f(x) = 3x^3 + 5x + lnx find f'(1)

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