# Find g''(x) if g(x) = 6x^3-4x^2 +8x -3

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### 3 Answers

We have the function g(x) = 6x^3-4x^2 +8x -3 and we have to find the second derivative.

The first derivative of g(x) = 6x^3-4x^2 +8x -3, is

g'(x) = 18x^2 - 8x + 8

The derivative of g'(x) = 18x^2 - 8x + 8 is

g''(x) = 36 x - 8

**Therefore for g(x) = 6x^3-4x^2 +8x -3, g''(x) = 36 x - 8.**

Given the function g(x) = 6x^3 - 4x^2 + 8x -3

We need to find the second derivative g''(x).

First we will differentitae g(x).

==> g'(x) = (6x^3)' - (4x^2)' + (8x)' -(3)'

= 18x^2 - 8x + 8 - 0

==> g'(x) = 18x^2 - 8x +8

Now we will differentiate again:

==> g''(x) = (18x^2)' -(8x)' + (8)'

= 36x - 8 + 0

**==> g''(x) = 36x -8**

g(x) = 6x^3-4x^2+8x-3. To find g"(x).

We use (x^n)' = n*x^(n-1).

To find the g"(x) , we get first g'(x) = (g(x)). and then ((g(x)')'

g'(x) = (6x^3-4x^2 +8x -3)'= (6*3x^2 -4*2x+8) = 18x^2 = 18x^2-8x+8.

g"(x) = (g'(x))' = (18x^2-8x+8)'

g"(x) = 18*2x -8 = 36x -8.

Therefore g"(x) = (6x^3-4x^2 +8x -3)" = 36x-8.