# How do you find the derivative of #g(x)= 3tan4xsin2xcos2x#?

First just remember

So the result is

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To find the derivative of ( g(x) = 3\tan(4x)\sin(2x)\cos(2x) ), you can use the product rule and chain rule.

Let's denote:

( u(x) = 3\tan(4x) )

( v(x) = \sin(2x)\cos(2x) )

Apply the product rule:

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

Now, differentiate ( u(x) ) and ( v(x) ) separately using the chain rule and product rule, respectively:

( u'(x) = 3 \cdot \frac{d}{dx}[\tan(4x)] )

( v'(x) = \frac{d}{dx}[\sin(2x)]\cos(2x) + \sin(2x)\frac{d}{dx}[\cos(2x)] )

After finding the derivatives, substitute back into the product rule formula and simplify to find ( g'(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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