91Ó°ÊÓ

Begin by using the differentials of the equation $w=f\left(x,y\right)$then substitute from one equation into the other to determine $\frac{∂w}{∂r}$as follows:

localid="1664331253335" $$\begin{gathered} w=f\left(x,y\right) \\ dw=\frac{∂f\left(x,y\right)}{∂x}dx+\frac{∂f\left(x,y\right)}{∂y}dy \\ \frac{∂w}{∂r}=\frac{∂w}{∂x}\frac{∂x}{∂r}+\frac{∂w}{∂y}\frac{∂y}{∂r}...\left(1\right) \end{gathered}$$

Differentiate the functions x and y partially as follows:

$$\begin{gathered} \frac{∂x}{∂r}=\frac{∂}{∂r}\left(r\cos \mathrm{θ}\right),\frac{∂y}{∂r}=\frac{∂}{∂r}\left(r\sin \mathrm{θ}\right) \\ \frac{∂x}{∂r}=\cos \mathrm{θ},               \frac{∂y}{∂r}=\sin \mathrm{θ}...\left(2\right) \end{gathered}$$

Substitute equation (2) into equation (1) as follows:

$$\begin{gathered} \frac{∂w}{∂r}=\frac{∂f\left(x,y\right)}{∂x}\cos \mathrm{θ}+\frac{∂f\left(x,y\right)}{∂y}\sin \mathrm{θ} \\ \frac{∂w}{∂r}=f_x\cos \left(\mathrm{θ}\right)+f_y\sin \left(\mathrm{θ}\right) \end{gathered}$$

Thus, the required answer is $\frac{∂w}{∂r}=f_x\cos \left(\mathrm{θ}\right)+f_y\sin \left(\mathrm{θ}\right)$.

Using the exact same method as in the preceding part, derive a formula for $\frac{∂w}{∂\mathrm{θ}}$as follows:

localid="1664331279875" $$\begin{gathered} w=f\left(x,y\right) \\ \frac{∂w}{∂\mathrm{θ}}=\frac{∂w}{∂x}\frac{∂x}{∂\mathrm{θ}}+\frac{∂w}{∂y}\frac{∂y}{∂\mathrm{θ}}...\left(1\right) \end{gathered}$$

Differentiate the functions x and y partially as follows:

$$\begin{gathered} \frac{∂x}{∂\mathrm{θ}}=\frac{∂}{∂\mathrm{θ}}\left(r\cos \mathrm{θ}\right),\frac{∂y}{∂\mathrm{θ}}=\frac{∂}{∂\mathrm{θ}}\left(r\sin \mathrm{θ}\right) \\ \frac{∂x}{∂\mathrm{θ}}=-r\sin \mathrm{θ},           \frac{∂y}{∂\mathrm{θ}}=r\cos \mathrm{θ}...\left(2\right) \end{gathered}$$

Substitute equation (2) into equation (1) as follows:

$$\begin{gathered} \frac{∂w}{∂\mathrm{θ}}=-r\frac{∂f\left(x,y\right)}{∂x}\sin \mathrm{θ}+r\frac{∂f\left(x,y\right)}{∂y}\cos \mathrm{θ} \\ \frac{∂w}{∂\mathrm{θ}}=r\left(f_y\cos \mathrm{θ}-f_x\sin \mathrm{θ}\right) \end{gathered}$$

Thus, the required answer is $\frac{∂w}{∂\mathrm{θ}}=r\left(f_y\cos \mathrm{θ}-f_x\sin \mathrm{θ}\right)$.

The same method can be used to obtain an expression for the second derivative with respect to r we have the first derivative of the expression with respect to r, denoted as$G(x,y)$.

$$\begin{gathered} \frac{∂w}{∂r}=f_x\cos \mathrm{θ}+f_y\sin \mathrm{θ}=G\left(x,y\right) \\ dG\left(x,y\right)=\frac{∂G\left(x,y\right)}{∂x}dx+\frac{∂G\left(x,y\right)}{∂y}dy \\ \frac{∂G}{∂r}=\frac{∂G}{∂x}\frac{∂z}{∂r}+\frac{∂G}{∂y}\frac{∂y}{∂r}...\left(1\right) \end{gathered}$$

Differentiate the functions G with respect to x and y partially as follows:

$$\begin{gathered} \frac{∂G}{∂x}=\frac{∂}{∂x}\left(f_x\cos \mathrm{θ}+f_y\sin \mathrm{θ}\right)=f_{xx}\cos \mathrm{θ}+f_{yx}\sin \mathrm{θ} \\ \frac{∂G}{∂y}=\frac{∂}{∂y}\left(f_x\cos \mathrm{θ}+f_y\sin \mathrm{θ}\right)=f_{xy}\cos \mathrm{θ}+f_{yy}\sin \mathrm{θ} \end{gathered}$$

Differentiate the function x and y as follows:

$$\begin{gathered} \frac{∂x}{∂r}=\frac{∂}{∂r}\left(r\cos \mathrm{θ}\right),\frac{∂y}{∂r}=\frac{∂}{∂r}\left(r\sin \mathrm{θ}\right) \\ \frac{∂x}{∂r}=\cos \mathrm{θ},               \frac{∂y}{∂r}=\sin \mathrm{θ}...\left(2\right) \end{gathered}$$

Substitute equation (2) into equation (1) as follows:

$$\begin{gathered} \frac{∂G}{∂r}=\left(f_{xx}\cos \mathrm{θ}+f_{yx}\sin \mathrm{θ}\right)\cos \mathrm{θ}+\left(f_{xy}\cos \mathrm{θ}+f_{yy}\sin \mathrm{θ}\right)\sin \mathrm{θ} \\ \frac{∂G}{∂r}=\frac{{∂}^{2}w}{∂r^2}=f_{xx}{\cos }^2\mathrm{θ}+f_{yx}\sin \mathrm{θ}\cos \mathrm{θ}+f_{xy}\cos \mathrm{θ}\sin \mathrm{θ}+f_{yy}{\sin }^2\mathrm{θ} \\ \frac{{∂}^{2}w}{∂r^2}=f_{xx}{\cos }^2\mathrm{θ}+2f_{xy}\cos \mathrm{θ}\sin \mathrm{θ}+f_{yy}{\sin }^2\mathrm{θ} \end{gathered}$$

Where we utilized the relation $f_{xy}=f_{yx}$to get the final expression.

Thus, the required answer is $\frac{{∂}^{2}w}{∂r^2}=f_{xx}{\cos }^2\mathrm{θ}+2f_{xy}\cos \mathrm{θ}\sin \mathrm{θ}+f_{yy}{\sin }^2\mathrm{θ}$.