How do you find the derivative of #arcsin(4x)#?

Question

12896 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-22T14:28:01

# 4 / (sqrt{1 - 16x^2})#

#y = arcsin(4x)#

#sin y = 4x#

#cos y y' = 4#

#y' = 4/ (cos y)#

#= 4 / (sqrt{1 - sin^2 y})#

#= 4 / (sqrt{1 - (4x)^2})#

#= 4 / (sqrt{1 - 16x^2})#