y=xtanx+(tanx)xLet
u=xtanx and v=(tanx)x
Now,
u=xtanx
Taking log both sides, we get
logu=tanxlogx
Differentiating above equation w.r.t. x, we have
1ududx=tanx(1x)+logx(sec2x)
dudx=u(tanxx+sec2xlogx)
dudx=xtanx(tanxx+sec2xlogx)
Similarly,
v=(tanx)x
Taking log both sides, we have
logv=xlog(tanx)
1vdvdx=log(tanx)⋅1+x(1tanxsec2x)
dvdx=v(log(tanx)+xsec2xtanx)
dvdx=(tanx)x(log(tanx)+xsec2xtanx)
Therefore,
dydx=dudx+dvdx
dydx=xtanx(tanxx+sec2xlogx)+(tanx)x(log(tanx)+xsec2xtanx)