Question

偏导数高数题？

第二问和第三问！

Answer 1

追问

太厉害了谢谢！！

Answer 2

（2） z = (1/3)^(-y/x) = 3^(y/x)
∂z/∂x = 3^(y/x)(-y/x^2)ln3 = (-yln3/x^2)3^(y/x)
∂z/∂y = 3^(y/x)(1/x)ln3 = (ln3/x)3^(y/x)
（3） z = [(x+y)/(x-y)]sin(x/y)
∂z/∂x = [-2y/(x-y)^2]sin(x/y) + (1/y)[(x+y)/(x-y)]cos(x/y)
∂z/∂y = [2x/(x-y)^2]sin(x/y) - (x/y^2)[(x+y)/(x-y)]cos(x/y)

Answer 3

Answer 4

令 u = xz，v = z-y，则 z = f(u,v), 两边对 x 求偏导，得
∂z/∂x = (∂f/∂u)(∂u/∂x) + (∂f/∂v)(∂v/∂x)
= (∂f/∂u)(z+x∂z/∂x) + (∂f/∂v)(∂z/∂x)
(1-x∂f/∂u-∂f/∂v)∂z/∂x = z∂f/∂u
∂z/∂x = z(∂f/∂u)/(1-x∂f/∂u-∂f/∂v)；
z = f(u,v), 两边对 y 求偏导，得
∂z/∂y = (∂f/∂u)(∂u/∂y) + (∂f/∂v)(∂v/∂y)
= (∂f/∂u)(x∂z/∂y) + (∂f/∂v)(∂z/∂y-1)
(1-x∂f/∂u-∂f/∂v)∂z/∂y = -∂f/∂v
∂z/∂y = (-∂f/∂v)/(1-x∂f/∂u-∂f/∂v).