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

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Answer 1 · 2017-10-26T17:01:54

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

we will need the chain rule

$(dy)/(dx)=(dy)/(du)(du)/(dx)$

$y=sin^2(6x)$

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

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

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

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

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