Question

What is the derivative of `f(x)=(log_6(x))^2` ?

Answer 1

Method 1:

We will begin by using the change-of-base rule to rewrite `f(x)` equivalently as:

`f(x) = (lnx/ln6)^2`

We know that `d/dx[ln x] = 1/x`.

(if this identity looks unfamiliar, check some of the videos on this page for further explanation)

So, we will apply the chain rule:

`f'(x) = 2*(lnx/ln6)^1 * d/dx[ln x / ln 6]`

The derivative of `ln x/6` will be `1/(xln6)`:

`f'(x) = 2*(lnx/ln6)^1 * 1 / (xln 6)`

Simplifying gives us:

`f'(x) = (2lnx)/(x(ln6)^2)`

Method 2:

The first thing to note is that only `d/dx ln(x) = 1/x` where `ln = log_e`. In other words, only if the base is `e`.

We must therefore convert the `log_6` to an expression having only `log_e = ln`. This we do using the fact

`log_a b = (log_{n}b)/(log_{n} a) = (ln b)/ln a` when `n=e`

Now, let `z = (ln x/ln 6)` so that `f(x) = z^2`

Therefore, `f'(x)=d/dx z^2 = (d/dz z^2)( dz/dx) = 2z d/dx (ln x/ln 6)`

`= (2z)/(ln 6) d/dx ln x = (2z)/(ln 6) 1/x`
`= (2/ln 6)(ln x/ln 6)(1/x) = (2 ln x)/(x*(ln 6)^2)`