<s>[INST] Using product and chain rule to find derivative. 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?
 [/INST] Everything is correct in your answer, except for the chain rule for $v$.
The derivative of $2-x^3$ is $-3x^2$.
So $v'=5(2-x^3)^4(-3x^2)$ and this is why the $15x^2$ becomes negative.
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