Question about second derivatives of parametric equations?

Question

Consider the parametric curve given by $$x=4+t^2,\, y=−10t^2−10t^3$$

For $\dfrac{dy}{dx}$, I found $-5(2+3t)$

For $\dfrac{d^2y}{dx^2}$, I keep getting $\dfrac{d}{dt}\dfrac{-5(2+3t)}{2t} = \dfrac{-15}{2t}$ It's wrong! Why? The answer is supposed to be $-10\left(\dfrac{3}{4t}\right)$

Answer 1

Note that $$10\frac{3}{4t}=\frac{30}{4t}=\frac{15}{2t}.$$ Thus, your answer agrees with the supposed one.