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

Mar 3, 2016

$\left(x + \sqrt{x}\right) ' = 1 + \frac{1}{2 \sqrt{x}}$

#### Explanation:

Recall the following:

$1$. $\textcolor{red}{\text{Sum Rule}}$: $\left[f \left(x\right) + g \left(x\right)\right] ' = f ' \left(x\right) + g ' \left(x\right)$
$2$. $\textcolor{b l u e}{\text{Power Rule}}$: $\left({x}^{n}\right) ' = \textcolor{\mathmr{and} a n \ge}{n} {x}^{\textcolor{\mathmr{and} a n \ge}{n} - 1}$

Finding the Derivative
$1$. Start by rewriting $\sqrt{x}$ as ${x}^{\frac{1}{2}}$. Recall that $x$ is ${x}^{1}$.

$\left(x + \sqrt{x}\right) '$

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

$2$. Using the $\textcolor{red}{\text{sum rule}}$, determine the derivative of each term in the expression. To do that, use the $\textcolor{b l u e}{\text{power rule}}$.

$= \textcolor{\mathmr{and} a n \ge}{1} {x}^{\textcolor{\mathmr{and} a n \ge}{1} - 1} + \textcolor{\mathmr{and} a n \ge}{\frac{1}{2}} {x}^{\textcolor{\mathmr{and} a n \ge}{\frac{1}{2}} - 1}$

$3$. Simplify.

$= \textcolor{\mathmr{and} a n \ge}{1} {x}^{0} + \textcolor{\mathmr{and} a n \ge}{\frac{1}{2}} {x}^{- \frac{1}{2}}$

$= \textcolor{\mathmr{and} a n \ge}{1} \left(1\right) + \textcolor{\mathmr{and} a n \ge}{\frac{1}{2}} {\left(\frac{1}{x}\right)}^{\frac{1}{2}}$

$= 1 + \textcolor{\mathmr{and} a n \ge}{\frac{1}{2}} \left(\frac{1}{\sqrt{x}}\right)$

$\textcolor{g r e e n}{= 1 + \frac{1}{2 \sqrt{x}}}$

$\therefore$, the derivative is $1 + \frac{1}{2 \sqrt{x}}$.