What is the first derivative of #sin^2(6x)#?

Question

↳Redirected from "Question #3fdac"

1873 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-26T17:01:54

#(dy)/(dx)=12sin6xcos6x#

we will need the chain rule

#(dy)/(dx)=(dy)/(du)(du)/(dx)#

#y=sin^2(6x)#

#u=sin6x=>(du)/(dx)=6cos6x#

#y=u^2=>(dy)/(du)=2u#

#:.=>(dy)/(dx)=2uxx6cos6x#

#(dy)/(dx)=12ucos6x=12sin6xcos6x#