Question

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

Answer 1

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