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

1 Answer
salamat
Mar 28, 2017

#= n*sin^n theta* cot theta#

Explanation:

#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 #

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