How do you find the derivatives of #y=log_2(3x)#?

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Answer 1 · 2017-07-16T17:50:08

#d/dx (log_2(3x)) = 1/(xln(2))#

Let us first derive the formula for the derivative of a function of the form #log_b(x)#:

#log_b(x) = ln(x)/ln(b)#
#d/dx (log_b(x)) = d/dx (ln(x)/ln(b))#

But #b# is a constant, so #ln(b)# would be a constant, and would therefore pull out of the differentiation.

#d/dx (log_b(x)) = 1/ln(b) * d/dx (ln(x))#

We know that #d/dx (ln(x)) = 1/x#
so, #d/dx (log_b(x)) = 1/ln(b) * 1/x = 1/(x*ln(b))#

Therefore we established that
#d/dx (log_b(x)) = 1/(x*ln(b))#

Moving on to your question,
#d/dx (log_2(3x)) = 1/(3x*ln(2)) * d/dx(3x)# by chain rule
#d/dx (log_2(3x)) = 1/(3x*ln(2)) * 3 = 3/(3xln(2))#

further simplifying by cancelling the #3# in the numerator and denominator...
#d/dx (log_2(3x)) = 1/(xln(2))#