frac{2}{3}\right),\frac{{∂}^{2}w}{∂x^2}=-4$

Now, $\frac{{∂}^{2}w}{∂x^2}=-3y^2-2x$

Implies that,

$\frac{{∂}^{2}w}{∂y^2}=-6y$

So,at (0,0);$\frac{{∂}^{2}w}{∂y^2}=0$

And at$\left(\frac{-2}{3},\frac{2}{3}\right);\frac{{∂}^{2}w}{∂y^2}$