# What is the derivative of #y = sin(tan 2x)#?

# dy/dx = 2cos(tan2x)sec^2 2x#

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To find the derivative of ( y = \sin(\tan(2x)) ), we'll apply the chain rule.

First, let's differentiate the outer function, (\sin(u)), where (u = \tan(2x)).

[ \frac{d}{du}(\sin(u)) = \cos(u) ]

Next, we differentiate the inner function, ( u = \tan(2x) ), using the chain rule:

[ \frac{d}{dx}(\tan(2x)) = \sec^2(2x) \cdot 2 ]

Now, combining both derivatives using the chain rule:

[ \frac{dy}{dx} = \cos(\tan(2x)) \cdot \sec^2(2x) \cdot 2 ]

So, the derivative of ( y = \sin(\tan(2x)) ) is:

[ \frac{dy}{dx} = 2\cos(\tan(2x))\sec^2(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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