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

May 18, 2017

$1 + \frac{\sqrt{x}}{2 x}$

#### Explanation:

Chain Rule is not needed here. Just use thee Power Rule.

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

May 18, 2017

Use algebra and the power rule.

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

#### Explanation:

When $f \left(x\right) = {x}^{n}$, the power rule states that the derivative follows the trend $f ' \left(x\right) = n {x}^{n - 1}$.

Applied to your specific problem...

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

(Use your knowledge of algebra to rewrite the root as an exponent...)

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

(Now just take the derivative by using the power rule...)

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