# How do you differentiate #f(x)=(2x^3)(ln3x)(sinx) # using the product rule?

# f'(x) = 6x^2 \ ln3x \ sinx + 2x^2 \ sinx + 2x^3 \ ln3x \ cosx #

We have:

We need to use the triple product rule, giving

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To differentiate ( f(x) = (2x^3)(\ln(3x))(sin(x)) ) using the product rule, you would differentiate each term separately and then apply the product rule. Here's the step-by-step process:

- Identify the terms: ( f(x) = u \cdot v \cdot w ), where ( u = 2x^3 ), ( v = \ln(3x) ), and ( w = \sin(x) ).
- Differentiate each term:
- ( u' = 6x^2 ) (derivative of ( 2x^3 ) with respect to ( x )).
- ( v' = \frac{1}{3x} ) (derivative of ( \ln(3x) ) with respect to ( x )).
- ( w' = \cos(x) ) (derivative of ( \sin(x) ) with respect to ( x )).

- Apply the product rule: ( f'(x) = u'vw + uv'w + uvw' ).
- Substitute the derivatives and original functions back into the formula:
- ( f'(x) = (6x^2)(\ln(3x))(\sin(x)) + (2x^3)\left(\frac{1}{3x}\right)(\sin(x)) + (2x^3)(\ln(3x))(\cos(x)) ).

So, the derivative of ( f(x) = (2x^3)(\ln(3x))(\sin(x)) ) is ( f'(x) = (6x^2)(\ln(3x))(\sin(x)) + \frac{2}{3}(\sin(x)) + (2x^3)(\ln(3x))(\cos(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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