(1) z=xsin(y/x), z'
(2) z=√ln(xy), z'
(3) z=(1+xy)^y, z'
lnz=yln(1+xy), z'
z'(y>=(1+xy)^y*[ln(1+xy)+xy/(1+xy)].
(1) z=arctan(y/x), z'
(2) z=√(3x^2+y^2), z'
f(x,y)=x^2+(y-1)arcsin√(x/y),
f'
或
f(x,1)=x^2, 得 f'u'
u''
=[y^2/(x^2+y^2)^(3/2)]f'(r)+[x^2/(x^2+y^2)]*f''(r);
由轮换性,得
u''
则 u''
=f''(r)+f'(r)/r,由题设得
f''(r)+f'(r)/r=4 为f'(r)的一阶线性微分方程, 则
f'(r)=e^(-∫dr/r)[∫4e^(∫dr/r)dr+C1] = (1/r)[∫4rdr+C1]
= (1/r)(2r^2+C1) = 2r+C1/r,
得 f(r)=r^2+C1lnr+C2