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

Mar 12, 2018

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

#### Explanation:

We note that derivative of sums and differences can be split apart individually. Meaning,

$\frac{d}{\mathrm{dx}} x$ + $\frac{d}{\mathrm{dx}} \sqrt{x}$

where $\frac{d}{\mathrm{dx}} x = 1$ from the general power rule which states,

$\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$

Continuing on this basis,

$\sqrt{x} = {x}^{\frac{1}{2}}$

Since we follow the same rules we first bring down the $\frac{1}{2}$ and subtract $1$ from $\frac{1}{2}$ leaving,

$\frac{d}{\mathrm{dx}} \sqrt{x} = \frac{1}{2} {x}^{- \frac{1}{2}}$

We further note that we can remove the negative sign of a number by moving it to the opposite of its current location of the form $\frac{n}{d}$ where we now move ${x}^{- \frac{1}{2}}$ to the denominator resulting in $\frac{1}{2 \sqrt{x}}$

Now we simply add the two derivatives together equaling,

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

Mar 12, 2018

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

#### Explanation:

Note the sum rule for derivatives and the power rule:

$\frac{d}{\mathrm{dx}} \left(f \left(x\right) + g \left(x\right)\right) = f ' \left(x\right) + g ' \left(x\right)$

As such, $\frac{d}{\mathrm{dx}} f \left(x\right)$, treating $x$ and $\sqrt{x}$ as their own functions in a sense, is...

$\frac{d}{\mathrm{dx}} \left(x + \sqrt{x}\right)$

$\frac{d}{\mathrm{dx}} x + \frac{d}{\mathrm{dx}} \sqrt{x}$

Using the power rule for the first term...

$1 + \frac{d}{\mathrm{dx}} {x}^{\frac{1}{2}}$

Using the power rule for the second...

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

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

Which can be rewritten as:

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