How do you find the derivative of arcsin(2x^2)?

1 Answer
maganbhai P.
Mar 9, 2018

(4x)/sqrt(1-4x^4

Explanation:

Here,
y=sin^(-1)(2x^2), take , u=2x^2
y=sin^(-1)u
(dy)/(du)=1/sqrt(1-u^2)and(du)/(dx)=4x
color(red)((dy)/(dx)=(dy)/(du)*(du)/(dx))=1/sqrt(1-u^2)*4x
=>(dy)/(dx)=1/(sqrt(1-(2x^2)^2))*4x=(4x)/sqrt(1-4x^4

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
5071 views around the world
You can reuse this answer
Creative Commons License