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

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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**

f(x) = 2x^3+1.

To find f'(1).

Solution:

f'(1) is the value of f'(x) at x = 1, where f'(x) is the derivative of x.

So we find first the derivative of f(x) or f'(x).

f(x) = 2x^3+1.

f'(x) = d/dx(2x^3+1) = (2x^3+1)'.

f'(x) = (2x^3)'+(1)' , as d/dx(u(x)+v(x)) = (u(x)+v(x))' = u'(x) +v'(x).

f'(x) = 2(x^3)' +0 , as (const)' = 0. and (k*g(x)' = k*g'(x).

f'(x) = 2*3x^(3-1), as (x^n) = nx^(n-1).

f'(x) =6x^2,

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