
Dayv O. answered 03/12/24
Caring Super Enthusiastic Knowledgeable Pre-Calculus Tutor
Domain cannot have x2=1, Domain is (-∞,-1)U(-1,1)U(1,∞)
|1-x2|=1-x2 when x2<1,,,x<1 and x>-1
|1-x2|=x2-1 when x2>1 ,,x>1 and x<-1
function 1
f(x)=ln(1-x2)1/2 ,,, for x domain (-1,1)
f'(x)=(1-x2)-1/2[(1/2)(1-x2)-1/2(-2x)=(-)x/(1-x2)
fuction 2 ,,, for x domain (-∞,-1)U(1,∞)
f(x)=ln(x2-1)1/2
f'(x)=(x2-1)-1/2[(1/2)(x2-1)-1/2(2x)=x(x2-1)