Question

设z=u/v,其中u=x∧2y,v=x+y∧2,求z对x的偏导数和z对y的偏导数

Answer 1

z = u/v，u = x^(2y) = e^(2ylnx), v = x+y^2
∂z/∂x = (v∂u/∂x - u∂v/∂x) / v^2
= [(x+y^2)e^(2ylnx)(2y/x) - e^(2ylnx)] / (x+y^2)^2
= [(2y/x)(x+y^2)-1]x^(2y) / (x+y^2)^2,
∂z/∂y = (v∂u/∂y - u∂v/∂y) / v^2
= [(x+y^2)e^(2ylnx)(2lnx) - 2ye^(2ylnx)] / (x+y^2)^2
= 2[(x+y^2)lnx-y]x^(2y) / (x+y^2)^2

若是 z = u/v，u = yx^2 , v = x+y^2
∂z/∂x = (v∂u/∂x - u∂v/∂x) / v^2
= [2xy(x+y^2) - yx^2] / (x+y^2)^2
= xy(x+2y^2) / (x+y^2)^2,
∂z/∂y = (v∂u/∂y - u∂v/∂y) / v^2
= [x^2(x+y^2) - 2x^2y^2] / (x+y^2)^2
= x^2(x-y^2) / (x+y^2)^2