How do you differentiate #arcsin(csc(x^3)) )# using the chain rule?

1 Answer
Feb 17, 2018

Answer:

#(dy)/dx=(3x^2csc(x^3)cot(x^3))/sqrt(1-(csc(x^3))^2)#

Explanation:

Given:
#y=arcsin(csc(x^3))#
#arcsin(csc(x^3))=sin^-1(csc(x^3))#

Let #t=csc(x^3)#
#y=sin^-1t#
#(dy)/dt=1/sqrt(1-t^2)=1/sqrt(1-(csc(x^3))^2#

#(dy)/dx=(dy)/(dt)(dt)/dx#

#t=csc(x^3)#

Let #u=x^3#

#t=cscu#

#(dt)/(du)=cscucotu=csc(x^3)cot(x^3)#

#(dt)/dx=(dt)/(du)(du)/(dx)#

#(du)/(dx)=3x^2#

#(dt)/dx=(csc(x^3)cot(x^3))(3x^2)=3x^2csc(x^3)cot(x^3)#

#(dy)/dx=1/sqrt(1-(csc(x^3))^2)(3x^2csc(x^3)cot(x^3))#

#(dy)/dx=(3x^2csc(x^3)cot(x^3))/sqrt(1-(csc(x^3))^2)#