# How do you differentiate x^2 + 8x + 3 / sqrtx using the quotient rule?

##### 1 Answer
May 17, 2016

$2 x + 8 - \frac{3}{2 {x}^{\frac{3}{2}}}$

#### Explanation:

The quotient rule is not required here as we have a sum of terms which can be differentiated using the $\textcolor{b l u e}{\text{ power rule}}$

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

Before applying this rule we need to rewrite the term$\frac{3}{\sqrt{x}}$
$\Rightarrow \frac{3}{\sqrt{x}} = \frac{3}{x} ^ \left(\frac{1}{2}\right) = 3 {x}^{- \frac{1}{2}}$

We now have: ${x}^{2} + 8 x + 3 {x}^{- \frac{1}{2}}$

$\Rightarrow \frac{d}{\mathrm{dx}} \left({x}^{2} + 8 x + 3 {x}^{- \frac{1}{2}}\right) = 2 x + 8 - \frac{3}{2} {x}^{- \frac{3}{2}}$

$= 2 x + 8 - \frac{3}{2 {x}^{\frac{3}{2}}}$