How do you find the derivative of $y = {sin}^{n} θ$ $y=sin^ntheta$ ?

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How do you find the derivative of $y = {sin}^{n} θ$ $y=sin^ntheta$ ?

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Answer 1 · 2017-03-28T07:12:55

$= n \cdot {sin}^{n} θ \cdot cot θ$ $= n*sin^n theta* cot theta$

Explanation:

$y = {sin}^{n} θ$ $y = sin ^n theta$

let say $x = sin θ$ $x = sin theta$ ,
$\frac{d x}{d θ} = cos θ$ $(dx)/(d theta) = cos theta$

$y = x^{n}$ $y = x^n$
$\frac{d y}{d x} = n \cdot x^{n - 1} = n {sin}^{n - 1} θ$ $(dy)/(dx) = n*x^(n-1) = n sin^(n-1) theta$

therefore,
$\frac{d y}{d θ} = \frac{d y}{d x} \cdot \frac{d x}{d θ}$ $(dy)/( d theta) =(dy)/(dx) * (dx)/(d theta)$

$= n \cdot {sin}^{n - 1} θ \cdot cos θ = n \cdot \frac{{sin}^{n} θ}{sin θ} \cdot cos θ$ $= n*sin^(n-1) theta* cos theta = n*sin^n theta / sin theta* cos theta$

$= n \cdot {sin}^{n} θ \cdot cot θ$ $= n*sin^n theta * cot theta$

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How do you find the derivative of $y = {sin}^{n} θ$ $y=sin^ntheta$ ?

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