In this example, we have a quotient, where the numerator is a product.

Once again, we let `y=u/v`, where `u=x^2(3x+1)` and `v=x^4+2`.

The quotient formula requires `(du)/(dx)` but u is a product.

Let `u = pq` where p = x² and q = 3x + 1.

` {: ((du)/(dx),=p(dq)/(dx)+q(dp)/(dx)),(,=(x^2)(3)+(3x+1)(2x)),(,=3x^2+6x^2+2x),(,=9x^2+2x) :}`

We also require

`(dv)/(dx)=4x^3`

So

` {: ((dy)/(dx),=(d(u/v))/(dx)=(v(du)/(dx)-u(dv)/(dx))/(v^2)),(,=((x^4+2)(9x^2+2x)-(x^2)(3x+1)(4x^3))/((x^4+2)^2)),(,=(9x^6+2x^5+18x^2+4x-12x^6-4x^5)/((x^4+2)^2)),(,=(-3x^6-2x^5+18x^2+4x)/((x^4+2)^2)) :}`