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I'm currently studying the product rule and have come across a section of questions that seems to make no sense. I'm sure there's just one little thing that I'm missing but I am unable to spot it. Anyhow, I was hoping someone could show me step-by-step how to solve the following, and hopefully I can get the rest:

Differentiate $(x^2 - 1)(x^3 - 1)$. You may need both the chain rule and the product rule

Thanks in advance

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$ \textbf{hint:} $

if you require the derivative with respect to x

then use

$$ \frac{d}{dx}u(x)v(x) = \frac{du}{dx}v + u\frac{dv}{dx} $$ using

$$ u = x^2-1,\\ v = x^3-1. $$

take the derivatives of the functions first and plug in and then simplify.

the result is

$$ 2x\left(x^3-1\right) + \left(x^2-1\right)3x^2 = 5x^4-3x^2-2x $$

taking the result $$ 5x^4-3x^2-2x = x\left(5x^3-3x-2\right) $$ I know that there is a root x = 1 using reminder theorem. therefore I know I can write the equation as

$$ 5x^4-3x^2-2x = x(x-1)P(x) $$

use long division to get P(x)?

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  • $\begingroup$ I have up to that point and i get to 2x*(x^3-1) + (x^2-1).3x^2 but no matter which way i try the simplifying whether multiplying first and then simplyifiying or the reverse i never seem to get the answer $\endgroup$ – Paul Aug 22 '14 at 9:29
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    $\begingroup$ What is the book answer? $\endgroup$ – David Aug 22 '14 at 9:30
  • $\begingroup$ x(x-1)(5x^2 + 5x + 2) $\endgroup$ – Paul Aug 22 '14 at 9:33
  • $\begingroup$ @Paul have you come across algebraic division? $\endgroup$ – Chinny84 Aug 22 '14 at 9:36
  • $\begingroup$ Yes, and i see where you'd add that in, it's incredible how simple something is once you know the answer. thanks for the help $\endgroup$ – Paul Aug 22 '14 at 9:39
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It is easier to go head on with simple problems like these: $$\frac{d}{dx} [(x^2 - 1)(x^3 - 1)] \\ = \frac{d}{dx} (x^5 - x^2 -x^3 + 1)\\ = \frac{d}{dx} (x^5) - \frac{d}{dx} (x^2) -\frac{d}{dx}(x^3) + \frac{d}{dx}(1)\\ = 5x^{5-1} - 2x^{2-1} - 3x^{3-1} + 0\\ = 5x^4 - 3x^2 - 2x\\ = x(5x^3 - 3x -2)\\ = x(x-1)(5x^2 + 5x + 2)$$

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