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

1 Answer
Jul 16, 2017

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

Explanation:

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))