You need to evaluate derivative of the function, such that:

`f'(x) = (3x^3 + 5x + ln x)' => f'(x) = (3x^3)' + (5x)' + (ln x)'`

`f'(x) = 9x^2 + 5 + 1/x`

You need to evaluate `f'(x)` at `x = 1` , such that:

`f'(1) = 9*1^2 + 5 + 1/1 => f'(1) = 9 + 5 + 1 => f'(1) = 15`

**Hence, evaluating derivative of the given function at `x = 1` , yields **`f'(1) = 15.`

We'll apply delta method to determine the instantaneous rate of change of y with respect to x.

dy/dx = lim [f(x + delta x) - f(x)]/delta x, delta x->0

We also can write:

dy/dx = lim [f(x + h) - f(x)]/h, h->0

f(x+h) = 3(x+h)^3 + 5(x+h) + ln(x+h)

We'll raise to cube x + h:

f(x+h) = 3x^3 + 9x^2h + 9xh^2 + 3h^3 + 5x + 5h + ln(x+h)

lim [f(x + h) - f(x)]/h = lim [3x^3 + 9x^2h + 9xh^2 + 3h^3 + 5x + 5h + ln(x+h) - 3(x)^3 - 5(x) - ln(x)]/h

We'll eliminate like terms:

lim [3x^3 + 9x^2h + 9xh^2 + 3h^3 + 5x + 5h + ln(x+h) - 3(x)^3 - 5(x) - ln(x)]/h = lim [9x^2h + 9xh^2 + 3h^3 + 5h + ln(x+h)/(x)]/h

lim [f(x) - f(1)]/(x-1) = lim (3x^3 + 5x + lnx - 8 - ln 1)/(x-1)

lim (3x^3 + 5x + lnx - 8)/(x-1) = (8-8)/(1-1) = 0/0

We'll apply L'Hospital rule:

lim (3x^3 + 5x + lnx - 8)/(x-1) = lim (3x^3 + 5x + lnx - 8)'/(x-1)'

lim (3x^3 + 5x + lnx - 8)'/(x-1)' = lim (9x^2 + 5 + 1/x)

We'll substitute x by 1:

lim (9x^2 + 5 + 1/x) = 9+5+1 = 15

But f'(1) = lim [f(x) - f(1)]/(x-1) = 15