Your presentation confused me at first, but I think you did it correctly. It might help to use two different sub variables, maybe u and v.
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?
Sorry for the confusion. It was difficult to put into the forums. (Much easier on paper)! And it turns out my solution is the correct solution. The program just had a glitch and kept telling me it was wrong. Thanks for the tip and verifying my work though!