# How do you differentiate #3sin^2(3x) # using the chain rule?

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To differentiate ( 3\sin^2(3x) ) using the chain rule, you apply the chain rule which states that ( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) ).

Let ( u = 3x ) and ( v = \sin(u) ).

Now, ( \frac{du}{dx} = 3 ) and ( \frac{dv}{du} = 2\sin(u)\cos(u) ) (using the chain rule for ( \sin^2(u) )).

Therefore, ( \frac{dv}{dx} = \frac{dv}{du} \cdot \frac{du}{dx} = 2\sin(3x)\cos(3x) \cdot 3 = 6\sin(3x)\cos(3x) ).

So, ( \frac{d}{dx}[3\sin^2(3x)] = 6\sin(3x)\cos(3x) ).

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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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