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

Nov 2, 2016

Derivative of the function, $f \left(x\right) = x + \sqrt{x} ,$ is $1 + \frac{1}{2 \sqrt{x}} .$

#### Explanation:

Let, $y = f \left(x\right) = x + \sqrt{x} .$

$\therefore$Differentiating $y$ w.r.t $x$ is,
$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(x + \sqrt{x}\right) = \frac{d}{\mathrm{dx}} \left(x + {x}^{\frac{1}{2}}\right) .$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \frac{1}{2} \cdot {x}^{\frac{1}{2} - 1} = 1 + \frac{1}{2} {x}^{- \frac{1}{2}} .$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \left(\frac{1}{2}\right) \frac{1}{x} ^ \left(\frac{1}{2}\right) = 1 + \frac{1}{2 \sqrt{x}} .$

Therefore, Derivative of the function, $f \left(x\right) = x + \sqrt{x} ,$ is $1 + \frac{1}{2 \sqrt{x}} .$ (answer).

Nov 2, 2016

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

#### Explanation:

$\textcolor{b l u e}{\text{Given: } f \left(x\right) = x + \sqrt{x}}$

Being of the old school I will use the Leibnitz notation

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Note the general rule of $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$

set$\text{ } y = x + \sqrt{x}$

Then $\text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(x\right) + \frac{d}{\mathrm{dx}} \left(\sqrt{x}\right)$

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$\textcolor{b l u e}{\text{Consider }} \frac{d}{\mathrm{dx}} \left(x\right) \to \frac{d}{\mathrm{dx}} \left({x}^{1}\right) = 1 \times {x}^{0} = 1$

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$\textcolor{b l u e}{\text{Consider }} \frac{d}{\mathrm{dx}} \left(\sqrt{x}\right) \to \frac{d}{\mathrm{dx}} \left({x}^{\frac{1}{2}}\right) = \frac{1}{2} {x}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}$

I do not like roots in the denominator so lets see if we can get rid of it.

Multiply by 1 but in the form of $1 = \frac{\sqrt{x}}{\sqrt{x}}$ giving:

$\frac{1}{2 \sqrt{x}} \times \frac{\sqrt{x}}{\sqrt{x}} = \frac{\sqrt{x}}{2 x}$
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$\textcolor{b l u e}{\text{Putting it all together}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) = 1 + \frac{\sqrt{x}}{2 x} \text{ }$ which is the same as $1 + \frac{1}{2 \sqrt{x}}$