y=(1 x^2)arc tan x得二阶导数
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由题意tany=x所以可得siny/cosy=x……1cosy/siny=1/x……2两式相加得到1/sinycosy=x+1/x整理得到sinycosy=x/1+(x的2次方
求函数y=(x-1)*e^(π/2+arctanx)的斜渐近线x→+∞lim[(x-1)*e^(π/2+arctanx)]/x=x→+∞lime^(π/2+arctanx)-[x→+∞lim[e^(π
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)]
手写不易 …………
在定义反正切函数时,规定值域为(-pi/2,pi/2)因为一个函数有反函数的充分必要条件是这个函数是一一映射.
y=(2x^3+x^(1/2)+4ArcTan[x])/xy'=(1/(2Sqrt[x])+6x^2+4/(1+x^2))/x-(Sqrt[x]+2x^3+4ArcTan[x])/x^2y''=(-(
∵(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'=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=
1.dy={arctanx+x/(1+x^2)-1/2*[2x/(1+x^2)]}dx2.y'=(6x)sec^2(3x^2+1)3.f'(x)=2cos(a^x+1/x)*[-sin(a^x+1/x
y=(arctanx)/(1+x)y'=[(arctanx)'(1+x)-(1+x)'arctanx]/(1+x)^2=[(1+x)/(1+x^2)-arctanx]/(1+x)^2
y=(1+x²)arctanxy'=(1+x²)*1/(1+x²)*(0+2x)arctanxy'=1+2xarctanxy''=0+2[arctanx*x*1/(1+x
=∫x/(1+x^2)dx-∫arctanx/(1+x^2)dx=0.5∫1/(1+x^2)d(1+x^2)-∫arctanxdarctanx=0.5ln(1+x^2)-0.5(arctanx)^2.
1.y=90+arctanx/(x-2)a)因为arctanx的定义域是R,所以要使函数有意义,只需x/(x-2)有意义,即定义域为x≠2b)令t=x/(x-2),反解得x=(2t)/(t-1)所以t
❶证明:tan(arctanX+arctanY)=(X+Y)/(1-XY)证明:tan(arctanx+arctany)=(tanarctanx+tanarctany)/[1-(tana
原式=∫xdx/(1+x^2)-∫arctanxdx/(1+x^2)=1/2*∫d(1+x^2)/(1+x^2)-∫arctanxdarctanx=1/2*ln(1+x^2)-1/2*(arctanx