Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

$$u = (1+x^2)^4$$ $$u' = 4 (1+x^2)^3(2x)$$

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


$$((1+x^2)^4)(5(2-x^3)^4(3x^2)) + ((2-x^3)^5 )(4 (1+x^2)^3(2x))$$

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?

share|cite|improve this question

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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.