第二章导数与微分测验题

一、设曲线yf(x与ysinx在原点相切，求limnfn 二、已知曲线的极坐标方程r1cos，求曲线上对应于

63x三、设yf 3x

f(x)arctanx2dydxx1、设ysin4xcos4x，求y(n)（n为正整数x22、设y为xx2

y确定的，求xxf(t) d3、设函数f(x)二阶可导，f(0)0，且 ，

|t

dx2|t04、设y ，求5yf(x)xg(y)互为反函数，yf(x)可导，且f(x0，f(3)5h(x)

f[3

23x1)]，求五、已知当x0时，f(x有定义且二阶可导，问a,b,cF(x)

f x

xf(x)ln[1sin(xaxa)arctan2ex在xa七、设函数f(x)[0,)f(1)a(a0)，又对x,y0,)f(xyf(xf(y)f(xe2x xf(x)

x

（1）a,F(x在x0F(f(x在x0t

f(xt)f(xt表示这个极限值，试证：存在ca,b使g(c)解 先两边取对数(假定x5)得lny1[ln(x5)1ln(x2 5

y' 2x 5x x2 x555x2y' [ x555x2x 5(x25yf(x)与xg(y)互为反函数，yf(x)可导，且f(x0，f(3)5h(x)f[3

23x1)]，求解h(x)

g2(x23x1)]2g(x23x1)g(x23x1)(2x f[gh(1)1 2g(5)g(55，又yf(f[g

故3g(5)f(3)g(51h(1)f(3)23g(5)53x0时，f(x有定义且二阶可导，问a,b,cF(x)

f x x

解F(x)在x0处连 则F(0)lim(ax2bxc)cF(0)f(0)

cf x

F(0)limF(x)F(0)limf(x)f(0)f x

x xF(x)

x ax2bxcF(0)x

x

x xF(x在x0处一阶可导，则有bff x 此时，F(x)f x2ax xF(0)limF(x)F(0)limf(x)f(0)f x x x x F(0)limF(x)F(0)lim2axbb x x x xF(x在x0处二阶可导，则有a1f2f(x)ln[1sin(xaxa)arctan2ex在f(a)0

af(a)limf(x)f(a)limln[1sin(xa)](xa)arctan2x x x xlimln[1sin(xa)]arctan2 limsin(xa)arctan2ea1arctan2x x x x七、设函数f(x)[0,)f(1)a(a0)，又对x,y0,)f(xyf(xf(y)f(xf(xf(xf(1)f(1)0f(1)limf(1xf(1)limf(1xx x f(x)limf(xxf(x)limf(x(1

x))fxx

x f(x)f(1 x)f lim f(1)axx 又d(f(x)) dx，即d(f(x)alnx)0f(x)alnxcf(10c0f(xalne2x xf(x)

x

（1）a,F(x在x0F(f(x在x0 b10b1，所以f(0)0，又f(xx0处可导，所f(0)f f(0)

f(x)f(0)lime2x1 x

x f(0)limf(xf(0)limsinax0 所以a2f(0 x x x x（2） F(f(x))limF[f(x)]F[f limF[f(x)]F[f(0)]f(x)fx x x f(x)f xlimF[f(x)]F[f(0)]limf(x)ff(x f(x)f x xlimF(y)F(0)limf(x)f(0)F(0)f(0)2Fy y x xf(xt)f(x t 表示这个极限值，试证：存在ca,b使g(c)证明用反证法若xa,b，有g(x而g(x)limf(xtf(xt)limf(xtf(x)limf(xf(xt)2ft t t )t当t0时，f(xt)f