How do you differentiate #f(x) = sin(1/sqrt(arcsinx)) # using the chain rule?

1 Answer
Dec 24, 2017

Answer:

#-cos(1/sin^-1x)/(2sqrt(1-x^2)(sin^-1 x)^(3/2)#

Explanation:

let #u = sin^-1x#

#rArr sinu =x #

#rArr (dsinu)/dx = 1 #

# cosu (du)/dx = 1 #
# (du)/dx = 1/cosu = 1/(sqrt(1-sin^2u) #= # 1/(sqrt(1-x^2) #

Now, take the derivative suing chain rule,
#(df(x))/dx= (df(u))/(du) (du)/(dx) #
# (dsin(1/sqrt(u)))/(dx) #=# d/(du)sin (1/sqrt(u)) (d u^(-½))/(du) (du)/(dx)#
# (df(x))/dx = cos (1/sqrt(u)) (-1/2u^(-3/2)) 1/(sqrt(1-x^2)#
# (df(x))/dx = -1/2 cos (1/sqrt(u))/(sqrt(1-x^2) u^(3/2)#
#(df(x))/dx=-cos(1/sin^-1x)/(2sqrt(1-x^2)(sin^-1 x)^(3/2)#