# How do you find the derivative of #(sin x + 2x) / (cos x - 2)#?

#f'(x) = (2xsinx-3)/(cosx-2)^2#

Utilizing the quotient rule, this can be distinguished.

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To find the derivative of ( \frac{\sin x + 2x}{\cos x - 2} ), you would use the quotient rule, which states that the derivative of a quotient of two functions is the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

So, applying the quotient rule:

[ \left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} ]

Let ( f(x) = \sin x + 2x ) and ( g(x) = \cos x - 2 ). Then:

[ f'(x) = \cos x + 2 ] [ g'(x) = -\sin x ]

Substituting these into the quotient rule formula:

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