y=arctanx^2 利用复合函数求导的链式法则记y=arctanuu=x^2y'=dy/dx=dy/du*du/dx=1/(1+u^2)*2x=2x/(1+x^4) y=(arcsinx)^2 记y=u^2u=arcsinxy'=dy/dx=dy/du*du/dx=2u*1/sqrt(1-x^2)=2(arcsinx)/sqrt(1-x^2)...