# Find the derivative of the function. (2x-5)^4(x^2+x+1)^5 ?

## I don't really understand whats going on in the third line of the solution to this problem. Can someone please give me a step by step explanation of the steps I need to do starting from the third line down?

Feb 19, 2018

$\text{Please see explanation below.}$

#### Explanation:

$\text{Yes, this is a simplification technique that may be described as}$
$\text{factoring out the" \ lowest \ "powers of common exponentiated}$
$\text{factors. It is very powerful, and makes some simplifications a}$
$\text{lot easier and faster, and can get you to the end result very}$
$\text{directly.}$

$\text{[It works for any exponents -- positive, negative, fractional, any.]}$

$\text{In the case you have, at the line prior to line 3, you had:}$

$\setminus q \quad f ' \left(x\right) \setminus = \setminus {\left(2 x - 5\right)}^{4} \setminus \cdot 5 {\left({x}^{2} + x + 1\right)}^{4} \left(2 x + 1\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad + {\left({x}^{2} + x + 1\right)}^{5} \setminus \cdot 4 {\left(2 x - 5\right)}^{3} \cdot 2.$

$\text{Maybe to illustrate the technique more easily, let me name}$
$\text{the common exponentiated factors here:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad A = 2 x - 5 , \setminus q \quad B = {x}^{2} + x + 1.$

$\text{Now we can rewrite" \ \ f'(x) \ \ "as:}$

$\setminus q \quad f ' \left(x\right) \setminus = \setminus {\left(2 x - 5\right)}^{4} \setminus \cdot 5 {\left({x}^{2} + x + 1\right)}^{4} \left(2 x + 1\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad + {\left({x}^{2} + x + 1\right)}^{5} \setminus \cdot 4 {\left(2 x - 5\right)}^{3} \cdot 2$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus {A}^{4} \setminus \cdot 5 {B}^{4} \left(2 x + 1\right) + {B}^{5} \setminus \cdot 4 {A}^{3} \cdot 2$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus {A}^{4} {B}^{4} \setminus \cdot 5 \left(2 x + 1\right) + {A}^{3} {B}^{5} \setminus \cdot 4 \cdot 2 .$

$\text{Now factor out the" \ lowest \ "powers of the quantites" \ A, B ":}$

$\setminus q \quad f ' \left(x\right) \setminus = \setminus {A}^{4} {B}^{4} \setminus \cdot 5 \left(2 x + 1\right) + {A}^{3} {B}^{5} \setminus \cdot 4 \cdot 2 .$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus {A}^{3} {B}^{4} \left[A \setminus \cdot 5 \left(2 x + 1\right) + B \setminus \cdot 4 \cdot 2\right] .$

$\text{Now let's return" \ \ A, B \ \ "to their original values, as we assigned}$
$\text{them above:}$

$\setminus q \quad f ' \left(x\right) \setminus = \setminus {A}^{4} {B}^{4} \setminus \cdot 5 \left(2 x + 1\right) + {A}^{3} {B}^{5} \setminus \cdot 4 \cdot 2$

 :. \qquad f'(x) \ = \ ( 2 x - 5 )^3 ( x^2 + x + 1 )^4 [ ( 2 x - 5 ) \cdot 5 ( 2 x + 1 )

 \qquad \qquad \qquad \qquad \qquad \qquad \qquad + ( x^2 + x + 1 ) \cdot 4 cdot 2 ] .

$\text{(Sorry for the line breaks, system can go no farther !!)}$

$\text{This is how you go from line (2) to line (3), in the example you}$
$\text{provided. I think this is what you wanted explained. But just}$
$\text{to be sure, let me continue, and finish the problem as you gave}$
$\text{it.}$

$\text{So, continuing:}$

 :. \qquad f'(x) \ = \ ( 2 x - 5 )^3 ( x^2 + x + 1 )^4 [ ( 2 x - 5 ) \cdot 5 ( 2 x + 1 )

 \qquad \qquad \qquad \qquad \qquad \qquad \qquad + ( x^2 + x + 1 ) \cdot 4 cdot 2 ] .

 \qquad \qquad \qquad \qquad \qquad = \ ( 2 x - 5 )^3 ( x^2 + x + 1 )^4 [ 5 ( 2 x - 5 ) ( 2 x + 1 )

 \qquad \qquad \qquad \qquad \qquad \qquad \qquad + 8 ( x^2 + x + 1 ) ]

$\setminus = \setminus {\left(2 x - 5\right)}^{3} {\left({x}^{2} + x + 1\right)}^{4} \left[5 \left(4 {x}^{2} - 8 x - 5\right) + 8 {x}^{2} + 8 x + 8\right]$

$\setminus = \setminus {\left(2 x - 5\right)}^{3} {\left({x}^{2} + x + 1\right)}^{4} \left[20 {x}^{2} - 40 x - 25 + 8 {x}^{2} + 8 x + 8\right]$

 \ = \ ( 2 x - 5 )^3 ( x^2 + x + 1 )^4 ( 28 x^2 - 32 x - 17 );

$\text{which is as your example, as given, ended.}$

$\text{I hope this helps !!}$