Answer
$y'=\dfrac{1}{\sqrt{(x^2+1)^3}}.$
Work Step by Step
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=x; u'(x)=1$
$v(x)=\sqrt{x^2+1}$
$v'(x)$ is found using the Chain Rule:
$u=(x^2+1)$; $\dfrac{du}{dx}=2x$
$\dfrac{d}{du}v(u)=\dfrac{1}{2\sqrt{u}}$
$\dfrac{d}{dx}v(x)=\dfrac{d}{du}v(u)\times\dfrac{du}{dx}=(2x)\dfrac{1}{2\sqrt{x^2+1}}=\dfrac{x}{\sqrt{x^2+1}}.$
$y'=\dfrac{(1)(\sqrt{x^2+1})-(\dfrac{x}{\sqrt{x^2+1}})(x)}{(\sqrt{x^2+1})^2}$
$=\dfrac{(\dfrac{1}{\sqrt{x^2+1}})(x^2+1-x^2)}{x^2+1}$
$=\dfrac{1}{\sqrt{(x^2+1)^3}}.$
