设函数z=y/[f(x2-y2)],其中f为可导函数,证明:1/x•z/x+1/y•z/y=z/y2.
证明:∂z/∂x=[-yf′(x2-y2)•2x]/[f2(x2-y2)] ∂z/∂y=[f(x2-y2)+yf′(x2-y2)•2y]/[f2(x2-y2)] ∴1/x•(∂z/∂x)+1/y•(∂z/∂y)=z/y2
设函数z=y/[f(x2-y2)],其中f为可导函数,证明:1/x•z/x+1/y•z/y=z/y2.
证明:∂z/∂x=[-yf′(x2-y2)•2x]/[f2(x2-y2)] ∂z/∂y=[f(x2-y2)+yf′(x2-y2)•2y]/[f2(x2-y2)] ∴1/x•(∂z/∂x)+1/y•(∂z/∂y)=z/y2