What is the derivative of ${sin}^{2} (4 x)$ $sin^2(4x)$ ?

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Answer 1 · 2017-11-08T14:11:03

$\frac{d y}{d x} = 8 sin (4 x) cos (4 x)$ $(dy)/(dx)=8sin(4x)cos(4x)$

$y = {(sin (4 x))}^{2} = {(f (x))}^{2}$ $y=(sin(4x))^2=(f(x))^2$

$y=(f(x))^n=>(dy)/(dx)=n*f'(x)*(f(x))^(n-1)$

So, $y=(sin(4x))^2$
$(dy)/(dx)=2*sin(4x)*d/(dx)sin(4x)$
$=2sin(4x)*4cos(4x)$
$=8sin(4x)cos(4x)$

What is the derivative of sin^2(4x)sin2(4x)sin^2(4x)?