How do you find the derivative of #arcsin(1/sqrt(x^2+1))#?

Question

Charles Anctil · Accepted Answer

#dy/dx=-1/(x^2+1); if x>0,#

#=1/(x^2+1); if x<0.#

N.B.: #y=arc sin(1/sqrt(x^2+1)), x inRR# is not differentiable at #x=0.# 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.!

Scarlett Allison · Answer

undefined To find the derivative of arcsin(1/sqrt(x^2+1)), you can use the chain rule. Let u = 1/sqrt(x^2+1). Then differentiate u with respect to x, and substitute u into the derivative of arcsin(u). The derivative is -(1/sqrt(x^2+1))/sqrt(1 - (1/sqrt(x^2+1))^2).