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

Question

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

1 Answer

Impact of this question

4719 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-09T16:34:46

#(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#

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question