## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 4 - Calculating the Derivative - 4.2 Derivatives of Products and Quotients - 4.2 Exercises - Page 216: 34

#### Answer

$y = 11x - 6$

#### Work Step by Step

$\begin{gathered} f\,\left( x \right) = \,\left( {2x - 1} \right)\,\left( {x + 4} \right) \hfill \\ Use\,\,the\,\,product\,\,rule\,\,to\,\,find\,\,{f^,}\,\left( x \right) \hfill \\ {f^,}\,\left( x \right) = \,\left( {2x - 1} \right)\,{\left( {x + 4} \right)^,} + \,\left( {x + 4} \right)\,{\left( {2x - 1} \right)^,} \hfill \\ Then \hfill \\ {f^,}\,\left( x \right) = \,\left( {2x - 1} \right)\,\left( 1 \right) + \,\left( {x + 4} \right)\,\left( 2 \right) \hfill \\ {f^,}\,\left( x \right) = 2x - 1 + 2x + 8 \hfill \\ {f^,}\,\left( x \right) = 4x + 7 \hfill \\ Evaluate\,\,{f^,}\,\left( x \right)\,\,at\,\,x = 1 \hfill \\ {f^,}\,\left( 1 \right) = 4\,\left( 1 \right) + 7 \hfill \\ {f^,}\,\left( 1 \right) = 11 \hfill \\ Then \hfill \\ m = 11 \hfill \\ Use\,\,y - {y_1} = m\,\left( {x - {x_1}} \right) \hfill \\ y - 5 = 11\,\left( {x - 1} \right) \hfill \\ Simplifying \hfill \\ y - 5 = 11x - 11 \hfill \\ y = 11x - 6 \hfill \\ \hfill \\ \end{gathered}$

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