Question

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（

Answer 1

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)