# How do you differentiate f(x) =sqrtx (x^2 - 2x + 1)^4  using the product rule?

Jun 6, 2016

The product rule states that $f ' \left(x\right) = \left(g \left(x\right) \times h ' \left(x\right)\right) + \left(g ' \left(x\right) \times h \left(x\right)\right)$, in a function where $f \left(x\right) = g \left(x\right) \times h \left(x\right)$

#### Explanation:

In your function $f \left(x\right)$, let $g \left(x\right) = \sqrt{x} = {x}^{\frac{1}{2}}$

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

Since the application of the product rule uses the derivatives of both functions, we need to differentiate both. $g \left(x\right)$ is relatively straight-forward. By the power rule we have that:

$g ' \left(x\right) = \frac{1}{2} \times {x}^{\frac{1}{2} - 1}$

$g ' \left(x\right) = \frac{1}{2} \times {x}^{- \frac{1}{2}}$

$g ' \left(x\right) = \frac{1}{2 {x}^{\frac{1}{2}}}$

Differentiating $h \left(x\right)$ is a little trickier. We will have to make use of the chain rule.

Let $y = {u}^{4}$ and $u = {x}^{2} - 2 x + 1$

The chain rule states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$, therefore we will have to differentiate both $y$ and $u$ and then find their product.

$y ' = 4 {u}^{4 - 1}$

$y ' = 4 {u}^{3}$

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$u ' = 2 x - 2$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$h ' \left(x\right) = 4 {u}^{3} \left(2 x - 2\right)$

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

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

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Finally, we can do our calculation using the product rule.

$f ' \left(x\right) = \left(g \left(x\right) \times h ' \left(x\right)\right) + \left(g ' \left(x\right) \times h \left(x\right)\right)$

f'(x) = (x^(1/2)((8x - 8)(x^2 - 2x + 1)^3) + (1/(2x^(1/2))(x^2 - 2x + 1)^4)

Simplifying:

$f ' \left(x\right) = \left(\sqrt{x} \left(8 \left(x - 1\right)\right) {\left({x}^{2} - 2 x + 1\right)}^{3}\right) + \frac{{\left({x}^{2} - 2 x + 1\right)}^{4}}{2 \sqrt{x}}$

$f ' \left(x\right) = \sqrt{x} \left(8 x - 8\right) {\left({x}^{2} - 2 x + 1\right)}^{3} + \frac{{\left({x}^{2} - 2 x + 1\right)}^{4}}{2 \sqrt{x}}$

Hopefully this helps!