91Ó°ÊÓ

Given:

$$\begin{gathered} z=(x^2+xy)e^y, \\ x=r\cos \mathrm{θ}and \\ y=r\sin \mathrm{θ} \end{gathered}$$

We have to find the indicated derivatives and express your answers as functions of a single variable.

By Theorem 12.33, we have

$\frac{∂z}{∂r}=\frac{∂z}{∂x}·\frac{∂x}{∂r}+\frac{∂z}{∂y}·\frac{∂y}{∂r}---\left(1\right)$

So first we find $\frac{∂z}{∂x},\frac{∂x}{∂r},\frac{∂z}{∂y}·\frac{∂y}{∂r}$

So we have

$$\begin{gathered} \frac{∂z}{∂x}=e^y(2x+y) \\ \frac{∂z}{∂y}=e^y·x^2+x{e}^y+xy{e}^y \\ \frac{∂x}{∂r}=\cos \mathrm{θ} \\ \frac{∂y}{∂r}=\sin \mathrm{θ} \end{gathered}$$

Use these values in (1) we get

$$\begin{gathered} \frac{∂z}{∂r}=e^y(2x+y)·\cos \mathrm{θ}+x{e}^y(x+1+y)·\sin \mathrm{θ} \\ \frac{∂z}{∂r}=e^y\left[(2x+y)·\cos \mathrm{θ}+x\sin \mathrm{θ}(x+y+1)\right] \end{gathered}$$

This result is correct, but it is preferable to write the function as a function of just $rand\mathrm{θ}$.

We use $x=r\cos \mathrm{θ}andy=r\sin \mathrm{θ}$to do so:

role="math" localid="1649765534715" $$\begin{gathered} \frac{∂z}{∂r}=e^{r\sin \mathrm{θ}}\left[\left(2\right(r\cos \mathrm{θ})+r\sin \mathrm{θ})·\cos \mathrm{θ}+r\cos \mathrm{θ}\sin \mathrm{θ}(r\cos \mathrm{θ}+r\sin \mathrm{θ}+1)\right] \\ \frac{∂z}{∂r}=r\cos \mathrm{θ}{e}^{r\sin \mathrm{θ}}\left[(2+\sin \mathrm{θ}+\sin \mathrm{θ}(r\cos \mathrm{θ}+r\sin \mathrm{θ}+1)\right] \end{gathered}$$