Find the derivative of $$y =(1+x^2)^4 (2-x^3)^5$$ To solve this I used the product rule and the chain rule.
$$u = (1+x^2)^4$$ $$u' = 4 (1+x^2)^3(2x)$$
$$v= (2-x^3)^5$$ $$v' = 5(2-x^3)^4(3x^2)$$
$$uv'+vu'$$
$$((1+x^2)^4)(5(2-x^3)^4(3x^2)) + ((2-x^3)^5 )(4 (1+x^2)^3(2x))$$
The answer I got is: $$(15x^2)(1-x^2)^4(2-x^3)^4 + 8x(2-x^3)^5(1+x^2)^3$$.
Why is the answer $$8x(x^2 +1)^3(2-x^3)^5-15x^2(x^2)(X^2+1)^4(2-x^3)4$$? How did the $15x^2$ become negative?