To find the derivatives, we need to use the quotient rule, which means that:

`f(x)=(x^2-6)^2/(x^2+3)^3`

so then

`f'(x)={2(x^2-6)(2x)(x^2+3)^3-(x^2-6)^2(3)(x^2+3)^2(2x)}/(x^2+3)^6` factor the numerator

`={2x(x^2-6)(x^2+3)^2(2(x^2+3)-3(x^2-6))}/(x^2+3)^6` simplify numerator

`={2x(x^2-6)(2x^2+6-3x^2+18)}/(x^2+3)^4` simplify further

`={2x(x^2-6)(12-x^2)}/(x^2+3)^4` expand the numerator

`={2x(12x^2-x^4+6x^2-72)}/(x^2+3)^4` once more

`={-2x^4+36x^2-144}/(x^2+3)^4`

Now we can differentiate again.

`f''(x)={(-8x^3+72x)(x^2+3)^4-(-2x^4+36x^2-144)(4)(x^2+3)^3(2x)}/(x^2+3)^8`

`={8x(x^2+3)^3((-x^2+9)(x^2+3)+2x^4-36x^2+144)}/(x^2+3)^8` simplify numerator

`={8x(-x^4+6x^2+27+2x^4-36x^2+144)}/(x^2+3)^8`

`={8x(x^4-30x^2+171)}/(x^2+3)^8`

** The second derivative is...**

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To find the derivatives, we need to use the quotient rule, which means that:

`f(x)=(x^2-6)^2/(x^2+3)^3`

so then

`f'(x)={2(x^2-6)(2x)(x^2+3)^3-(x^2-6)^2(3)(x^2+3)^2(2x)}/(x^2+3)^6` factor the numerator

`={2x(x^2-6)(x^2+3)^2(2(x^2+3)-3(x^2-6))}/(x^2+3)^6` simplify numerator

`={2x(x^2-6)(2x^2+6-3x^2+18)}/(x^2+3)^4` simplify further

`={2x(x^2-6)(12-x^2)}/(x^2+3)^4` expand the numerator

`={2x(12x^2-x^4+6x^2-72)}/(x^2+3)^4` once more

`={-2x^4+36x^2-144}/(x^2+3)^4`

Now we can differentiate again.

`f''(x)={(-8x^3+72x)(x^2+3)^4-(-2x^4+36x^2-144)(4)(x^2+3)^3(2x)}/(x^2+3)^8`

`={8x(x^2+3)^3((-x^2+9)(x^2+3)+2x^4-36x^2+144)}/(x^2+3)^8` simplify numerator

`={8x(-x^4+6x^2+27+2x^4-36x^2+144)}/(x^2+3)^8`

`={8x(x^4-30x^2+171)}/(x^2+3)^8`

**The second derivative is `{8x(x^4-30x^2+171)}/(x^2+3)^8` .**