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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?

2 Answers 2

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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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The problem is in your differentiation of $v= (2-x^3)^5$ You have: $v'= 5(2-x^3)^4(3x^2)$

However, the derivative of $2-x^3$ is $-3x^2$. Thus, $v' = 5(2-x^3)^4(-3x^2)=-15x^2(2-x^3)^4$