Given the function:
[tex]y=(-2x^4-3)(-2x^2+1)[/tex][tex]\begin{gathered} \text{let u=-2x}^4-3 \\ \frac{du}{dx}=-8x^3-0=-8x^3 \\ \\ \text{let v=-2x}^2+1 \\ \frac{dv}{dx}=-4x+0=-4x \end{gathered}[/tex]By applying the product rule, we have:
[tex]y^{^{\prime}}=V\frac{du}{dx}+U\frac{dv}{dx}[/tex]Thus, we have:
[tex]\begin{gathered} y^{^{\prime}}=(-2x^2+1)(-8x^3)+(-2x^4-3)(-4x) \\ y^{^{\prime}}=16x^5-8x^3+8x^5+12x \\ y^{^{\prime}}=16x^5+8x^5-8x^3+12x \\ y^{^{\prime}}=24x^5-8x^3+12x \end{gathered}[/tex]