由于z=f(u,x,y),u=xey,因此 ?z ?x = ?f ?u ?u ?x + ?f ?x =eyf′1+f′2∴ ?2z ?x?y = ? ?y (eyf′1+f′2)=eyf′1+ey(xeyf″11+f″13)+(xeyf″21+f″23)=eyf′1+xe2yf″11+eyf″13+xeyf″12+f″23