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

Jul 26, 2017

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

Explanation:

We have:

$g \left(y\right) = \left({x}^{2} - 1\right) {\left({x}^{2} - 2 x + 1\right)}^{4}$

In addition to the product rule we will also require the chain rule. Differentiating wrt $x$ we get:

$g ' \left(y\right) = \left({x}^{2} - 1\right) \left(\frac{d}{\mathrm{dx}} {\left({x}^{2} - 2 x + 1\right)}^{4}\right) + \left(\frac{d}{\mathrm{dx}} \left({x}^{2} - 1\right)\right) {\left({x}^{2} - 2 x + 1\right)}^{4}$

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

$\text{ } = \left({x}^{2} - 1\right) \left(4 {\left({x}^{2} - 2 x + 1\right)}^{3} \left(2 x - 2\right)\right) + \left(2 x\right) {\left({x}^{2} - 2 x + 1\right)}^{4}$

$\text{ } = 2 {\left({x}^{2} - 2 x + 1\right)}^{3} \left(2 \left(2 x - 2\right) \left({x}^{2} - 1\right) + x \left({x}^{2} - 2 x + 1\right)\right)$

$\text{ } = 2 {\left({x}^{2} - 2 x + 1\right)}^{3} \left(2 \left(2 {x}^{3} - 2 x - 2 {x}^{2} + 2\right) + \left({x}^{3} - 2 {x}^{2} + x\right)\right)$

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