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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.

Answer and Explanation:

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


Learn more about this topic:

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When to Use the Quotient Rule for Differentiation

from Math 104: Calculus

Chapter 8 / Lesson 8
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