Answer
$(x^3+x) \sqrt {x^3+2} (\dfrac{3x^2+1}{x^3+x} +\dfrac{3x^2}{2x^3+4}) $
Work Step by Step
We have: $y=(x^3+x) \sqrt {x^3+2}$
We differentiate both sides with respect to $x$.
$\ln y=\ln (x^3+x) +\dfrac{1}{2}\ln (x^3+2) \\ \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{3x^2+1}{x^3+x} +\dfrac{1}{2}(\dfrac{3x^2}{x^3+2}) \\ \dfrac{dy}{dx}=y(\dfrac{3x^2+1}{x^3+x} +\dfrac{1}{2}(\dfrac{3x^2}{x^3+2}) \\ \dfrac{dy}{dx}=(x^3+x) \sqrt {x^3+2} (\dfrac{3x^2+1}{x^3+x} +\dfrac{3x^2}{2x^3+4}) $
