Given that $f^{\prime }(x)=\frac{x}{x^2+1}$ and $g(x)=\sqrt{3x-1},$ find $F^{\prime }(x)$ if $F(x)=f(g(x))$

Key Concepts: Chain Rule, Composite Functions Explanation: To find $F^{\prime }(x),$ we will use the chain rule, which states that if $F(x)=f(g(x)),$ then $F^{\prime }(x)=f^{\prime }(g(x))\cdot g^{\prime }(x).$ Step by Step Solution: Step 1. First, we need to find $g^{\prime }(x).$ Using the power rule, we have $g^{\prime }(x)=\frac{d}{dx}(\sqrt{3x-1})=\frac{1}{2\sqrt{3x-1}}.$ Step 2. Next, we need to find $f^{\prime }(g(x)).$ We can substitute $g(x)$ for $x$ in the given expression for $f^{\prime }(x):$ $f^{\prime }(g(x))=\frac{g(x)}{g(x{)}^2+1}.$ Substituting the expression for $g(x),$ we get $f^{\prime }(g(x))=\frac{\sqrt{3x-1}}{(3x-1)+1}=\frac{\sqrt{3x-1}}{3x}.$ Step 3. Now, we can apply the chain rule to find $F^{\prime }(x).$ $\begin{array}{rl}{F}^{\prime }(x)& =f^{\prime }(g(x))\cdot g^{\prime }(x)\\ & =\frac{\sqrt{3x-1}}{3x}\cdot \frac{1}{2\sqrt{3x-1}}\\ & =\frac{1}{2x}\end{array}$ Final Answer: $F^{\prime }(x)=\frac{1}{2x}$