Question

∂z/∂x - ∂z/∂y da função z= sen(x+y)

Answer 1

$z = sen(x + y)\\\frac{dz}{dx} = sen(u) * \frac{du}{dx}\\ \frac{dz}{dx} = cos(u) * (x+y)'\\ \frac{dz}{dx} = cos(x+y) * 1\\ \frac{dz}{dx} = cos(x+y) \\\\\frac{dz}{dy} = sen(u) * \frac{du}{dy}\\ \frac{dz}{dx} = cos(u) * (x+y)'\\ \frac{dz}{dx} = cos(x+y) * 1\\ \frac{dz}{dx} = cos(x+y) \\\\\frac{dz}{dx} - \frac{dz}{dy} = cos(x+y) - cos(x+y) = 0$