7.设 =arctan dfrac (1)(x) ,则 dfrac ({d)^2y}(d{x)^2}(x)_(x=1)= __

3、设 (x,y)=arctan dfrac (x)(y), 则 (1,1)=-|||-(A)1; (B)0; (C) dfrac {1)(2),dfrac

设(X,Y)的分布函数为(x,y)=dfrac (1)({pi )^2}(dfrac (pi )(2)+arctan dfrac (x)(2))(dfrac (

7.设 (x,y)=dfrac (1)(xy),r=sqrt ({x)^2+(y)^2} _(1)= (x,y)|(x,y)in {R)^2 dfrac (1)

dfrac (1)(5)(X)_(1)+dfrac (2)(5)(X)_(2)+dfrac (2)(5)(X)_(3)-|||-D .dfrac (1)(7)(

设函数 y=y(x) 由方程 -x(e)^y=1 所确定,求 dfrac ({d)^2y}(d{x)^2}(|)_(x=0) 的值.

7.已知函数 (x+dfrac (1)(x))=(x)^2+dfrac (1)({x)^2} ,则 f(x)=-|||-__

7.不定积分 int dfrac (arctan x)({x)^2(1+(x)^2)}dx= __

7.设 =dfrac (y)(f({x)^2-(y)^2)} 其中f为可导函数,验证: dfrac (1)(x)dfrac (partial z)(partia

(5)设 (x,y)=ln (x+dfrac (y)(2x)) ,则 _(y)(1,0)= () .-|||-(A)1 (B) dfrac (1)(2) (C)

2.设D |x|+|y|leqslant 1, 则 iint (|x|+y)dxdy= () .-|||-(A)0 (B) dfrac (1)(3) (C) d