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

Use the quotient rule:

If we let

Then, to find the derivative using the quotient rule would generally look like:

Therefore,

Find the derivatives and simplify.

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

- Identify ( u(x) = \cot x ) and ( v(x) = 1 - \sin x ).
- Apply the quotient rule, which states that the derivative of ( \frac{u(x)}{v(x)} ) is given by: [ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} ]
- Differentiate ( u(x) ) and ( v(x) ) separately: [ u'(x) = -\csc^2 x ] [ v'(x) = -\cos x ]
- Substitute the derivatives and the original functions into the quotient rule formula: [ y' = \frac{(-\csc^2 x)(1 - \sin x) - (\cot x)(-\cos x)}{(1 - \sin x)^2} ]
- Simplify the expression: [ y' = \frac{-\csc^2 x + \csc^2 x \sin x + \cot x \cos x}{(1 - \sin x)^2} ] [ y' = \frac{\csc^2 x(\sin x - 1) + \cot x \cos x}{(1 - \sin x)^2} ]

This is the derivative of ( y = \frac{\cot x}{1 - \sin x} ) using the quotient rule.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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