# How do you find the derivative of x+sqrt(x)?

##### 1 Answer
Mar 21, 2018

$\textcolor{b l u e}{1 + \frac{1}{2} {x}^{- \frac{1}{2}}}$ or $\textcolor{b l u e}{1 + \frac{1}{2 \sqrt{x}}}$

#### Explanation:

$f \left(x\right) = x + \sqrt{x}$

Rewriting as:

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

We can differentiate this using the power rule:

The Power Rules states that:

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

$\frac{\mathrm{dy}}{\mathrm{dx}}$ is distributive over the sum, so:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(a {x}^{2} + b x + c\right) = \frac{\mathrm{dy}}{\mathrm{dx}} \left(a {x}^{2}\right) + \frac{\mathrm{dy}}{\mathrm{dx}} \left(b x\right) + \frac{\mathrm{dy}}{\mathrm{dx}} \left(c\right)$

Differentiating $f \left(x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(x + {x}^{\frac{1}{2}}\right) = 1 \cdot {x}^{1 - 1} + \frac{1}{2} \cdot {x}^{\frac{1}{2} - 1}$

$= {x}^{0} + \frac{1}{2} {x}^{- \frac{1}{2}} = \textcolor{b l u e}{1 + \frac{1}{2} {x}^{- \frac{1}{2}}}$ or $\textcolor{b l u e}{1 + \frac{1}{2 \sqrt{x}}}$