# how would I find the second derivative of x^3+4x^2+4x/x^2-1 ?

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### 1 Answer

You need to find the first derivative and then you may find the second derivative.

You should notice that the function you need to differentiate is rational, hence you need to use quotient rule such that:

`f'(x) = ((x^3+4x^2+4x)'*(x^2-1) - (x^3+4x^2+4x)*(x^2-1)')/((x^2 - 1)^2)`

`f'(x) = ((3x^2 + 8x + 4)(x^2 - 1) - 2x(x^3+4x^2+4x))/((x^2 - 1)^2)`

You need to open the brackets such that:

`f'(x) = (3x^4 - 3x^2+ 8x^3 - 8x+ 4x^2- 4 - 2x^4 - 8x^3 - 8x^2)/((x^2 - 1)^2)`

Collecting like terms yields:

`f'(x) = (x^4 - 7x^2 - 8x- 4)/((x^2 - 1)^2)`

You need to differentiate f'(x) with respect to x to find f"(x) such that:

`f''(x) = ((x^4 - 7x^2 - 8x- 4)'*(x^2 - 1)^2 - (x^4 - 7x^2 - 8x- 4)*((x^2 - 1)^2)')/((x^2 - 1)^4)`

`f''(x) = ((4x^3 - 14x - 8)*(x^2 - 1)^2 - 4x(x^4 - 7x^2 - 8x- 4)(x^2 - 1))/((x^2 - 1)^4)`

You need to factor out `x^2 - 1` such that:

`f''(x) = (x^2-1)((4x^3 - 14x - 8)*(x^2 - 1) - 4x(x^4 - 7x^2 - 8x- 4))/((x^2 - 1)^4)`

Reducing like terms yields:

`f''(x) = ((4x^3 - 14x - 8)*(x^2 - 1) - 4x(x^4 - 7x^2 - 8x- 4))/((x^2 - 1)^3)`

You need to open the brackets:

`f''(x) = (4x^4 - 4x^3- 14x^3 + 14x- 8x^2 + 8 - 4x^5 + 28x^3+ 32x^2 + 16x))/((x^2 - 1)^3)`

Collecting like terms yields:

`f''(x) = ( - 4x^5 + 4x^4 + 10x^3 + 24x^2+ 30x + 8)/((x^2 - 1)^3)`

**Hence, evaluating the second derivative yields `f''(x) = ( - 4x^5 + 4x^4 + 10x^3 + 24x^2 + 30x + 8)/((x^2 - 1)^3).` **