# How do you find the derivative of #y=ln(sin2x)#?

The derivative is

We'll need to use the derivative chain rule:

Plugging in the functions:

That's the derivative. Hope this helped!

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To find the derivative of y = ln(sin(2x)), you can use the chain rule.

Let u = sin(2x).

Then, y = ln(u).

Now, apply the chain rule, which states that if y = ln(u), then y' = (1/u) * u'.

First, find u':

u' = d/dx [sin(2x)] = cos(2x) * d/dx [2x] = 2cos(2x)

Now, substitute u' back into the chain rule:

y' = (1/u) * u' = (1/sin(2x)) * 2cos(2x) = 2cos(2x) / sin(2x)

Therefore, the derivative of y = ln(sin(2x)) is y' = 2cos(2x) / sin(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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