已知y＝x²arctanx-搜答案-智慧作业帮

y=arctanx,求y'

y'=1/(1+x²)

1.f(x)+f(-x)=2(arccosx+arccos-x)+arctanx+arctan-x-2pi=2pi+0-2pi=0,得证.2.arctanx+arctan1/y=arctan3tan(

1/(1+x^2)

y=f[(x-1)/(x+1)],f'(x)=arctanx^2,求dy/dx,dy两边对x求导：dy/dx=f'[(x-1)/(x+1)]*2/（x+1)^2=arctan[(x-1)/(x+1)]

lettana=xthenarctanx=acota=1/xarctan(1/x)=90°-aarctanx+arctan（1/x）=90°

已知y＝x

∵x2−4+4−x2x+2有意义，∴x2−4≥04−x2≥0x+2≠0，解得：x=2，∴y=0+1=1，∴2x+y=5．故答案为：5．

令u=x+arctanx,则u'=1+1/(1+x^2)则y=f^2(u)dy/dx=2f(u)f'(u)u'=2f(u)f'(u)[1+1/(x+x^2)]

∵(1+x^2)y'+y=arctanx==>[(1+x^2)y'+y]e^(arctanx)/(1+x^2)=arctanx*e^(arctanx)/(1+x^2)(等式两端同乘e^(arctanx

y=arctanx/x*x+1为有界函数因为|arctanx|

y=arctanx+arctan(1-x/1+x)tany=tan[arctanx+arctan(1-x/1+x)]=[x+(1-x)/(1+x)]/[1-x*(1-x)/(1+x)]=1∴y=kπ+

y'=2xarctanx+1y''=2arctanx+2x/(1+x^2)y''/x=1=π/2+1

(1+x^2)y'=arctanxy'=arctanx/(1+x^2)两边积分：y=∫arctanx/(1+x^2)dx=∫arctanxd(arctanx)=1/2(arctanx)^2+C

首先结果是1/(1+x^2)推导过程x=tany对x求导1=y'*sec^2y=>y'=1/sec^2y=1/(tan^2y+1)=1/(x^2+1）觉得好请采纳不懂可以追问再问：为什么sec^2y=

已知：y＝x

证明：y＝x2+6x+9x2−9÷x+3x2−3x−x+3=(x+3)2(x+3)(x−3)×x(x−3)x+3−x+3=x-x+3=3．故不论x为任何有意义的值，y值均不变．

y=(arctanx)/(1+x)y'=[(arctanx)'(1+x)-(1+x)'arctanx]/(1+x)^2=[(1+x)/(1+x^2)-arctanx]/(1+x)^2

是tany=x,那么arctanx=y,

❶证明：tan(arctanX+arctanY)=(X+Y)/(1-XY)证明：tan(arctanx+arctany)=(tanarctanx+tanarctany)/[1-(tana

结果得3/4计算过程如下：（1）：令2x-t=ut:0->x则u:2x->x且dt=-du∫(上限x下限0)tf(2x-t)dt=∫(上限x下限2x)(u-2x)f(u)dtu=∫(上限x下限0)(u

f'(x)=2(arctanx)*1/(1+x^2)

不用推导,直接就是公式啊,=1/(1+x^2)