Find the differential process of y = ln (cosx + LNX) more carefully and faster

Find the differential process of y = ln (cosx + LNX) more carefully and faster

y'=1/（cosx+ lnx）+(-sinx)+1/x
dy=[1/（cosx+ lnx）+(-sinx)+1/x]dx

How to use Cauchy's mean value theorem to find the limit of (x's square-x) / SiNx

Is it when x tends to zero, if so
Let f (x) = x ^ 2-x,
G(x)=sinx,
F'(x) =2x-1,F'(0)=-1,
G'(x)=cosx,G'(0)=1,
When x tends to zero,
(x^-x)/sinx=(F(x)-F(0))/(G(x)-G(0))=F'(0)/G'(0)=-1

Differential SiNx and cosx infinitely close to 0 with F (x + H) - f (x) / h

Is the derivation process, using sum difference product formula and SiNx and X are equivalent infinitesimals s in'x = LIM (sin (x + H) - SiNx) / h = 2limcos (x + H / 2) sin (H / 2) / h = 2limcos (x + H / 2) * (H / 2) / h = limcos (x + H / 2) = cosxcos'x = Lim [cos (x + H) - cosx] / h = - 2limsin (x + H / 2) sin (H / 2) / h =