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

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

First we will find the first derivative f'(x);

f'(x) = 3*3x^2 + 5 + 1/x

= 9x^2 + 1/x + 5

Now we will substiute with x = 1 to calculate f'(1):

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

= 9 + 1 + 5

= 15

**=> 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**