# How do you differentiate #log_2(3x-1)#?

For this problem, the following formula will be used

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To differentiate log base 2 of (3x - 1), use the chain rule:

d/dx [log₂(3x - 1)] = (1 / ln(2)) * (1 / (3x - 1)) * d/dx(3x - 1) = (1 / ln(2)) * (1 / (3x - 1)) * 3 = 3 / (ln(2) * (3x - 1))

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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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