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

derivative pls

Question

(x^4 + x)^3 (x^2 - 1)^2

derivative pls

mathslover · Answer

http://www.wolframalpha.com/input/?i=derivative+of+%28x%5E4+%2B+x%29%5E3+%28x%5E2+-+1%29%5E2+ That will be too long to explain .. try to go through that

hartnn · Answer

@JayCT u know product rule ?

anonymous · Answer

no not really

hartnn · Answer

$\huge \frac{d}{dx}f(x)g(x)=f(x)g\prime (x)+g(x)f\prime(x)$
note this down.

anonymous · Answer

ok how do i apply it here. can you demonstrate pls

hartnn · Answer

take f(x) as (x^4+x)^3
and g(x) as (x^2-1)^2

anonymous · Answer

(x^4 + x)^3 (x^2 - 1)^2
wow. chain and product rule.
3(x^4 + x)^2(x^2 - 1)^2(4x^3 + 1) + (x^4+x)^3(2)(x^2 - 1)(2x)

anonymous · Answer

francis is that final?

hartnn · Answer

^^ was next step.....did u get it ?

hartnn · Answer

just little simplification

anonymous · Answer

im confused. is francis answer final or just a step

anonymous · Answer

the final step is missing. it is not simplified. i simply took out the fun in solving :D

anonymous · Answer

ok you skipped a step basically? but this is the final answer, 3(x^4 + x)^2(x^2 - 1)^2(4x^3 + 1) + (x^4+x)^3(2)(x^2 - 1)(2x)?

hartnn · Answer

the final answer would be the next step after simplification of above:
$4x(x^2-1)(x^4+x)^3+3(x^2-1)^2(4x^3+1)(x^4+x)^2$

anonymous · Answer

no...i didnt skip a step...thats just it. it looks like i skipped a step because it is long but i didnt. its a long one. product rule and chain rule is always fun :D

anonymous · Answer

why is wolfram saying something different http://www.wolframalpha.com/input/?i=derivative+of+%28x%5E4+%2B+x%29%5E3+%28x%5E2+-+1%29%5E2

hartnn · Answer

same thing just simplified more.