Question

For $\mathit{w}\mathbf{=}\mathbf{8}{\mathit{x}}^{\mathbf{4}}\mathbf{+}{\mathit{y}}^{\mathbf{4}}\mathbf{-}\mathbf{2}\mathit{x}{\mathit{y}}^{\mathbf{2}}$, find${\mathbf{\partial }}^{\mathbf{2}}\mathit{w}\mathbf{/}\mathbf{\partial }{\mathit{x}}^{\mathbf{2}}$and${\mathbf{\partial }}^{\mathbf{2}}\mathit{w}\mathbf{/}\mathbf{\partial }{\mathit{y}}^{\mathbf{2}}$at the points where$\mathbf{\partial }\mathit{w}\mathbf{/}\mathbf{\partial }\mathit{x}\mathbf{=}\mathbf{\partial }\mathit{w}\mathbf{/}\mathbf{\partial }\mathit{y}\mathbf{=}\mathbf{0}$.

Short answer

The value of provided equation at (0,0) both zero and at $\left(\frac{1}{4},\frac{1}{2}\right)\frac{{\partial }^{2}w}{\partial x^2}=6$and$\frac{{\partial }^{2}w}{\partial y^2}=2$

Step-by-step solution

Explanation of solution

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

Partial differentiation

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.

Calculation

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

Consider,

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

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,

$\frac{\partial w}{\partial x}=32x^3-2x{y}^2$

Since,

$\frac{{\partial }^{2}w}{\partial x^2}=96x^2$

So, at (0,0), ${\partial }^{2}w/\partial x^2=0$

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

Now,

$\frac{\partial w}{\partial y}=4y^3-4xy$

Implies that,

$\frac{{\partial }^{2}w}{\partial y^2}=12y^2-4x$

So, at (0,0)$\frac{\partial w}{\partial y}=0$

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

$\frac{{\partial }^{2}w}{\partial y^2}=12\times \frac{1}{4}-4\times \frac{1}{4}$

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