(5)z=lntan(x)/(y); (6)z=(1+xy)^y; (7)u=x^(y)/(2); (8)u=arctan(x-y)^2.

(5)z=lntan(x)/(y); (6)z=(1+xy)^y; (7)u=x^(y)/(z); (8)u=arctan(x-y)^z.(5)$z=ln\ta

证明u(x,y)=(x-y)(x^2+4 xy+y^2)u(x,y)=(x-y)(x^2+4 xy+y^2)u(x,y)=(x-y)(x^2+4 xy+y^2)

2.设z=u^2ln v，而u=(x)/(y)，v=3x-2y，求Z_(x)，Z_(y).2.设$z=u^{2}\ln v$，而$u=\frac{x}{y}$，

2.9 由下列条件求解析函数 (z)=u+iv.-|||-(1) =(x-y)((x)^2+4xy+(y)^2);-|||-(2) =2xy+3x;-|||-(

九、设 f(z)=u+iv 解析,且 -v=(x-y)((x)^2+4xy+(y)^2), 求f(z)·

求下列函数的偏导数:-|||-(5) =ln tan dfrac (x)(y);-|||-(6) =((1+xy))^y;-|||-(7) =(x)^dfrac

分解因式：x^2(y-z)+y^2(z-x)+z^2(x-y).分解因式：$x^{2}(y-z)+y^{2}(z-x)+z^{2}(x-y)$.

设z=sin(uv)，u=x+y，v=x-y，则(partial z)/(partial y)=【】 设$z=\sin(uv)$，$u=x+y$，$v=x-y

【单选题】若 f (z)= u (x,y)+i v (x,y)在Z平面上解析, u (x,y)=x 2 -y 2 +x,则 v (x,y)=A. xy+xB.

30.由下列各已知调和函数求解析函数f(z)=u+iv.1)u=(x-y)(x²+4xy+y²);2v=(y)/(x^2)+y^(2),f(2)=0;30.由下