91Ó°ÊÓ

If any function $f\left(x,y\right)$is having two variables, then its derivative with respect to x nd y will be considered as partial derivatives of $f\left(x,y\right)$.

The given equations are:

$$\begin{gathered} \frac{∂F}{∂r}=\cos \mathrm{θ}\frac{∂F}{∂x}+\sin \mathrm{θ}\frac{∂F}{∂y}...\left(1\right) \\ \frac{∂F}{∂\mathrm{θ}}=-r\sin \mathrm{θ}\frac{∂F}{∂x}+r\cos \mathrm{θ}\frac{∂F}{∂y}...\left(2\right) \end{gathered}$$

Now, multiply eq. (1) by $r\sin \mathrm{θ}$and eq. (2) by $\cos \mathrm{θ}$. Then, we have:

$$\begin{gathered} r\sin \mathrm{θ}\frac{∂F}{∂r}=r\sin \mathrm{θ}\cos \mathrm{θ}\frac{∂F}{∂x}+r{\sin }^2\mathrm{θ}\frac{∂F}{∂y}...\left(3\right) \\ \cos \mathrm{θ}\frac{∂F}{∂\mathrm{θ}}=-r\sin \mathrm{θ}\cos \mathrm{θ}\frac{∂F}{∂x}+r{\cos }^2\mathrm{θ}\frac{∂F}{∂y}...\left(4\right) \end{gathered}$$

Adding eq. (3) and (4), we get:

$$\begin{gathered} r\frac{∂F}{∂y}=r\sin \mathrm{θ}\frac{∂F}{∂r}+\cos \mathrm{θ}\frac{∂F}{∂\mathrm{θ}} \\ \frac{∂F}{∂y}=\sin \mathrm{θ}\frac{∂F}{∂r}+\frac{1}{r}\cos \mathrm{θ}\frac{∂F}{∂\mathrm{θ}} \end{gathered}$$

Again, multiply eq. (1) by $-r\cos \mathrm{θ}$ and eq. (2) by $\sin \mathrm{θ}$. Then, we have:

$$\begin{gathered} -r\cos \mathrm{θ}\frac{∂F}{∂r}=r{\cos }^2\mathrm{θ}\frac{∂F}{∂x}-r\sin \mathrm{θ}\cos \mathrm{θ}\frac{∂F}{∂y}...\left(5\right) \\ \sin \mathrm{θ}\frac{∂F}{∂\mathrm{θ}}=-r{\sin }^2\mathrm{θ}\frac{∂F}{∂x}+r\sin \mathrm{θ}\cos \mathrm{θ}\frac{∂F}{∂y}...\left(6\right) \end{gathered}$$

Adding eq. (5) and (6), we get:

$\frac{∂F}{∂x}=\cos \mathrm{θ}\frac{∂F}{∂r}-\frac{1}{r}\sin \mathrm{θ}\frac{∂F}{∂\mathrm{θ}}$

Hence, the required derivatives are:

$\frac{∂F}{∂x}=\cos \mathrm{θ}\frac{∂F}{∂r}-\frac{1}{r}\sin \mathrm{θ}\frac{∂F}{∂\mathrm{θ}}$and$\frac{∂F}{∂y}=\sin \mathrm{θ}\frac{∂F}{∂r}+\frac{1}{r}\cos \mathrm{θ}\frac{∂F}{∂\mathrm{θ}}$