# How do you find the derivative of #cosx/(sinx-2)#?

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To find the derivative of ( \frac{\cos(x)}{\sin(x) - 2} ), you can use the quotient rule, which states that if you have a function in the form ( \frac{u}{v} ), then the derivative is given by:

[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} ]

Where ( u = \cos(x) ) and ( v = \sin(x) - 2 ), and their derivatives are ( \frac{du}{dx} = -\sin(x) ) and ( \frac{dv}{dx} = \cos(x) ) respectively.

Substitute these values into the quotient rule formula:

[ \frac{d}{dx} \left( \frac{\cos(x)}{\sin(x) - 2} \right) = \frac{(\sin(x) - 2)(-\sin(x)) - \cos(x)(\cos(x))}{(\sin(x) - 2)^2} ]

Simplify the expression:

[ \frac{d}{dx} \left( \frac{\cos(x)}{\sin(x) - 2} \right) = \frac{-\sin^2(x) + 2\sin(x)\cos(x) - \cos^2(x)}{(\sin(x) - 2)^2} ]

Thus, the derivative of ( \frac{\cos(x)}{\sin(x) - 2} ) is ( \frac{-\sin^2(x) + 2\sin(x)\cos(x) - \cos^2(x)}{(\sin(x) - 2)^2} ).

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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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