# How do you use the Product Rule to find the derivative of y = x(x^2 - 2x + 1)^4?

Aug 1, 2015

${y}^{'} = \left(9 x - 1\right) \cdot {\left(x - 1\right)}^{7}$

#### Explanation:

The product rule allows you to differentiate functions that can be written as the product of two other functions

$y = f \left(x\right) \cdot g \left(x\right)$

by using this formula

color(blue)(d/dx(y) = f^'(x) * g(x) + f(x) * g^'(x)

In your case, you can write $y$ as

$y = x \cdot {\left({x}^{2} - 2 x + 1\right)}^{4} = x \cdot {\left[{\left(x - 1\right)}^{2}\right]}^{4} = x \cdot {\left(x - 1\right)}^{8}$

This means that its derivative can be found by using the product rule

${y}^{'} = \left[\frac{d}{\mathrm{dx}} \left(x\right)\right] \cdot {\left(x - 1\right)}^{8} + x \cdot \frac{d}{\mathrm{dx}} {\left(x - 1\right)}^{8}$

${y}^{'} = 1 \cdot {\left(x - 1\right)}^{8} + x \cdot 8 \cdot {\left(x - 1\right)}^{7}$

${y}^{'} = {\left(x - 1\right)}^{7} \left(x - 1 + 8 x\right) = \textcolor{g r e e n}{\left(9 x - 1\right) \cdot {\left(x - 1\right)}^{7}}$