Question

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

Answer 1

`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))`