How do you differentiate f(x) = x(x^2 - 2)^(3/2)  using the product rule?

Dec 12, 2015

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

Explanation:

According to the product rule:

$f ' \left(x\right) = {\left({x}^{2} - 2\right)}^{\frac{3}{2}} \frac{d}{\mathrm{dx}} \left[x\right] + x \frac{d}{\mathrm{dx}} \left[{\left({x}^{2} - 2\right)}^{\frac{3}{2}}\right]$

Find each derivative separately.

$\frac{d}{\mathrm{dx}} \left[x\right] = 1$

$\frac{d}{\mathrm{dx}} \left[{\left({x}^{2} - 2\right)}^{\frac{3}{2}}\right] = \frac{3}{2} {\left({x}^{2} - 2\right)}^{\frac{1}{2}} \frac{d}{\mathrm{dx}} \left[{x}^{2} - 2\right]$

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

The previous rule required the chain rule.

Plug both derivatives back in.

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

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

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