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)

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
2241 views around the world
You can reuse this answer
Creative Commons License