Question

Помогите решить задачу по алгебре, очень надо.

Answer 1

$z=arctg(4^{\frac{x^2}{y}})\\\\z'_{x}=\frac{\partial z}{\partial x}=\frac{1}{1+\left (4^{\frac{x^2}{y}}\right )^2}\cdot \left (4^{\frac{x^2}{y}}\right )'_{x}=\frac{1}{1+4^{\frac{2x^2}{y}}}\cdot 4^{\frac{x^2}{y}}\cdot ln4\cdot \left(\frac{x^2}{y}\right )'_{x}=\\\\= \frac{1}{1+4^{\frac{2x^2}{y}}}\cdot 4^{\frac{x^2}{y}}\cdot ln4\cdot \frac{1}{y}\cdot 2x=\frac{2x\cdot ln4}{y\cdot \left (1+4^{\frac{2x^2}{y}}\right )}\cdot 4^{\frac{x^2}{y}}\; ;$

$z'_{y}=\frac{\partial z}{\partial y}=\frac{1}{1+4^{\frac{2x^2}{y}}}\cdot 4^{\frac{x^2}{y}}\cdot ln4\cdot \left (\frac{x^2}{y}\right )'_{y}=\frac{ln4}{1+4^{\frac{2x^2}{y}}}\cdot 4^{\frac{x^2}{y}}\cdot x^2\cdot \frac{-1}{y^2}=\\\\=-\frac{x^2\cdot ln4}{y^2\cdot \left (1+4^{\frac{2x^2}{y}}\right )}\cdot 4^{\frac{x^2}{y}}\; ;\\\\\\dz=\frac{2x\cdot ln4}{y\cdot \left (1+4^{\frac{2x^2}{y}}\right )}\cdot 4^{\frac{x^2}{y}}\cdot dx-\frac{x^2\cdot ln4}{y^2\cdot \left (1+4^{\frac{2x^2}{y}}\right )}\cdot 4^{\frac{x^2}{y}}\cdot dy$