What is the derivative of sqrt(x - 1)/sqrtx?

Jan 27, 2016

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

Explanation:

To find the derivative of $\frac{\sqrt{x - 1}}{\sqrt{x}}$ we can try the following
$\frac{\sqrt{x - 1}}{\sqrt{x}} = \sqrt{\frac{x - 1}{x}}$
on simplifying further

$= \sqrt{1 - \frac{1}{x}}$

Let us use chain rule.

Let $y = \sqrt{u}$ and $u = 1 - \frac{1}{x}$

By chain rule

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

$y = \sqrt{u}$
$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2 \sqrt{u}}$
Substituting back for $u$ we get
$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2 \sqrt{1 - \frac{1}{x}}}$

Now we shall find our $\frac{\mathrm{du}}{\mathrm{dx}}$

$u = 1 - \frac{1}{x}$
$\frac{\mathrm{du}}{\mathrm{dx}} = 0 - \left(- \frac{1}{x} ^ 2\right)$
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{1 - \frac{1}{x}}} \times \frac{1}{x} ^ 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{\frac{x - 1}{x}}} \times \frac{1}{x} ^ 2$

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