# Chapter 4 - Calculating the Derivative - 4.2 Derivatives of Products and Quotients - 4.2 Exercises - Page 216: 27

$f{\,^,}\left( x \right) = \frac{{60{x^3} + 57{x^2} - 24x + 13}}{{{{\left( {5x + 4} \right)}^2}}}$

#### Work Step by Step

$\begin{gathered} f\,\left( x \right) = \frac{{\,\left( {3{x^2} + 1} \right)\,\left( {2x - 1} \right)}}{{5x + 4}} \hfill \\ Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,the\,\,derivative \hfill \\ f{\,^,}\left( x \right) = \frac{{\,\left( {5x + 4} \right)\,\,{{\left[ {\,\left( {3{x^2} + 1} \right)\,\left( {2x - 1} \right)} \right]}^,} - \,\left( {3{x^2} + 1} \right)\,\left( {2x - 1} \right)\,{{\left( {5x + 4} \right)}^,}}}{{\,{{\left( {5x + 4} \right)}^2}}} \hfill \\ Use\,\,the\,\,product\,\,rule\,\,to\,\,find\,\,{\left[ {\,\left( {3{x^2} + 1} \right)\,\left( {2x - 1} \right)} \right]^,} \hfill \\ f{\,^,}\left( x \right) = \frac{{\left( {5x + 4} \right)\left[ {\,\left( {3{x^2} + 1} \right)\,\,\left( 2 \right)\left( {2x - 1} \right)\,\left( {6x} \right)} \right] - \,\left( {3{x^2} + 1} \right)\,\left( {2x - 1} \right)\,\left( 5 \right)\,\,}}{{\left( {5x + 4} \right){\,^2}\,}} \hfill \\ Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\ f{\,^,}\left( x \right) = \frac{{\left( {5x + 4} \right)\,\left( {6{x^2} + 2 + 12{x^2} - 6x} \right) - \,\left( {3{x^2} + 1} \right)\,\left( {10x - 5} \right)}}{{{{\left( {5x + 4} \right)}^2}}} \hfill \\ f{\,^,}\left( x \right) = \frac{{\,\left( {5x + 4} \right)\,\left( {18{x^2} - 6x + 2} \right) - \,\left( {30{x^3} - 15{x^2} + 10x - 5} \right)}}{{{{\left( {5x + 4} \right)}^2}}} \hfill \\ f{\,^,}\left( x \right) = \frac{{90{x^3} - 30{x^2} + 10x + 72{x^2} - 24x + 8 - 30{x^3} + 15{x^2} - 10x + 5}}{{{{\left( {5x + 4} \right)}^2}}} \hfill \\ f{\,^,}\left( x \right) = \frac{{60{x^3} + 57{x^2} - 24x + 13}}{{{{\left( {5x + 4} \right)}^2}}} \hfill \\ \end{gathered}$

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