What is f'(1) if f(x)=2x^3+1 ?

To calculate the value of the first derivative in a given
point, x0 = 1, we'll have to apply the limit of the ratio [f(x) -
f(x0)]/(x-x0):

limit [f(x) - f(1)]/(x-1), when x tends to
1.

We'll substitute f(x) and we'll calculate the value of
f(1):

f(1) = 2*1^3 + 1

f(1) =
2+1

f(1) = 3

limit [f(x) -
f(1)]/(x-1) = lim (2*x^3 + 1 - 3)/(x - 1)

We'll combine
like terms:

lim (2*x^3 + 1 - 3)/(x - 1) = lim (2*x^3 -
2)/(x - 1)

We'll factorize the numerator by
2:

lim (2*x^3 - 2)/(x - 1) = lim
2(x^3-1)/(x-1)

We'll write the difference of cubes as a
product:

x^3 - 1 = (x-1)(x^2 + x +
1)

lim 2(x^3-1)/(x-1) = 2 lim (x-1)(x^2 + x +
1)/(x-1)

We'll simplify the ratio and we'll
get:

2 lim (x-1)(x^2 + x + 1)/(x-1) = 2 lim (x^2 + x +
1)

We'll substitute x by 1 and we'll
get:

2 lim (x^2 + x + 1) = 2(1^2 + 1 +
1)

2 lim (x^2 + x + 1) = 2*3

2
lim (x^2 + x + 1) = 6

But f'(x0) =
f'(1)

f'(1) = limit [f(x) -
f(1)]/(x-1)

f'(1) =
6