# How do you differentiate g(y) =(x^2 + 6) (x^2 - 2)^(3/2  using the product rule?

Apr 9, 2017

#### Answer:

$\left(2 x\right) {\left({x}^{2} - 2\right)}^{\frac{1}{2}} \left[\left({x}^{2} - 2\right) + \frac{3}{2} \left({x}^{2} + 6\right) \left(2 x\right)\right]$

#### Explanation:

For product rule if it is in the form where f(x)g(x) like above it becomes: f'(x)(g(x) + f(x)g'(x)

For chain rule you take the derivative normally by bringing the exponent to the front and subtracting the exponent by 1. However the additional step requires you to find the derivative of the inside and multiply.

$\left(2 x\right) {\left({x}^{2} - 2\right)}^{\frac{3}{2}} + \left({x}^{2} + 6\right) \frac{3}{2} {\left({x}^{2} - 2\right)}^{\frac{1}{2}} \left(2 x\right)$

$\left(2 x\right) {\left({x}^{2} - 2\right)}^{\frac{3}{2}} + \frac{3}{2} \left({x}^{2} + 6\right) {\left({x}^{2} - 2\right)}^{\frac{1}{2}} \left(2 x\right)$

Simplify this to:
$\left(2 x\right) {\left({x}^{2} - 2\right)}^{\frac{1}{2}} \left[\left({x}^{2} - 2\right) + \frac{3}{2} \left({x}^{2} + 6\right) \left(2 x\right)\right]$