# How do you find the derivative of y= x/sqrt(x^2+1) ?

Sep 4, 2014

$y ' = \frac{1}{{x}^{2} + 1} ^ \left(\frac{3}{2}\right)$

Solution :

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

Using Quotient Rule, which is

$y = \frac{f}{g}$, then $y ' = \frac{g f ' - f g '}{{g}^{2}}$

similarly following for the given problem, yields

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

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

$y ' = \frac{\left({x}^{2} + 1\right) - {x}^{2}}{{x}^{2} + 1} ^ \left(\frac{3}{2}\right)$

$y ' = \frac{1}{{x}^{2} + 1} ^ \left(\frac{3}{2}\right)$