1、lim（x趋近于0）xcotx-1/x平方.2、设y=ln根号1-x/arccosx,求y'（0）.3、设y=ln（

lim(x->0)(xcotx-1/x^2)
=lim(x->0)(cosx*(x/sinx)-1/x^2) lim(x->0)(x/sinx)=lim(x->0)1/(sinx/x)=1
=-∞
y=ln√(1-x)/arccosx
y'=[ [-1/2√(1-x)]/√(1-x) ] /arccosx +ln√(1-x)*(-1/√(1-x^2))
=(-1/2)(1/(1-x))(1/arccosx) +ln√(1-x)*(-1/√(1-x^2)
y'(0)=(-1/2)(1/(π/2))+0=π
y=ln(1+x+y)
e^y=(1+x+y)
y'e^y=1+y'
y'(e^y-1)=1
y'=1/(e^y-1)
dy/dx=1/(x+y)