# Indicate derivative g(x)=(3x^2-2x+3)^4, g'(x) and g"(x)

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### 2 Answers

if `g(x) = (3x^2-2x+3)^4`

`g'(x) = 4*(6x-2)*(3x^2-2x+3)^3`

`g'(x) = 8(3x-1)(3x^2-2x+3)^3`

To find the second derivative, g''(x), you need to remember the derivative of a product,

(UV)' = U'V + V'U

so, g''(x) is,

`g''(x) = 8[3*(3x^2-2x+3)^3 -(3x-1)*(6x-2)*(3x^2-2x+3)^2]`

`g''(x)= 24(3x^2-2x+3)^2[(3x^2-2x+3) -(3x-1)(3x-1)]`

`g''(x)= 24(3x^2-2x+3)^2[3x^2-2x+3 -9x^2+6x-1]`

`g''(x)= 24(3x^2-2x+3^2)[-6x^2+4x+2]`

`g''(x)= -48(3x^2-2x+3)^2(3x^2-2x-1)`

The function g(x) = `(3x^2-2x+3)^4` . The derivative can be calculated using the chain rule.

g'(x) = `8*(3x^2-2x+3)^3*(3x - 1)`

g''(x) = `8*3*(3x^2-2x+3)^2*(3x - 1)^2 + 24*(3x^2-2x+3)^3`

=> `24*(3x^2-2x+3)^2*(3x - 1)^2 + 24*(3x^2-2x+3)^3`

**The derivative g'(x) = `12*(3x^2-2x+3)^3*(3x - 1)` and g''(x) = `24*(3x^2-2x+3)^2*(3x - 1)^2 + 24*(3x^2-2x+3)^3`**