# Question #d0701

When dealing with derivatives, $n$-roots ($n \in m a t h \boldsymbol{N}$) are better treated as rational powers: ${x}^{\frac{1}{n}}$.
In case of square roots:

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

To compute the derivative of an $n$-root, we can apply the derivation formula for real powers (see the proof at the end of this post):

$\frac{d}{\mathrm{dx}} \left({x}^{\alpha}\right) = \alpha {x}^{\alpha - 1}$, where $\alpha \in m a t h \boldsymbol{R}$, $\forall x > 0$

Note that in some cases, the validity of this formula can be extendend to $x \ne 0$ or to $x \in m a t h \boldsymbol{R}$.
In case of square roots $\alpha = \frac{1}{2}$ the domain is $x > 0$. We get:

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

In our specific case, $f \left(x\right) = 2 \sqrt{x}$. We get:

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

To prove that $\frac{d}{\mathrm{dx}} {x}^{\alpha} = \alpha {x}^{\alpha - 1}$ for $\alpha \in m a t h \boldsymbol{R}$ and $x > 0$, we can use implicit differentiation and logarithms.

Let $g \left(x\right) = {x}^{\alpha}$. Note that $x > 0 \implies {x}^{\alpha} > 0 \implies g \left(x\right) > 0$, so we can write the equality using logarithms and rewrite it using the "logarithm of a power" property:

$\ln g \left(x\right) = \ln {x}^{\alpha} \iff \ln g \left(x\right) = \alpha \ln x$

Now, we differentiate on both sides (implicit differentiation):

$\frac{d}{\mathrm{dx}} \left(\ln g \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(\alpha \ln x\right) \iff \frac{\frac{d}{\mathrm{dx}} g \left(x\right)}{g \left(x\right)} = \alpha \frac{1}{x}$

So we get the following result (recall that $g \left(x\right) = {x}^{\alpha}$):

$\frac{d}{\mathrm{dx}} g \left(x\right) = \frac{\alpha}{x} g \left(x\right) = \frac{\alpha}{x} {x}^{\alpha} = \alpha {x}^{- 1} {x}^{\alpha} = \alpha {x}^{\alpha - 1}$

That's the result we were looking for.