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

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Answer 1 · 2017-10-26T17:01:54

$\frac{d y}{d x} = 12 sin 6 x cos 6 x$ $(dy)/(dx)=12sin6xcos6x$

we will need the chain rule

$\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)(du)/(dx)$

$y = {sin}^{2} (6 x)$ $y=sin^2(6x)$

$u = sin 6 x \Rightarrow \frac{d u}{d x} = 6 cos 6 x$ $u=sin6x=>(du)/(dx)=6cos6x$

$y = u^{2} \Rightarrow \frac{d y}{d u} = 2 u$ $y=u^2=>(dy)/(du)=2u$

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

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

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