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# How do you differentiate g(y) =(x^2 - 2x + 1)^4 (4x^6 + 5)  using the product rule?

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#### Explanation

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#### Explanation:

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Ben Share
Jun 13, 2018

Treat the two terms as a normal product rule, but then use the chain rule on the first function when differentiating to get the full solution: $g ' \left(x\right) = 8 {\left(x - 1\right)}^{7} \left(7 {x}^{6} - 3 {x}^{5} + 5\right)$

#### Explanation:

First, let's designate the two pieces of the function as ${f}_{1} \left(x\right)$ and ${f}_{2} \left(x\right)$:

${f}_{1} \left(x\right) = {\left({x}^{2} - 2 x + 1\right)}^{4}$
${f}_{2} \left(x\right) = 4 {x}^{6} + 5$

For a given function, the derivative using the product rule is:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left({f}_{1} \left(x\right) {f}_{2} \left(x\right)\right) = {f}_{1} ' \left(x\right) \cdot {f}_{2} \left(x\right) + {f}_{1} \left(x\right) \cdot {f}_{2} ' \left(x\right)$

This means we'll need to know the derivatives of each $f$ function. The first function requires a chain rule expansion:

$\frac{\mathrm{dy}}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

Deriving ${f}_{1} ' \left(x\right)$:

$\frac{\mathrm{dy}}{\mathrm{dx}} {\left({x}^{2} - 2 x + 1\right)}^{4} = 4 {\left({x}^{2} - 2 x + 1\right)}^{3} \cdot \left(2 x - 2\right)$

${f}_{1} ' \left(x\right) = {\left({x}^{2} - 2 x + 1\right)}^{3} \left(8 x - 8\right)$

Deriving ${f}_{2} ' \left(x\right)$:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(4 {x}^{6} + 5\right) = 24 {x}^{5}$

Now, we reassemble (and simplify):

$g ' \left(x\right) = {\left({x}^{2} - 2 x + 1\right)}^{3} \left(8 x - 8\right) \left(4 {x}^{6} + 5\right) + 24 {x}^{5} {\left({x}^{2} - 2 x + 1\right)}^{4}$

$g ' \left(x\right) = 8 \left({\left({x}^{2} - 2 x + 1\right)}^{3} \left(x - 1\right) \left(4 {x}^{6} + 5\right) + 3 {x}^{5} {\left({x}^{2} - 2 x + 1\right)}^{4}\right)$

Note that the factored form of ${x}^{2} - 2 x + 1$ is ${\left(x - 1\right)}^{2}$

$g ' \left(x\right) = 8 \left({\left({\left(x - 1\right)}^{2}\right)}^{3} \left(x - 1\right) \left(4 {x}^{6} + 5\right) + 3 {x}^{5} {\left({\left(x - 1\right)}^{2}\right)}^{4}\right)$

$g ' \left(x\right) = 8 \left({\left(x - 1\right)}^{6} \left(x - 1\right) \left(4 {x}^{6} + 5\right) + 3 {x}^{5} {\left(x - 1\right)}^{8}\right)$

$g ' \left(x\right) = 8 \left({\left(x - 1\right)}^{7} \left(4 {x}^{6} + 5\right) + 3 {x}^{5} {\left(x - 1\right)}^{8}\right)$

$g ' \left(x\right) = 8 {\left(x - 1\right)}^{7} \left(\left(4 {x}^{6} + 5\right) + 3 {x}^{5} \left(x - 1\right)\right)$

$g ' \left(x\right) = 8 {\left(x - 1\right)}^{7} \left(4 {x}^{6} + 5 + 3 {x}^{6} - 3 {x}^{5}\right)$

color(green)(g'(x)=8(x-1)^7(7x^6-3x^5+5)

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