# Question #84eab

Dec 4, 2014

It is somewhat difficult to read the question as written. Thus, to prevent confusion, it will be listed that the answer provided below will help to find f'(x) if $f \left(x\right) = 2 x \sqrt{x} - 3 \sqrt{x} + \frac{1}{\sqrt{x}}$.

For the sake of simplicity, one will wish to recall that $x = {x}^{1} , \sqrt{x} = {x}^{\frac{1}{2}}$, $\frac{1}{x} = {x}^{- 1}$ and thus that $\frac{1}{\sqrt{x}} = {x}^{- \frac{1}{2}}$. Further, when multiplying together two expressions which have the same base but different exponents, one simply adds the exponents. (For example, ${x}^{1} {x}^{\frac{1}{2}} = {x}^{\left(\frac{1}{2} + 1\right)} = {x}^{\frac{3}{2}}$). Thus...

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

Via use of the power rule (stating that, for $g \left(x\right) = {x}^{n} , g ' \left(x\right) = n {x}^{n - 1}$)...

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

Returning to the initial notation, which utilized $\sqrt{x}$...

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