Question

What is the derivative of `log_2(3x-1)`?

Answer 1

`d/dx log_2(3x-1) = 3/(3xln 2-ln2)`

Explanation:

You might already know that:

`d/dx ln x = 1/x`

In order to use this, we have to apply the change of base rule:

`log_a(b) = log_c(b)/log_c(a)`

Therefore:

`log_2(3x-1)`

`= ln (3x-1)/ ln 2`

Now we can differentiate the expression:

`d/dx ln (3x-1)/ ln 2`

`1/ln 2` is a constant, so we can bring it out:

`d/dx ln (3x-1)/ ln 2`
`= 1/ln 2 * d/dx ln (3x-1)`

Next, we can use the chain rule:

`(f @ g)'(x) = f'(g(x)) * g'(x)`

1) Take the derivative of the outer function..
2) ...plug in the inner function.
3) Multiply by the derivative of the inner function.

So:

`1/ln 2 * d/dx ln (3x-1)`

`= 1/ln 2 * 1/(3x-1) * d/dx(3x-1)`

`= 1/ln 2 * 1/(3x-1) * (3)`

`= 3/((ln 2)(3x-1))`

`= 3/(3xln 2-ln2)`