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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?

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2 Answers 2

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, $$v' = 5(2-x^3)^4(-3x^2)=-15x^2(2-x^3)^4$$

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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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