xsin+x+2+的不定积分

胡质诸4321求教几个高等数学题1.求f(x,y)=xsin(x+y)+ycos(x+y)的二级偏导数2.求Z=xsin(x2+y2)的全微分3.求函数Z=2x+3y2,当x=10,y=8,△x=2,△y=0.3的全增量△z和全微... -
丘蚂性18790757004 ______[答案] 1 Z=xsin(x+y)+ycos(x+y)Zx=sin(x+y)+xcos(x+y)-ysin(x+y)Zy=xcos(x+y)+cos(x+y)-ysin(x+y)所以Zxx=cos(x+y)+cos(x+y)-xsin(x+y)-ycos(x+y)=2cos(x+y)-xsin(x+y)-ycos(x+y)Zxy=Zyx=cos(x+y)-xsin(x+y)-sin(x+y)-ycos...

胡质诸4321d dx ∫x0sin(t2+1)dt=( ) -
丘蚂性18790757004 ______[选项] A. 2xsin(x2+1)cos(x2+1) B. sin(x2+1) C. 2xsin(x2+1) D. sin(x2+1)cos(x2+1)

胡质诸4321已知0≤x≤π/2,函数y=4倍根号下2*正弦x*余弦x+2倍正弦(x+π/4)*正弦(π/4 - x)的极值?y=(4倍根号2)乘以(正弦x*余弦x)加.... -
丘蚂性18790757004 ______[答案] y=4X2^0.5Xsinxcosx+2Xsin(x+π/4)sin(π/4-x)=2X2^0.5sin(2x)+2Xsin(x+π/4)sin[π/2-(π/4+x)]=2X2^0.5sin(2x)+2Xsin(x+π/4)cos(π/4+x)=2X2^0.5sin(2x)+sin(2x+π/2)=2X2^0.5sin(2x)+cos(2x)=[(2X2^0.5)...

胡质诸4321x→0时,从下列无穷小中找出阶最低的无穷小( )A.sin(sinx)B.xsinxC.ln(1+x2) -
丘蚂性18790757004 ______[答案] ①选项A.x→0时,sin(sinx)~sinx~x,因此其阶为1; ②选项B.x→0时,xsinx~x2,因此其阶为2; ③选C.x→0时,ln(1+x2)~x2,因此其阶为2 故阶最低的无穷小为1 故选:A.

胡质诸4321证明以下函数为有界函数?1.y=sin3X/1+X2 2.y=sin2 1/x+cosX 3.sin1/Xsin2X 备注:3,2都为次方请写下证明的过程, -
丘蚂性18790757004 ______[答案] 均为初等函数,定义域为R,且1,2两函数当x趋于无穷时,极限为0故有界 3.为有界函数与有界函数的乘积,当然有界

胡质诸4321函数y=x2cos2x的导数为( )A.y′=2xcos2x - x2sin2xB.y′=2xcos2x - 2x2sin2xC.y′=x2cos2x - 2xsin2xD -
丘蚂性18790757004 ______ y′=(x2)′cos2x+x2(cos2x)′=2xcosx-2x2sin2x. 故选B.

胡质诸4321根号下1+x2除以x2的不定积分 -
丘蚂性18790757004 ______ 令 x=tant 原式= ∫cscx^2 * sect dt =- ∫ sect t cot = -sect* cot +∫ sect dt = -√(1+x^2) / x +ln Ix+√(1+x^2)I +c

胡质诸4321下列函数中( )是基本初等函数 A y=e - x+1 - --1- - B y=1+x2 C y= sin x D y= xsin x -
丘蚂性18790757004 ______ c: 基本初等函数包括以下几种:(1)常数函数y = c( c 为常数)(2)幂函数y = x^a( a 为常数)(3)指数函数y = a^x(a>0, a≠1)(4)对数函数y =log(a) x(a>0, a≠1,真数x>0)(5)三角函数:主要有以下 6 个:正弦函数y =sin x余弦函数y =cos x 正切函数y =tan ...

胡质诸4321函数y=x2sinx的导数为( ) -
丘蚂性18790757004 ______[选项] A. y′=x2cosx-2xsinx B. y′=2xsinx+x2cosx C. y′=2xsinx-x2cosx D. y′=xcosx-x2sinx

胡质诸4321∫  ln(x+1+x2)1+x2dx. -
丘蚂性18790757004 ______[答案] 因为(ln(x+1+x2))′=1x+1+x2•(1+x1+x2)=1x+1+x2x+1+x21+x2=11+x2, 所以, ∫ln(x+1+x2)1+x2dx=∫ln(x+x2+1)dln(x+x2+1)=12ln2(x+x2+1)+c.