Find $y^\prime$ if $y=\dfrac{x^{2}+1}{2x-3}$.

Solution The function $y$ is the quotient of $f( x) =x^{2}+1$ and $g( x) =2x-3$. Using the Quotient Rule, we have $$\begin{eqnarray*} y^\prime &=&\dfrac{d}{dx}\!\left(\! \dfrac{x^{2}+1}{2x-3}\!\right) =\dfrac{\left[ \dfrac{d}{dx}( x^{2}+1) \right]\! ( 2x-3) -(x^{2}+1) \left[ \dfrac{d}{dx} \! ( 2x-3) \right] }{( 2x-3) ^{2}} \\[5pt] &=&\dfrac{( 2x) ( 2x-3) -( x^{2}+1) (2) }{( 2x-3) ^{2}}=\dfrac{4x^{2}-6x-2x^{2}-2}{( 2x-3) ^{2}}=\dfrac{2x^{2}-6x-2}{( 2x-3) ^{2}} \end{eqnarray*}$$

provided $x\neq \dfrac{3}{2}$.