Find the derivative of the function $f(x)=\frac{\sqrt{x^3+1}}{x^2+2}$ and simplify your answer by rationalizing the radical expression.

Step 1 ：Step 1: Apply the quotient rule for derivatives, which states that if we have a function in the form of $f(x)=\frac{g(x)}{h(x)}$, then its derivative is $f^{\prime }(x)=\frac{g^{\prime }(x)h(x)-g(x)h^{\prime }(x)}{[h(x){]}^2}$. In this case, $g(x)=\sqrt{x^3+1}$ and $h(x)=x^2+2$.

Step 2 ：Step 2: Find $g^{\prime }(x)$ and $h^{\prime }(x)$. The derivative of $g(x)$ is $\frac{1}{2}(x^3+1{)}^{-\frac{1}{2}}\cdot 3x^2$, simplifying to $\frac{3x^2}{2\sqrt{x^3+1}}$. The derivative of $h(x)$ is $2x$.

Step 3 ：Step 3: Substitute $g^{\prime }(x)$, $h(x)$, $g(x)$ and $h^{\prime }(x)$ into the quotient rule to obtain $f^{\prime }(x)=\frac{(\frac{3x^2}{2\sqrt{x^3+1}})(x^2+2)-(\sqrt{x^3+1})(2x)}{(x^2+2{)}^2}$.

Step 4 ：Step 4: Simplify this expression to get $f^{\prime }(x)=\frac{3x^4+6x^2-2x\sqrt{x^3+1}}{2(x^2+2{)}^2\sqrt{x^3+1}}$.

Step 5 ：Step 5: To rationalize the expression, multiply the numerator and the denominator by $\sqrt{x^3+1}$, we get $f^{\prime }(x)=\frac{3x^4\sqrt{x^3+1}+6x^2\sqrt{x^3+1}-2x(x^3+1)}{2(x^2+2{)}^2(x^3+1)}$.

Step 6 ：Step 6: Simplify this expression to get $f^{\prime }(x)=\frac{3x^4\sqrt{x^3+1}+6x^2\sqrt{x^3+1}-2x^4-2x}{2(x^2+2{)}^2(x^3+1)}$.