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$$ y = (2x - 3)^4 \cdot (x^2 + x + 1)^5$$

I know that it should be the chain rule and product rule used together to get the answer

$$ y = \frac{dx}{dy}((2x - 3)^4) \cdot (x^2 + x + 1)^5 + \frac{dx}{dy}(x^2 + x + 1)^5 \cdot (2x - 3)^4 $$

this gives me something ridiculous like this

$$8(2x-3)^3 \cdot (x^2 + x + 1)^5 + (x^2 + x + 1)^4 \cdot (2x+1) (2x-3)^4$$

This is wrong and I keep getting it, I don't know how to simplify it without expanding everything.

The book Houdini's out $(2x -3)^3 (x^2 + x + 1)^4 (28x^2 - 12x - 7)$

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  • $\begingroup$ Let me reassure you: It's not wrong (but it can be simplified using algebra). What makes you think it is? $\endgroup$
    – Lord_Farin
    Jun 18, 2013 at 20:45
  • $\begingroup$ I just assumed it was wrong since the book got something totally different and I can't convert my answer into book answer. $\endgroup$
    – user138246
    Jun 18, 2013 at 20:47
  • $\begingroup$ Please edit your question to include the answer your book gives. This will help identifying and resolving your problems. $\endgroup$
    – Lord_Farin
    Jun 18, 2013 at 20:48
  • $\begingroup$ When you write $y = \frac{dx}{dy}((2x - 3)^4) * (x^2 + x + 1)^5 + \frac{dx}{dy}(x^2 + x + 1)^5 * (2x - 3)^4$ this s not correct. The left side shoulod be $\frac {dy}{dx}$ and the $\frac {dx}{dy}$'s on the right should be $\frac d{dx}$'s $\endgroup$ Jun 18, 2013 at 20:55
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    $\begingroup$ You have taken the derivative of both sides with respect to $x$. The left side is no longer $y$, it is $\frac {dy}{dx}.$ On the right side, where you write $\frac {dx}{dy}$ you want the derivative operator, so the first term becomes $(x^2+x+1)^5\frac d{dx}((2x-3)^4)$. Moving the other term in front of the $\frac d{dx}$ makes it clear what should have its derivative taken. This step has applied the product rule and you are ready for the chain rule. The third line is missing a factor $5$ in the second term from the power of $(x^2+x+1)$ $\endgroup$ Jun 18, 2013 at 21:18

3 Answers 3

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You're derivative is close: $$\frac d{dx}\left(x^2 + x + 4)^5\right) = \color{blue}{\bf 5}(x^2 + x + + 4)^4\cdot (2x + 1)$$ This gives us:

$$f'(x) = 8(2x-3)^3 \cdot (x^2 + x + 1)^5 + \color{blue}{\bf 5}(x^2 + x + 1)^4 (2x+1) (2x-3)^4$$

Then we can factor out common factors of each term of the sum:

$$ = (2x - 3)^3(x^2 + x + 1)^4\left(8(x^2 + x + 1) + 5(2x+1)(2x - 3)\right)$$ And then expand the factors where needed, and combine like terms in the right-most factor: $$ = (2x - 3)^3(x^2 + x + 1)^4\left(8x^2 + 8x + 8 + 5(4x^2 - 4x - 3)\right)$$

$$\bf = (2x - 3)^3(x^2 + x + 1)^4(28x^2 - 12x - 7)$$

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  • $\begingroup$ Is it? Shouldn't there be a factor of 5 in your second term? $\endgroup$ Jun 18, 2013 at 21:04
  • $\begingroup$ @Amzoti (could be...it's nice to use, short of having a black or white board to work with in front of both me and the OP!) $\endgroup$
    – amWhy
    Jun 19, 2013 at 1:58
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We have $$8(2x-3)^3 * (x^2 + x + 1)^5 + 5(x^2 + x + 1)^4 (2x+1) (2x-3)^4\\= (2x-3)^3(x^2+x+1)^4(8(x^2+x+1)+5(2x+1)(2x-3)) \\=(2x-3)^3(x^2+x+1)^4(8x^2+8x+8+20x^2-20x-15)\\= (2x-3)^3(x^2+x+1)^4(28x^2-12x-7)$$

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  • $\begingroup$ OP may have mistyped that term in the post. Also, there is a factor of 5 from Chain Rule omitted in the second set of factors. $\endgroup$ Jun 18, 2013 at 21:03
  • $\begingroup$ I fixed it, the 28x is now 28x^2 which was in the book but I couldn't see. $\endgroup$
    – user138246
    Jun 18, 2013 at 21:03
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Your result is not quite correct, and also not the form that it is conventionally left in. You can factor out $ \ (2x+3)^3 \ $ and $ \ (x^2+x+1)^4 \ $ from both terms, and then consolidate the remaining factors in both terms algebraically; you will have to simplify $ \ 8 \cdot (x^2 + x + 1) \ + \ 5 \cdot (2x + 1) \cdot (2x-3) \ $ .

(Currently, you have an error in your use of the Chain Rule, and a typo in the book's answer...)

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  • $\begingroup$ There is no error in the book's answer. $\endgroup$
    – user138246
    Jun 18, 2013 at 20:58
  • $\begingroup$ The first term in the third parentheses should be $ \ 28x^2 \ . $ $\endgroup$ Jun 18, 2013 at 21:00

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