设=dfrac (2x)({x)^2-(y)^2} ，则 =dfrac (2x)({x)^2-(y)^2}=dfrac (2x)({x)^2-(y)^2}____________.

2x(1+x)-(x)^2-|||-7.设函数f(x)可导且 (x)=dfrac (2x)(sqrt {{a)^2-(x)^2}} 求 dfrac (d)(dx

设=dfrac (y)(f({x)^2-(y)^2)},其中f为可导函数.验证=dfrac (y)(f({x)^2-(y)^2)}.设,其中f为可导函数.验证.

(B) dfrac (x)(y)((y+1))^2-|||-(C) ^2((x+dfrac {1)(x))}^2. (D) dfrac (y)(x)((y+1)

2.已知 dfrac (x)(x+y)=dfrac (1)(3) ,求 dfrac ({x)^2-(y)^2}(2xy+{y)^2} 的值.

7.函数 (x,y)=dfrac (arcsin (3-{x)^2-(y)^2)}(x-{y)^2} 的定义域为7.函数 (x,y)=dfrac (arcsin

设((sin )^2x)=dfrac (x)(sin x)，求((sin )^2x)=dfrac (x)(sin x) .设，求.

(x,y)=(x)^2-(y)^2-x-|||-__ B . (x,y)=(x)^2-(y)^2-x-|||-__C . (x,y)=(x)^2-(y)^2-x

6.讨论下列函数的连续性:-|||-(1) (x,y)=dfrac ({x)^2-(y)^2}({x)^2+(y)^2}-|||-(2) (x,y)=dfrac

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

(sin x)=dfrac (1)({cos )^2x} in (0,dfrac (pi )(2)),则(sin x)=dfrac (1)({cos )^2x}