How do you find the derivative of $y=sin^ntheta$ ?

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Answer 1 · 2017-03-28T07:12:55

$= n*sin^n theta* cot theta$

$y = sin ^n theta$

let say $x = sin theta$ ,
$(dx)/(d theta) = cos theta$

$y = x^n$
$(dy)/(dx) = n*x^(n-1) = n sin^(n-1) theta$

therefore,
$(dy)/( d theta) =(dy)/(dx) * (dx)/(d theta)$

$= n*sin^(n-1) theta* cos theta = n*sin^n theta / sin theta* cos theta$

$= n*sin^n theta * cot theta$

How do you find the derivative of y=sin^nthetay=sin^ntheta?