# How do you differentiate # g(x) =(1+cosx)/(1-cosx) #?

Use the quotient rule :

Substituting into the quotient rule:

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By simplification

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To differentiate the function ( g(x) = \frac{1 + \cos x}{1 - \cos x} ), you can use the quotient rule. The quotient rule states that for a function ( f(x) = \frac{u(x)}{v(x)} ), its derivative is given by ( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} ). Applying this rule to ( g(x) ), where ( u(x) = 1 + \cos x ) and ( v(x) = 1 - \cos x ), we have:

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

Simplify the expression:

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

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

So, the derivative of ( g(x) ) with respect to ( x ) is ( \frac{2\sin x}{(1 - \cos x)^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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