91Ó°ÊÓ

The given data are listed below

$$\begin{gathered} z=x^2+2y^2.....\left(1\right) \\ x=r\cos \mathrm{θ}.....\left(2\right) \\ y=r\sin \mathrm{θ}.....\left(3\right) \end{gathered}$$

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.

From equation 2 and 3,

$$\begin{gathered} \frac{y}{x}=\frac{\sin \mathrm{θ}}{\cos \mathrm{θ}} \\ y=x\tan \mathrm{θ} \end{gathered}$$

Substitute the value of y in equation 1.

role="math" localid="1659088558350" $$\begin{gathered} z=x^2+2{(x\tan \mathrm{θ})}^2 \\ z=x^2(1+2{\tan }^2\mathrm{θ}) \end{gathered}$$

Take partial derivative of z with respect to x,

$$\begin{gathered} \frac{∂z}{∂x}=\frac{∂}{∂x}{x}^2(1+2{\tan }^2\mathrm{θ}) \\ \frac{∂z}{∂x}=2x(1+2{\tan }^2\mathrm{θ}) \end{gathered}$$

Take partial derivative of $\frac{∂z}{∂x}$with respect to ,

$$\begin{gathered} \frac{{∂}^{2}z}{∂x∂\mathrm{θ}}=\frac{{∂}^2}{∂x∂\mathrm{θ}}2x(1+2{\tan }^2\mathrm{θ}) \\ \frac{{∂}^{2}z}{∂x∂\mathrm{θ}}=8x\tan \mathrm{θ}se{c}^2\mathrm{θ} \end{gathered}$$

Hence,

$\frac{{∂}^{2}z}{∂x∂\mathrm{θ}}=8x\tan \mathrm{θ}se{c}^2\mathrm{θ}$