Let, y=arc sin (1/sqrt(x^2+1)), x in RR.
Let x=cottheta," so that, "theta in (0,pi), &, theta=arc cotx.
Observe that, the Range of cot fun. is RR, so we can take,
x=cottheta.
We will consider the following 2 Cases :
Case (1) : x>0.
x=cottheta, theta in (0,pi), and x>0 rArr theta in (0,pi/2).
Also, y=arc sin (1/sqrt(x^2+1))=arc sin (1/sqrt(cot^2theta+1))
=arc sin(1/csctheta)=arc sin(sintheta).
:. y=arc sin(sintheta), where, theta in (0,pi/2)sub(-pi/2,pi/2).
:.," by Defn. of "arc sin" fun., "y=theta=arc cotx; if x>0
:. dy/dx=d/dx arc cotx=-1/(x^2+1), if x>0.
Case (2) : x<0.
Here, because x<0, theta in (0,pi)-(0,pi/2)=(pi/2,pi), i.e.,
pi/2ltthetaltpi rArr -pi/2gt-thetagt-pi
rArr pi-pi/2>pi-theta>pi-pi, or, (pi-theta) in (-pi/2,0)
Also, sin(pi-theta)=sintheta.
Thus, y=arc sin(sintheta)=arc sin(sin(pi-theta),
where, (pi-theta) in (-pi/2,0) sub (-pi/2,pi/2).
Hence, by the Defn. of arc sin" fun., "y=pi-theta=pi-arc cotx, if x<0.
:., dy/dx=0-(-1/(x^2+1))=1/(x^2+1), if x<0.
Altogether, dy/dx=-1/(x^2+1); if x>0,
=1/(x^2+1); if x<0.
N.B.: From the above discussion, we conclude that,
y=arc sin(1/sqrt(x^2+1)), x inRR is not differentiable at x=0.
Enjoy Maths.!
