# How do you derive #y = (1 + sin(2x)]/[1 - sin(2x))# using the quotient rule?

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To derive ( y = \frac{1 + \sin(2x)}{1 - \sin(2x)} ) using the quotient rule, follow these steps:

The quotient rule is given by (\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2}), where (f = 1 + \sin(2x)) and (g = 1 - \sin(2x)).

First, find the derivatives of (f) and (g):

- (f' = \frac{d}{dx}(1 + \sin(2x)) = 2\cos(2x))
- (g' = \frac{d}{dx}(1 - \sin(2x)) = -2\cos(2x))

Now, apply the quotient rule:

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

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

[= \frac{4\cos(2x)}{(1 - \sin(2x))^2}]

That is the derivative of (y) with respect to (x).

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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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