Question

Помогите решить, буду очень благодарна. Тема частные производные.

Answer 1

$z=2^{x\cdot cosy}+tg\sqrt{y}\; \; ;\; \; \; x=tg\frac{u}{v}\; \; ;\; \; y=\frac{1}{u}\; ;\\\\\\\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\cdot \frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\cdot \frac{\partial y}{\partial u}=2^{x\cdot cosy}\cdot ln2\cdot cosy\cdot \frac{1}{cos^2\frac{u}{v}}\cdot \frac{1}{v}+\\\\+\Big (2^{x\cdot cosy}\cdot ln2\cdot (-x\cdot siny)+\frac{1}{cos^2\sqrt{y}}\cdot \frac{1}{2\sqrt{y}}\Big )\cdot (-\frac{1}{u^2})$

$\frac{\partial z}{\partial v}=\frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}\cdot \frac{\partial y}{\partial v}=2^{x\cdot cosy}\cdot ln2\cdot cosy\cdot \frac{1}{cos^2\frac{u}{v}}\cdot (-\frac{u}{v^2})+0=\\\\=-2^{x\cdot cosy}\cdot ln2\cdot \frac{u}{v^2\cdot cos^2\frac{u}{v}}\; \; ,\; \; (\frac{\partial y}{\partial v}=0\; )$