Derivative of $ y = (2x - 3)^4 \cdot (x^2 + x + 1)^5$ $$ y = (2x - 3)^4 \cdot (x^2 + x + 1)^5$$
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 $(2x -3)^3 (x^2 + x + 1)^4 (28x^2 - 12x - 7)$
 A: 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:
$$ = (2x - 3)^3(x^2 + x + 1)^4\left(8x^2 + 8x + 8 + 5(4x^2 - 4x - 3)\right)$$
$$\bf = (2x - 3)^3(x^2 + x + 1)^4(28x^2 - 12x - 7)$$
A: 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...)
A: We have $$8(2x-3)^3 * (x^2 + x + 1)^5 + 5(x^2 + x + 1)^4 (2x+1) (2x-3)^4\\=
(2x-3)^3(x^2+x+1)^4(8(x^2+x+1)+5(2x+1)(2x-3))
\\=(2x-3)^3(x^2+x+1)^4(8x^2+8x+8+20x^2-20x-15)\\=
(2x-3)^3(x^2+x+1)^4(28x^2-12x-7)$$  
