# find the second derivative $(x^3+x-1)(x^3+1)$

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

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!

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simplify $-x^3+x^3$ in the second line – janmarqz Jan 13 '14 at 5:26
Yes, you should cancel. It is technically correct without cancellation. – André Nicolas Jan 13 '14 at 5:27

## 2 Answers

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

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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...

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