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How do you differentiate #log_2 (x)#?

Answer 1

William Abdalla

#d/dx log_2(x) = 1/(x*ln(2))#

This follows from the general formula: #color(white)("XXX")d/dx(log_a(x)) = 1/(x*ln(a))#

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Answer 2

Christopher Agresta

#"d"/("d"x) [log_2(x)] = 1/(xln(2))#

As we know how to differentiate #ln(x)#, we should change the base of the logarithm first.

The according formula to change a logarithmic expression from the base #a# to the base #b# is

#log_color(red)(a)(color(blue)(x)) = log_b(color(blue)(x)) / log_b(color(red)(a)) #

You can apply the formula as follows:

#log_2(x) = ln(x) / ln(2)#

As #1/ln(2)# is just a constant and the derivative of #ln(x)# is #1/x#, our derivative is:

#"d"/("d"x) [log_2(x)] = "d"/("d"x) [ln(x) / ln(2) ] = 1/ln(2) * 1/x = 1/(xln(2))#

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Answer 3

Elijah Abram

To differentiate ( \log_2(x) ), you can use the chain rule. The derivative of ( \log_b(x) ) with respect to ( x ) is ( \frac{1}{x\ln(b)} ). So, the derivative of ( \log_2(x) ) is ( \frac{1}{x\ln(2)} ).

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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.