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?