I can't seem to figure out the following derivative:

Find D_{x}y

y = (3x-3)^{2}(2-x^{5})^{3}

Here is the work I have done so far..

Let u = 3x-3

Let y = u^{2}

= (3)(2u) = (6(3x-3))

Let u = 2-x^{5}

Let y = u^{3}

= (-5x^{4})(3u^{2}) = -15x^{4}(2-x^{5})^{2}

Now I used the Product Rule:

(6(3x-3))(2-x^{5})^{3}+ (-15x^{4}(2-x^{5})^{2})(3x-3)^{2}

Here is the Factoring/Simplifying:

3 (3x-3) (2-x^{5})^{2}[2 (2-x^{5}) - 5x^{4 }(3x-3)]

3 (3x-3) (2-x^{5})^{2}[4 - 2x^{5}- 15x^{5}+ 15x^{4}]

3 (3x-3) (2-x^{5})^{2}[-17x^{5}+ 15x^{4}+ 4]

The bolded portion above is the answer I am getting. Anyone see anything I did wrong?