设z=u^2v,而u=xcosy,v=xsiny 求偏z⼀偏x和偏z⼀偏y

2024-12-26 04:22:01
推荐回答(1个)
回答1:

∂u/∂x=cosy,∂u/∂y=-xsiny;
∂v/∂x=siny,∂v/∂y=xcosy;
∂z/∂u=2v.u^(2v-1);
∂z/∂v=2(lnu)u^2v
∂z/∂x=∂z/∂u.∂u/∂x+∂z/∂v.∂v/∂x
=2v.u^(2v-1).cosy+2(lnu)u^2v.siny
=2[(v/u)u^2v.cosy+(lnu)u^2v.siny]
=2z[(v/u)cosy+(lnu)siny]
=2z[siny+ln(xcosy)siny]
=2zsiny[1+ln(xcosy)]
=2zsinyln(excosy)
∂z/∂y=∂z/∂u.∂u/∂y+∂z/∂v.∂v/∂y
=2v.u^(2v-1).(-xsiny)+2(lnu)u^2v.xcosy
=2zx[-v/usiny+(lnu)cosy]
=2zx[-tanysiny+ln(xcosy)cosy]