Question

If $\mathit{z}\mathit{=}{\mathit{x}}^{\mathit{2}}\mathit{+}\mathit{2}{\mathit{y}}^{\mathit{2}}$, $\mathit{x}\mathbf{=}\mathit{r}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$, $\mathit{y}\mathbf{=}\mathit{r}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$, find the following partial derivatives.

${\left(\frac{\partial z}{\partial y}\right)}_{\mathbf{x}}$.

Short answer

The value of provided equation is${\left(\frac{\partial z}{\partial y}\right)}_x=4y$ .

Step-by-step solution

Given data

The given data are listed below

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

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.

 Step 3: Calculation

Consider that,

$z=x^2+2y^2$

Take the partial derivative of with respect to y ,

$$\begin{gathered} {\left(\frac{\partial z}{\partial y}\right)}_x=\frac{\partial }{\partial y}\left(x^2+2y^2\right) \\ =\frac{\partial }{\partial y}\left(x^2\right)+\frac{\partial }{\partial y}\left(2y^2\right) \\ {\left(\frac{\partial z}{\partial y}\right)}_x=0+4y \end{gathered}$$

Therefore,

${\left(\frac{\partial z}{\partial y}\right)}_x=0+4y$

Therefore,

${\left(\frac{\partial z}{\partial y}\right)}_x=4y$