# How do you differentiate #g(x) = [1 + sin(2x)]/[1 - sin(2x)] # using the product rule?

In order to forcedly use the product rule, we'll rewrite it as

Chain rule will also be required in order to distinguish the second term.

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To differentiate ( g(x) = \frac{1 + \sin(2x)}{1 - \sin(2x)} ) using the product rule, first, identify the two functions as ( u(x) = 1 + \sin(2x) ) and ( v(x) = 1 - \sin(2x) ). Then, apply the product rule, which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

The derivative of ( u(x) ) is ( u'(x) = 2\cos(2x) ), and the derivative of ( v(x) ) is ( v'(x) = -2\cos(2x) ).

Now, apply the product rule formula:

[ g'(x) = u'(x)v(x) + u(x)v'(x) ]

Substitute the derivatives and the original functions into the formula:

[ g'(x) = (2\cos(2x))(1 - \sin(2x)) + (1 + \sin(2x))(-2\cos(2x)) ]

Simplify the expression:

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

Combine like terms:

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

[ g'(x) = -4\sin(2x)\cos(2x) ]

So, the derivative of ( g(x) ) with respect to ( x ) using the product rule is ( -4\sin(2x)\cos(2x) ).

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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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