This is a composed function and we'll differentiate from the last function to the first.

We'll re-write the function:

f(x) = (ln x)^1/9

The last function is the power function and the first function is the logarithmic function.

The function f(x) is the result of composition between u(x) = u^1/9 and v(x) = ln x

u(v(x)) = (lnx)^1/9

u'(x) = (u^1/9)' = (1/9)*u^(1/9 - 1)

u'(x) = (1/9)*u^(1-9)/9

u'(x) = 1/9*u^8/9

v'(x) = (ln x)' = 1/x

**[u(v(x))]' = 1/9x*(ln x)^8/9**

We'll use the power property of logarithms:

a*ln b = ln (b^a)

f'(x) = 1/(ln x)^8*9x/9

We'll simplify and we'll get:

**f'(x) = 1/(ln x)^8x/9**

To diferentiate the function f(x) = (ln(x))^(1/9).

The right side is written in the index form.

Or

ninth root of (lnx) = (ln(x))^(1/n).

We know d/dx {f(x)}^n = n {(f(x))^(n-1)}{f'(x)}.

Let f(x) = lnx.

Then f'(x) = 1/x.

Therefore d/dx (ln(x))^(1/9) ={ (1/9)(lnx)^((1/9) -1) }(ln(x))'.

d/dx (ln(x))^(1/9) = (1/9)(ln(x))^(-8/9) * (1/x).

d/dx (ln(x))^(1/9) = (1/9x) / (ln(x))^(8/9) , as a^-m = 1/a^m by index law.

d/dx(ln(x))^(1/9) = 1/{9x(ln(x))^(8/9)}.

Hope this helps.