设z=(y-x)/(x+y)ln(y/x)，求x(∂z/∂x)+y(∂z/∂y)．
设u=(y-x)/(x+y)，υ=y/x，z=ulnυ， ∂u/∂x=[-(x+y)-(y-x)]/(x+y)2=-[2y(x+y)2]，∂u/∂y=[(x+y)-(y-x)]/(x+y)3=2x/(x+y)． ∂υ/∂x=-y/x2 ∂υ/∂，y =1/x。 ∂z/∂x=∂z∂u/∂u∂x+∂u∂υ/∂υ∂x=lnυ[-2y/(x+y)2]+u/υ[-(y/x2)] =-[2y/(x+y)2+ ln(y/x)+[(x-y)/x(x+y)]， ∂z/∂y=∂z∂u/∂u∂y+∂z∂υ/∂υ∂y=lnυ[2x/(x+y)2]+(u/υ)•(1/x) =[2x/(x+y)2 ]ln(y/x)+[(y-x)/y(x+y)], X(∂z/∂x)+y(∂z/∂y)=0．
