求下列函数的全微分z=(In(x^2+y^2))^xy

两边即对数得：lnz=xy*ln(lnu),不妨记u=x^2+y^2
z'x/z=yln(lnu)+2x^2y/lnu,z'x=z[yln(lnu)+2x^2y/lnu]
z'y/z=xln(lnu)+2y^2x/lnu,z'y=z[xln(lnu)+2y^2x/lnu]
dz=z'x dx+z'y dy=z[yln(lnu)+2x^2y/lnu] dx+z[xln(lnu)+2y^2x/lnu]dy
=(lnu)^(xy){[yln(lnu)+2x^2y/lnu] dx+[xln(lnu)+2y^2x/lnu]dy}lnz=xy*ln(lnu)这是怎样得出的??就是两边同时取对数。但xy是次方~z=(In(x^2+y^2))^xy如何得出lnz=xy*ln(lnu)??你这样来看：z=A^B则lnz=BlnA这里A=ln(u), B=xy