求微分y＝arctane^x+arctane^-x设y＝arctane^x+arctane^-x求dy

Question

求微分y＝arctane^x+arctane^-x
设 y＝arctane^x+arctane^-x
求dy

Answer 1

d(arctanx)=dx/(1+x^2)
d(e^x)=e^xdx
d[e^(-x)]=-e^xdx
dy=d(arctane^x)+d[arctane^(-x)]
=d(e^x)/(1+e^2x)+d[e^(-x)]/[1+e^(-2x)]
=e^xdx/(1+e^2x)-e^(-x)dx/[1+e^(-2x)]
=e^x/(1+e^2x)-e^x/(e^2x+1)=0