# What is the derivative of f(x) = (2x - 3)^4 (x^2 + x + 1)^5?

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Jake L. Share
Sep 9, 2016

$f ' \left(x\right) = {\left(2 x - 3\right)}^{3} {\left({x}^{2} + x + 1\right)}^{4} \left(28 {x}^{2} - 12 x - 7\right)$

#### Explanation:

Looking at the equation,
$f \left(x\right) = {\left(2 x - 3\right)}^{4} {\left({x}^{2} + x + 1\right)}^{5}$

we first notice a couple patterns.
1. The function is a product of two terms
2. Each of the terms is a term with an exponent .

Since the function is a product of two terms, we know that we have to use the Product Rule to find the first derivative. The Product Rule states,

For an equation,
$f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$
$f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$

In our case,
$f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$
where,
$g \left(x\right) = {\left(2 x - 3\right)}^{4}$
$h \left(x\right) = {\left({x}^{2} + x + 1\right)}^{5}$

Now we need to calculate g'(x) and h'(x) for the product rule. For this, we need the Chain Rule. It states,
If $F \left(x\right) = f \left(g \left(x\right)\right)$,
$F ' \left(x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

so, in our specific case,
$g \left(x\right) = {\left(2 x - 3\right)}^{4}$
$g ' \left(x\right) = 4 {\left(2 x - 3\right)}^{3} \cdot \left(2\right)$
$h \left(x\right) = {\left({x}^{2} + x + 1\right)}^{5}$
$h ' \left(x\right) = 5 {\left({x}^{2} + x + 1\right)}^{4} \cdot \left(2 x + 1\right)$

Therefore,
$f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$
$= 8 \cdot {\left(2 x - 3\right)}^{3} \cdot {\left({x}^{2} + x + 1\right)}^{5} + {\left(2 x - 3\right)}^{4} \cdot 5 {\left({x}^{2} + x + 1\right)}^{4} \cdot \left(2 x + 1\right)$
$= {\left(2 x - 3\right)}^{3} \left(8 \cdot {\left({x}^{2} + x + 1\right)}^{5} + \left(2 x - 3\right) \cdot 5 {\left({x}^{2} + x + 1\right)}^{4} \cdot \left(2 x + 1\right)\right)$
$= {\left(2 x - 3\right)}^{3} \cdot {\left({x}^{2} + x + 1\right)}^{4} \cdot \left(8 \left({x}^{2} + x + 1\right) + 5 \left(2 x - 3\right) \left(2 x + 1\right)\right)$
$= {\left(2 x - 3\right)}^{3} \cdot {\left({x}^{2} + x + 1\right)}^{4} \cdot \left(8 {x}^{2} + 8 x + 8 + 5 \left(4 {x}^{2} - x - 3\right)\right)$
$= {\left(2 x - 3\right)}^{3} \cdot {\left({x}^{2} + x + 1\right)}^{4} \cdot \left(8 {x}^{2} + 8 x + 8 + 20 {x}^{2} - 5 x - 15\right)$
$= {\left(2 x - 3\right)}^{3} {\left({x}^{2} + x + 1\right)}^{4} \left(28 {x}^{2} - 12 x - 7\right) \square$

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