# Question #36b8c

Oct 15, 2014

By multiplying out,

$H \left(x\right) = \left(x - \sqrt{x}\right) \left(x + \sqrt{x}\right) = {x}^{2} - x$

By Power Rule,

$H ' \left(x\right) = 2 x - 1$.

I hope that this was helpful.

Nov 19, 2014

If you notice that $H \left(x\right)$ is the difference of perfect squares then the problem is much easier.

If you do not then you can use the Product Rule .

$H ' \left(x\right) = u v ' + v u '$

$H \left(x\right) = u v = \left(x - \sqrt{x}\right) \left(x + \sqrt{x}\right) = \left(x - {x}^{\frac{1}{2}}\right) \left(x + {x}^{\frac{1}{2}}\right)$

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

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

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

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

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

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

$H ' \left(x\right) = x + x - 1$

$H ' \left(x\right) = 2 x - 1$