How do you differentiate y=sqrtx?

Apr 4, 2015

The answer is $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x}}$. This is valid for $x > 0$.

If you know the power rule $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$, then you can derive this by recalling that $\sqrt{x} = {x}^{\frac{1}{2}}$ so that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {x}^{- \frac{1}{2}} = \frac{1}{2 {x}^{\frac{1}{2}}} = \frac{1}{2 \sqrt{x}}$.

If you don't know the power rule, it can be derived from the limit definition of the derivative as follows:

$\frac{d}{\mathrm{dx}} \left({x}^{\frac{1}{2}}\right) = {\lim}_{h \to 0} \frac{\setminus \sqrt{x + h} - \setminus \sqrt{x}}{h}$

$= {\lim}_{h \to 0} \frac{x + h - x}{h \left(\setminus \sqrt{x + h} + \setminus \sqrt{x}\right)} = {\lim}_{h \to 0} \frac{1}{\sqrt{x + h} + \setminus \sqrt{x}} = \setminus \frac{1}{2 \setminus \sqrt{x}}$ when $x > 0$.

When $x = 0$, $f \left(x\right) = \setminus \sqrt{x}$ is defined, but $f ' \left(0\right)$ is undefined. Technically, it's the non-existence of the following limit that confirms this most rigorously:

${\lim}_{h \to {0}^{+}} \frac{\setminus \sqrt{h} - \setminus \sqrt{0}}{h} = {\lim}_{h \to {0}^{+}} \frac{1}{\setminus} \sqrt{h} \setminus m b \otimes \left\{D N E\right\}$.