Question

z=ln(x^2+y^2), x=uv, y=3u−v. Найти ∂z/∂u .

Answer 1

$z=ln(x^2+y^2)\; ,\; \; x=uv\; ,\; \; y=3u-v\\\\\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\cdot \frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\cdot \frac{\partial y}{\partial u}=\frac{2x}{x^2+y^2}\cdot v+\frac{2y}{x^2+y^2}\cdot 3=\\\\=\frac{2xv}{x^2+y^2}+\frac{6y}{x^2+y^2}$