# How do you use the Product Rule to find the derivative of y = x(x^2 - 2)^(3/2)?

Aug 16, 2015

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

#### Explanation:

Your function can be written as the product of two other functions

$y = {\underbrace{x}}_{\textcolor{\mathmr{and} a n \ge}{f \left(x\right)}} \cdot {\underbrace{{\left({x}^{2} - 2\right)}^{\frac{3}{2}}}}_{\textcolor{red}{g \left(x\right)}}$

which means that you can differentiate it by using the product rule

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \left[\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)\right] \cdot g \left(x\right) + f \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)}$

In your case, the derivative of $y$ will be

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

Now, to find $\frac{d}{\mathrm{dx}} {\left({x}^{2} - 2\right)}^{\frac{3}{2}}$, you can use the chain rule for ${u}^{\frac{3}{2}}$, with $u = \left({x}^{2} - 2\right)$. This will get you

$\frac{d}{\mathrm{dx}} \left({u}^{\frac{3}{2}}\right) = \frac{d}{\mathrm{du}} {u}^{\frac{3}{2}} \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

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

$\frac{d}{\mathrm{dx}} {\left({x}^{2} - 2\right)}^{\frac{3}{2}} = \frac{3}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \cdot {\left({x}^{2} - 2\right)}^{\frac{1}{2}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x$

Plug this back into your target derivative to get

${y}^{'} = 1 \cdot {\left({x}^{2} - 2\right)}^{\frac{3}{2}} + x \cdot \left[3 x \cdot {\left({x}^{2} - 2\right)}^{\frac{1}{2}}\right]$

This is equivalent to

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

${y}^{'} = {\left({x}^{2} - 2\right)}^{\frac{1}{2}} \cdot \left(4 {x}^{2} - 2\right)$

If you want, you can write this as

${y}^{'} = \textcolor{g r e e n}{2 \cdot \left(2 {x}^{2} - 1\right) \cdot {\left({x}^{2} - 2\right)}^{\frac{1}{2}}}$