# What is the derivative of y=1/2(x^2-x^-2)?

Feb 29, 2016

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

#### Explanation:

$\setminus \frac{d}{\mathrm{dx}} \left(\setminus \frac{1}{2} \left({x}^{2} - {x}^{- 2}\right)\right)$

Taking the constant out as: ${\left(a \setminus \cdot f\right)}^{'} = a \setminus \cdot {f}^{'}$

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

Applying sum/difference rule s: ${\left(f \setminus \pm g\right)}^{'} = {f}^{'} \setminus \pm {g}^{'}$

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

We have,
$\frac{d}{\mathrm{dx}} \left({x}^{2}\right)$ = $2 x$

d/dx((x^(-2)) = $- 2 {x}^{- 3}$ =$- \frac{2}{x} ^ 3$

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

Simplifying it,we get,

$x + \setminus \frac{1}{{x}^{3}}$