# Find the derivative y' using the derivative rules: y=\frac{6x^4 +9}{ 9x^4 +81 }

## Question:

Find the derivative {eq}y' {/eq} using the derivative rules: {eq}y=\frac{6x^4 +9}{ 9x^4 +81 } {/eq}

## Quotient Rule for Differentiation:

You have to find the derivative of the given rational function. You will use quotient rule of differentiation and power rule of differentiation, then simplify it to get the desired result.

Use the quotient rule of differentiation and you have \begin{align*} y &= \frac{{6{x^4} + 9}}{{9{x^4} + 81}}\\ \frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {\frac{{6{x^4} + 9}}{{9{x^4} + 81}}} \right)\\ y' &= \frac{{\left( {9{x^4} + 81} \right) \cdot \frac{d}{{dx}}\left( {6{x^4} + 9} \right) - \frac{d}{{dx}}\left( {9{x^4} + 81} \right) \cdot \left( {6{x^4} + 9} \right)}}{{{{\left( {9{x^4} + 81} \right)}^2}}}\\ y' &= \frac{{\left( {9{x^4} + 81} \right) \cdot \left[ {6 \cdot 4{x^3} + 0} \right] - \left[ {9 \cdot 4{x^3} + 0} \right] \cdot \left( {6{x^4} + 9} \right)}}{{{{\left( {9{x^4} + 81} \right)}^2}}}\\ y' &= \frac{{24{x^3}\left( {9{x^4} + 81} \right) - 36{x^3}\left( {6{x^4} + 9} \right)}}{{{{\left( {9{x^4} + 81} \right)}^2}}}\\ y' &= \frac{{20{x^3}}}{{{{\left( {{x^4} + 9} \right)}^2}}}. \end{align*} 