设y=(1+x^2)arctanx-ln(x+根号(1+x^2))求微积分..

设y=(1+x^2)arctanx-ln(x+根号(1+x^2))求微积分.. 设y=[(x^2+1)/2]*(arctanx)^2-xarctanx+(1/2)ln(1+x^2),求y'以及dy 设y=(ln(arctanx))^2,求导数 求函数y=arctanx-1/2乘以ln(1+x^2)的极值. 设x>0,证明ln(1+x)>arctanx/1+x 3.设y=(1+x^2)arctanx,求y ,y/x=1 . 求导 y=0.5(0.5*ln((x+1)/(x-1))+arctanx) 设y=ln(1-2x),则y''(0). 设y=ln(1+2x),则y(0) 设y=ln(1-2x),则y''(0)... y=(1+x^2)arctanx微积分 设z=f(x,y)=arctanx/y ,y=√(x^2+1) ,求dz/dx 设y=ln(1+x^2)则dy= 已知f（x）=x arctanx – 1/2 ln(1+x^2),求dy lim(x→0)=（arctanx-x）/ln(1+2x^3) arctanx-ln((1+x^2)^1/2)的幂级数展开 微分方程y'-y/[(1+x^2)arctanx]=arctanx的通解? 求导y=ln ln ln(x^2+1)