# What is the derivative of the function f(x) = 2x^3+1 in the point x=1?

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

To find the derivative of the function at x = 1.

We know derivative x^n = d/dx(k*x^n) = (k*x^n0' = k * nx^(n-1).

Using this we get d/dx (k) = d/dx(kx^0) = 0. Or the drivative of a constantant = 0.

We s use the above results to fibd the derivative of 2x^3+1.

d/dxf(x) = d/dx { 2x^3 +1}.

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

d/dx{f(x) } = 2d/dx{x^3}+ derivative of a constant.

d/dx {f(x)} = 2*3x^(3-1) + 0.

d/dx{f(x) } =f'(x)= 6x^2.

Therefore f'(1) = 6*1^2 = 6.

Therefore f'(1) = 6.

Therefore the derivative 2x^3+1 is 6x^2 and at x = 1 , the derivative of (2x^3+1) = 6x^2 = 6*1^2 = 1.

To calculate the value of the first derivative in a given point, x = 1, we'll have to apply the limit of the ratio:

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

**The value of derivative of the function for x = 1 is:**

**f'(1) = 6**