y等于1加sinx除以1加cosx的导数请不要乱回答

已知y=(1+sinx)/(1+cosx)，求dy/dx
dy/dx=[(1+cosx)cosx+(1+sinx)sinx]/(1+cosx)²=(1+cosx+sinx)/(1+cosx)²

Y＝[（1+sinx）/（1+cosx）]′
=[（1+cosx）（1+sinx）′-（1+cosx）′（1+sinx）]/[（1+cosx）²]
=[（1+cosx）cosx-（-sinx）（1+sinx）]/[（1+cosx）²]
=（1+sinx+cosx）/[（1+cosx）²]

y=(1+sinx)/(1+cosx)
=1/(1+cosx)+sinx/(1+cosx)
=1/(2cos(x/2)^2)+tan(x/2)
=(1/2)sec(x/2)^2 +tan(x/2)
y'=(1/2)* (secx/2)' *(2sec(x/2)) +(x/2)' *sec(x/2)^2
=(1/2)*(1/2)*sec(x/2)tan(x/2)*(2sec(x/2) +(1/2)sec(x/2)^2
=(1/2)sec(x/2)^2*tan(x/2)+(1/2)sec(x/2)^2
=(1/(2cos(x/2)^2))(tan(x/2)+1)
=[1/(1+cosx)]*(sinx/(1+cosx)+1)
=sinx/(1+cosx)^2+1/(1+cosx)

y=cos x /(1-sin x) y1=[-sinx(1-sinx)-cosx(-cosx)]/(1-sinx1/(1-sinx) (x不等于2kπ+π/2) y=cos x /(1-sin x) 所以

y=u/v
y'=(u'v-v'u/v^2
y=(1+sinx)/(1+cosx)
y'=(cosx(1+cosx)+sinx(1+sinx))/(1+cosx)^2=(1+cosx+sinx)/(1+cosx)^2