# How do you find the derivative of  (3x^2 + 3x + 4) / sqrt(x)?

Mar 22, 2016

$f ' \left(x\right) = \frac{9 {x}^{2} + 3 x - 4}{2 {x}^{\frac{3}{2}}}$

#### Explanation:

The first to step is to rewrite this so that it looks more like a polynomial. Use numerical exponents:

$f \left(x\right) = \frac{3 {x}^{2} + 3 {x}^{1} + 4}{x} ^ \left(\frac{1}{2}\right)$

Split up the numerator:

$f \left(x\right) = \frac{3 {x}^{2}}{x} ^ \left(\frac{1}{2}\right) + \frac{3 {x}^{1}}{x} ^ \left(\frac{1}{2}\right) + \frac{4}{x} ^ \left(\frac{1}{2}\right)$

Divide the terms:

$f \left(x\right) = 3 {x}^{2 - \frac{1}{2}} + 3 {x}^{1 - \frac{1}{2}} + 4 {x}^{- \frac{1}{2}}$

$f \left(x\right) = 3 {x}^{\frac{3}{2}} + 3 {x}^{\frac{1}{2}} + 4 {x}^{- \frac{1}{2}}$

Now, differentiate each term individually:

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

This gives us a derivative of:

$f ' \left(x\right) = 3 \left(\frac{3}{2}\right) {x}^{\frac{3}{2} - 1} + 3 \left(\frac{1}{2}\right) {x}^{\frac{1}{2} - 1} + 4 \left(- \frac{1}{2}\right) {x}^{- \frac{1}{2} - 1}$

$f ' \left(x\right) = \frac{9}{2} {x}^{\frac{1}{2}} + \frac{3}{2} {x}^{- \frac{1}{2}} - 2 {x}^{- \frac{3}{2}}$

This could be your final answer, but it's easier to work with functions like this when they resemble the original function. To do this, multiply $f ' \left(x\right)$ by $\frac{2 {x}^{\frac{3}{2}}}{2 {x}^{\frac{3}{2}}}$.

$f ' \left(x\right) = \left(\frac{\frac{9}{2} {x}^{\frac{1}{2}} + \frac{3}{2} {x}^{- \frac{1}{2}} - 2 {x}^{- \frac{3}{2}}}{1}\right) \frac{2 {x}^{\frac{3}{2}}}{2 {x}^{\frac{3}{2}}}$

$f ' \left(x\right) = \frac{9 {x}^{2} + 3 x - 4}{2 {x}^{\frac{3}{2}}}$