91Ó°ÊÓ

In this question, it is provided that $w=x^4+y^4-2x{y}^2$ and to find$\frac{{∂}^{2}w}{∂x^2}$ and $\frac{{∂}^{2}w}{∂y^2}$.

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.

Put $\frac{{∂}^{2}w}{∂y^2}$and $c\frac{∂w}{∂x}=0$

Consider,

$$\begin{gathered} 32x^3-2y^2=0→| \\ 4y^3-4xy=0→|| \end{gathered}$$

Solve equation I and II two points, namely (0,0) and $\left(\frac{1}{4},\frac{1}{2}\right)$

Now,

$w=8x^4+y^4-2x{y}^2$

Implies that,

$\frac{∂w}{∂x}=32x^3-2x{y}^2$

Since,

$\frac{{∂}^{2}w}{∂x^2}=96x^2$

So, at (0,0), ${∂}^{2}w/∂x^2=0$

And at role="math" localid="1659090872615" $\left(1/4,1/2\right);{∂}^{2}w/∂x^2=6$

Now,

$\frac{∂w}{∂y}=4y^3-4xy$

Implies that,

$\frac{{∂}^{2}w}{∂y^2}=12y^2-4x$

So, at (0,0)$\frac{∂w}{∂y}=0$

And at the point $(1/4,1/2);$

$\frac{{∂}^{2}w}{∂y^2}=12×\frac{1}{4}-4×\frac{1}{4}$

Therefore, at$\left(\frac{1}{4},\frac{1}{2}\right);\frac{{∂}^{2}w}{∂y^2}=2$