What is the first derivative of sin^2(6x)sin2(6x)?

1 Answer
Oct 26, 2017

(dy)/(dx)=12sin6xcos6xdydx=12sin6xcos6x

Explanation:

we will need the chain rule

(dy)/(dx)=(dy)/(du)(du)/(dx)dydx=dydududx

y=sin^2(6x)y=sin2(6x)

u=sin6x=>(du)/(dx)=6cos6xu=sin6xdudx=6cos6x

y=u^2=>(dy)/(du)=2uy=u2dydu=2u

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

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