What is the derivative of this function #y=sin^-1(3x^5+1)^3#?

1 Answer
areebMaster
Jan 1, 2018

So, y = #sin^-1(3x^5+1)^3#
#(dy)/(dx) #= #(15x^4)/sqrt(1-(3x^5+1)^6)#

Explanation:

You have to use chain rule and a substitution(to make is simpler). You don't have to use substitution but I did. You should know that the derivative of arcsin x equals #1/(sqrt(1-x^2)# .
#dy/(dx)=dy/(du)*(du)/(dx)# From this you can see the du will cancel and #dy/(dx)# is obtained.
So, y = #sin^-1(3x^5+1)^3#
use u = #3x^5 +1 #
#(du)/(dx)# = #15x^4#

Substitute u in:
y= #sin^-1 (u)^3#
#(dy)/(du)#= #1/sqrt(1-(u^3)^2#
= #1/sqrt(1-u^6)#
Now to obtain #(dy)/(dx)# we need to multiply #(dy)/(du)# by #(du)/(dx)#

#(dy)/(dx)# = #(15x^4)/sqrt(1-u^6)#
Substitute #3x^5 +1# back in to get rid of u
#(dy)/(dx) #= #(15x^4)/sqrt(1-(3x^5+1)^6)#

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