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

Answer 1

#3/((3x-1)*ln2)#

For this problem, the following formula will be used

#color(white)(aaaaaaaaaaa)##color(green)(d/dxlog_a[f(x)] = 1/(f(x)*lna) *f'(x)#

What this basically means, is that the derivative of a log base a of some function equals 1 over that function times the natural log of base a times the derivative of #f(x)#.

#d/dxlog_2(3x-1)#

#=1/((3x-1)*ln2)*3#

#=3/((3x-1)*ln2)#

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Answer 2

Luke Alvarez

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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Answer from HIX Tutor

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.