# How do you differentiate y=sqrt(x)/(1+sqrt(x))?

Aug 29, 2016

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

#### Explanation:

$y = \frac{\sqrt{x}}{1 + \sqrt{x}} = \frac{1 + \sqrt{x} - 1}{1 + \sqrt{x}} = \frac{1 + \sqrt{x}}{1 + \sqrt{x}} - \frac{1}{1 + \sqrt{x}}$.

$\therefore y = 1 - \frac{1}{1 + \sqrt{x}}$

Knowing that, $\frac{d}{\mathrm{dt}} \left(\frac{1}{t}\right) = - \frac{1}{t} ^ 2$, and, using Chain Rule, we have,

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

$= 0 - \left\{- \frac{1}{{\left(1 + \sqrt{x}\right)}^{2}}\right\} \cdot \frac{d}{\mathrm{dx}} \left(\left(1 + \sqrt{x}\right)\right)$

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

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