设函数y=arctan(1+x^2),求dy/dx.

函数y=arctan(1+x^2)求dy/dx 设函数y=arctan(1+x^2),求dy/dx. 设x=t+arctan t+1,y=t的立方+6t－2,求dy／dx 设参数方程式 {x=ln(1+t^3) y=t-arctan t ;确定y是X的函数,求 dy/dx? 设ln(x^2+y^2)=arctan(y/x),则dy/dx= 已知函数arctan(y/x)=ln√((x∧2)+(y∧2)),求dy/dx 设y=1/a arctan x/a ,则 dy/dx 设y=arctan根号(x^2-1)-lnx/根号(x^2-1)求dy 设方程x-y+arctan y=0确定了y=y(x),求dy/dx. 设y=sin2x/(1+x^2),求dy/dx, 设y=√(x^2-1)求dy/dx x^2+y^2=e^(arctan（y/x）),求dy/dx