Question Number 142408 by mohammad17 last updated on 31/May/21
Commented by mohammad17 last updated on 31/May/21
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Answered by qaz last updated on 31/May/21
$$\frac{\partial\mathrm{w}}{\partial\mathrm{r}}=\frac{\partial}{\partial\mathrm{r}}\left(\mathrm{x}+\mathrm{2y}+\mathrm{z}^{\mathrm{2}} \right)=\frac{\partial\mathrm{x}}{\partial\mathrm{r}}+\mathrm{2}\frac{\partial\mathrm{y}}{\partial\mathrm{r}}+\frac{\partial\left(\mathrm{z}^{\mathrm{2}} \right)}{\partial\mathrm{r}}=\frac{\partial\mathrm{x}}{\partial\mathrm{r}}+\mathrm{2}\frac{\partial\mathrm{y}}{\partial\mathrm{r}}+\mathrm{2z}\frac{\partial\mathrm{z}}{\partial\mathrm{r}}=\frac{\mathrm{1}}{\mathrm{s}}+\mathrm{4r}+\mathrm{8r} \\ $$$$\mathrm{same}\:\:.....\frac{\partial\mathrm{w}}{\partial\mathrm{s}}=−\frac{\mathrm{r}}{\mathrm{s}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{s}} \\ $$