2. Find the derivative of $\displaystyle y = 3 sin^{3} (2x^{4} + 1) $

I used the product rule and this is my working, however I don't end up with the correct answer, can someone please tell me where I am going wrong, thanks.

$\displaystyle u = 3sin^{3} $

$\displaystyle v = 2x^{4} +1 $

$\displaystyle t = sin , y= 3t^{3} $

$\displaystyle \frac{dt}{dx} = cos $

$\displaystyle \frac{dy}{dt} = 9t^{2} $

so now I multiply both together to get the derivative of $\displaystyle 3sin^{3} $

$\displaystyle 9sin^{2} \times cos $

$\displaystyle \frac{dy}{dx} = 8x^{3} $

Now using the product rule and putting this together.

$\displaystyle 9sin^{2} \times cos \times (2x^{4} +1) + 8x^{3} \times 3sin^{3} $

the correct answer is.

$\displaystyle 72x^{3} sin^{2}(2x^{4} +1)cos(2x^{4} +1) $