已知函数f(u,v)具有二阶连续偏导数,f(1,1)=2是f(u,v)的极值,z=f(x+y,f(x,y))求z在(

Question

1,1)的混合二阶偏导

Answer 1

令u=x-y,v=y/x az/ax＝az/au×au/ax＋az/av×av/ax＝fu-y/x^2×fv a^2z/axay＝a(az/ax)/ay＝a(fu-y/x^2×fv)/ay＝a(fu)/ay-a(y/x^2×fv)/ay＝a(fu)/ay-1/x^2×fv-y/x^2×a(fv)/ay a(fu)/ay＝a(fu)/au×au/ay＋a(fu)/av×av/ay＝-fuu+1/x×fuv a(fv)/ay＝a(fv)/au×au/ay＋a(fv)/av×av/ay＝-fvu+1/x×fvv 回代，a^2z/axay＝a(fu)/ay-1/x^2×fv-y/x^2×a(fv)/ay ＝-fuu+1/x×fuv-1/x^2×fv-y/x^2×(-fvu+1/x×fvv) ＝-fuu+(x+y)/x^2×fuv-y/x^3×fvv-1/x^2×fv