Take the 2-minute tour ×
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.

$$(x^3+x-1)(x^3+1)$$ $$=x^6+x^4-x^3+x^3+x-1$$ $$f'(x)= 6x^5+4x^3-3x^2+3x^2+1$$

Am I suppose to cancel out $-3x^2+3x^2$?

$$f''(x)= 30x^4+12x^2-6x+6x$$

Can someone check my work please? Thank you so much!

share|improve this question
1  
simplify $-x^3+x^3$ in the second line –  janmarqz Jan 13 at 5:26
1  
Yes, you should cancel. It is technically correct without cancellation. –  André Nicolas Jan 13 at 5:27

2 Answers 2

You can cancel the 6x, but yes, that is correct according to wolfram alpha.

share|improve this answer

You could also work it out this way:

The Product Rule gives us $ \ [ fg ]' \ = \ f' \cdot g \ + \ f \cdot g' $ . Applying it again gives us

$$ [fg]'' \ = \ f'' \cdot g \ + \ 2 \cdot f' \cdot g' \ + \ f \cdot g'' \ . $$

(The "higher-derivative" Product Rule has a resemblance to the Binomial Theorem. [It's more like a relation to the Theorem if you use derivative operators.])

For the functions $ \ f(x) \ = \ x^3 \ + \ x \ - \ 1 \ \ $ and $ \ g(x) \ = \ x^3 \ + \ 1 \ \ , $ we have

$$ f'(x) \ = \ 3x^2 \ + \ 1 \ \ , \ \ f''(x) \ = \ 6x \ \ , \ \ g'(x) \ = \ 3x^2 \ \ , \ \ g''(x) \ = \ 6x \ \ ,$$

making the second derivative of the product $ \ fg \ $

$$ 6x \ \cdot \ (x^3 + 1 ) \ + \ 2 \ \cdot \ (3x^2 + 1) \ \cdot \ 3x^2 \ + \ (x^3 \ + \ x \ - \ 1 ) \ \cdot \ 6x $$

$$ = \ 6x^4 \ + \ 6x \ + \ 18x^4 \ + \ 6x^2 \ + \ 6x^4 \ + \ 6x^2 \ - \ 6x $$

$$ = \ 30x^4 \ + \ 12x^2 \ \ . $$

Granted, it's not much of a saving in calculation effort here, but it can be handy for more complicated functions...

share|improve this answer

Your Answer

 
discard

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.