# What is the derivative f'of the function f? f(x)=1+x^(1/2)

Feb 3, 2018

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

#### Explanation:

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

$f ' \left(x\right) = 0 + \frac{1}{2} {x}^{\frac{1}{2} - 1} = {x}^{- \frac{1}{2}} / 2 = \frac{1}{2 {x}^{\frac{1}{2}}} = \frac{1}{2 \sqrt{x}} = \frac{\sqrt{x}}{2 x}$

Feb 3, 2018

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

#### Explanation:

Since the derivative of a sum is the sum of the derivatives we have $f ' \left(x\right) = \left(1\right) ' + \left({x}^{\frac{1}{2}}\right) '$

Since the derivative of a constant is $0$ we can say $f ' \left(x\right) = \left({x}^{\frac{1}{2}}\right) '$

Then by the power rule

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