Question

求下列函数的偏导数：设z=(3x²+y²)^4x+2y，求∂z/∂x和∂z/∂y

Answer 1

z=(3x^2+y^2)^(4x+2y)
lnz =(4x+2y)ln(3x^2+y^2)
∂/∂x(lnz)=∂/∂x[(4x+2y)ln(3x^2+y^2)]
(1/z).∂z/∂x =(4x+2y).∂/∂x[ln(3x^2+y^2)] +ln(3x^2+y^2).∂/∂x(4x+2y)
(1/z).∂z/∂x =(4x+2y).[6x/(3x^2+y^2)] +ln(3x^2+y^2).(4)
∂z/∂x =[6x(4x+2y)/(3x^2+y^2) +4ln(3x^2+y^2)].(3x^2+y^2)^(4x+2y)
同样的
∂z/∂y =[2y(4x+2y)/(3x^2+y^2) +2ln(3x^2+y^2)].(3x^2+y^2)^(4x+2y)

Answer 2

u = 3x²+y²
v = 4x+2y
z = u^v
∂z/∂(u, v) = (v*u^(v-1), u^v * ln(u))
∂(u, v)/∂(x, y) = (6x, 2y; 4, 2)
∂z/∂(x, y) = (∂z/∂(u, v) )·(∂(u, v)/∂(x, y))
= (v*u^(v-1)*6x + u^v * 4ln(u), v*u^(v-1)*2y + u^v * 2ln(u))
= u^(v-1) (6xv + 4u ln(u), 2yv + 2u ln(u))
= 2*(3x²+y²)^(4x+2y-1) (3x(4x+2y)+2(3x²+y²)ln(3x²+y²), y(4x+2y)+(3x²+y²)ln(3x²+y²))