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.!
