已知y=x* arctanx,求y-搜答案-神马作文网

y=arctanx,求y'

y'=1/(1+x²)

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

求y''+arctanx=0通解

∵y''+arctanx=0==>y''=-arctanx==>y'=-∫arctanxdx=(1/2)ln(1+x^2)-xarctanx+C1*(应用分部积分法,C1*是常数)∴y=∫[(1/2)

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)]

手写不易 …………

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+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

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

如图.

f'(x)=(1+x²)'*arctanx+(1+x²)*(arctanx)'=2xarctanx+(1+x²)*1/(1+x²)=2xarctanx+1所以f

y'=1/(x^2+1)=1-x^2+x^4-x^6+...+(-1)^nx^(2n)+...所以y'|(x=0)=1y^(2n)|(x=0)=(-1)^n*(2n)!y^(2n+1)|(x=0)=0

y'=1/(1+x^2)

tany=1y可以有无穷多个值但是前面几步（arctanx∈(-π/2,-π/4)∪(-π/4,π/2）,arctan(1-x)/(1+x)∈(-π/2,-π/4)∪(-π/4,π/2)）限制y大于-

令m=arctanx,n=arctan(1-x)/(1+x)那么x=tanm,(1-x)/(1+x)=tann,y=m+n那么tany=tan(m+n)=(tanm+tann)/(1-tanntanm

因为arcsinx+arccosx=π/2(公式)arcsinx+arctanx=π/2所以arccosx=arctanx令arccosx=arctanx=BcosB=xtanB=xcosBtanB=

上下分别求导,arctanx求导=1/(1+x²）,分母求导为1,所以f(x)=arctanx/x的极限就等于1/(1+x²）的极限,当x趋于无穷大时1/(1+x²）趋于