设函数y=f(x)由方程e^2x+y-cos(xy)=e-1所确定，则曲线y=f(x)在点(0,1)处的法线方程为______．

在下列各题中,求由所给函数构成的复合函数,并求这函数分别对应于给定自-|||-变量值x1和x2的函数值:-|||-(1) =(u)^2, =sin x _(1)=dfrac (pi )(6) _(2)=dfrac (pi )(3);-|||-(2) =sin u =2x, _(1)=dfrac (pi )(8) _(2)=dfrac (pi )(4);-|||-(3) =sqrt (u), =1+(x)^2 _(1)=1 _(2)=2;-|||-(4) =(e)^u, =(x)^2, _(1)=0 _(2)=1;-|||-(5) =(u)^2 =(e)^x _(1)=1, _(2)=-1.

讨论函数f(x)= { ,xneq 0 0, x=0 .在x=0处的连续性与可导性

当x→0时 (x)=x-sin ax 与 (x)=(x)^2ln (1-bx) 是等价无穷小,-|||-则 ()-|||-(A) =1, =-dfrac (1)(6) (B) =1, =dfrac (1)(6)-|||-(C) =-1, =-dfrac (1)(6) (D) a=-1 =dfrac (1)(6)

(2) int xsqrt (x-1)dx;

3.已知 (x)=(x)^2-2x+3 , (x)=3(x)^2-x+1, 求下列-|||-(1) (x)+2g(x);-|||-(2) (x)-g(x);-|||-(3)f(x)·g(x );-|||-(4) dfrac (2f(x))(3g(x))

设 ,xlt 0, dfrac {x(x+1))({x)^2-1},xgeqslant 0, . 有两个第二类间断点

(8)(int )_(1)^2sqrt (2x-{x)^2}dx=__________________.

设=u(√x^2+ y^2)有二阶连续偏导数，且满足=u(√x^2+ y^2)，=u(√x^2+ y^2)。（I）求=u(√x^2+ y^2)；（II）求=u(√x^2+ y^2)的极值。

设幂级数 sum _(n=0)^infty (a)_(n)(x)^n 的收敛半径为 (0lt Rlt +infty ), 则 sum _(n=0)^infty (a)_(n)((dfrac {x)(2))}^n 的收敛半径为[ ].-|||-(A)2R (B) dfrac (R)(2) (C)R (D) dfrac (2)(R)

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