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

Question

Ezra Adams · Accepted Answer

#= 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 #

Emma Aldridge · Answer

undefined To find the derivative of $ y = \sin^n(\theta) $, where $ n $ is a constant:

1. Apply the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

2. Set $ u = \sin(\theta) $.

3. Rewrite the original function as $ y = u^n $.

4. Use the power rule for differentiation, which states that the derivative of $ x^n $ with respect to $ x $ is $ nx^{n-1} $.

5. Substitute $ u = \sin(\theta) $ back into the result to obtain the derivative in terms of $ \theta $.

So, the derivative of $ y = \sin^n(\theta) $ with respect to $ \theta $ is $ ny^{(n-1)}\cos(\theta) $.