$\displaystyle

\frac{f(x)}{g(x)} = \frac{\sqrt{x^2+1}cosx}{x} = \frac{(x^2+1)^\frac{1}{2}cosx}{x}

$

$\displaystyle

f ' (x) = \frac{1}{2}(x^2+1)^\frac{-1}{2}cosx - \sqrt{x^2+1}sinx

$

so

$\displaystyle

y ' = \frac{x \frac{1}{2}(x^2+1)^\frac{-1}{2}cosx - \sqrt{x^2+1}sinx - \sqrt{x^2+1}cosx}{x^2}

$

correct? thank you