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

Question

Elijah Abram · Accepted Answer

#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))#

Axel Alvarez · Answer

undefined To find the derivative of $ y = \log_2(3x) $, you can use the chain rule for differentiation.

First, differentiate the outer function $\log_2(u)$, treating the inner function $u = 3x$ as a whole. Then, multiply it by the derivative of the inner function.

The derivative of $\log_2(u)$ with respect to $u$ is $\frac{1}{u \ln 2}$.

The derivative of $u = 3x$ with respect to $x$ is $3$.

So, applying the chain rule, the derivative of $y$ with respect to $x$ is:

$$ y' = \frac{1}{3x \ln 2} \times 3 = \frac{1}{x \ln 2} $$