Determine the acceleration given the position function of a particle 
The position function of a particle in a test laboratory is $s(t) = \frac{10t}
{t^2+3}$.
  Determine the acceleration of the particle of particle after 4 seconds.

The speed $v(t)$ of the particle is the derivative of the position $\frac{ds(t)}{dt}$.
I got $v(t) = \frac{10(-t^2+3)}{(t^2+3)^2}$
For the acceleration $a(t)$ I tried taking the derivative of $v(t)$ but got the wrong answer at $a(4)$.
$$a(t)=\frac{-2t\cdot (t^3+3)^2-10(-t^2+3)\cdot 2(t^2+3)\cdot(2t)}{(t^2+3)^4}$$

$$s(t) = \frac{10t}{t^2+3}$$
$$\implies v(t) = \frac{ds}{dt} = \frac{(t^2+3)\cdot 10 - 10t\cdot 2t}{(t^2+3)^2} $$
$$v(t) = \frac{10t^2+30-20t^2}{(t^2+3)^2} = \frac{30-10t^2}{(t^2+3)^2}$$
$$\implies a(t) = \frac{dv}{dt} = \frac{(t^2+3)^2\cdot -20t -(30-10t^2)\cdot 2(t^2+3)\cdot 2t}{(t^2+3)^4}$$
$$a(t) = \frac{-20t(t^2+3)+4t(10t^2-30)}{(t^2+3)^3}$$
$$\implies a(4) = \frac{-80\cdot 19 +16\cdot 130}{19^3} = \frac{560}{6859}$$
