第4个回答 2021-04-21
u=[f(x,xy)]/x;求∂²u/∂x∂y=?
解:∂u/∂x={x[f '₁(x,xy)+yf '₂(x,xy)]-f(x,xy)•1}/x²
=(1/x)[f '₁(x,xy)]+(y/x²)f '₂(x,xy)-(1/x²)f(x,xy);
∂²u/∂x∂y=(1/x)(xf''₁₂)+(1/x²)f '₂(x,xy)+(y/x²)xf''₂₂-(1/x²)xf'₂
=f''₁₂+(1/x²)f '₂(x,xy)+(y/x)f''₂₂-(1/x)f₂';