# How do you find the derivative of # y =sinx/(1-cosx)#?

This function can be differentiated using the "quotient rule":

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To find the derivative of ( y = \frac{\sin x}{1 - \cos x} ), we use the quotient rule.

Let ( u(x) = \sin x ) and ( v(x) = 1 - \cos x ).

The derivative of ( u(x) ) is ( u'(x) = \cos x ), and the derivative of ( v(x) ) is ( v'(x) = \sin x ).

Now, applying the quotient rule ( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} ), we plug in these derivatives:

[ y' = \frac{(\cos x)(1 - \cos x) - (\sin x)(-\sin x)}{(1 - \cos x)^2} ]

Simplify the expression if necessary. This yields the derivative of ( y ) with respect to ( x ).

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