Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - Chapter Review - Review Exercises - Page 244: 26

Answer

$$\frac{{dy}}{{dx}} = 4x{\left( {3x - 2} \right)^4}\left( {21x - 4} \right)$$

Work Step by Step

$$\eqalign{ & y = 4{x^2}{\left( {3x - 2} \right)^5} \cr & {\text{differentiate both sides}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {4{x^2}{{\left( {3x - 2} \right)}^5}} \right] \cr & {\text{use product rule}} \cr & \frac{{dy}}{{dx}} = 4{x^2}\frac{d}{{dx}}\left[ {{{\left( {3x - 2} \right)}^5}} \right] + {\left( {3x - 2} \right)^5}\frac{d}{{dx}}\left[ {4{x^2}} \right] \cr & {\text{use the power rule with the chain rule }}\frac{d}{{dx}}\left[ {g{{\left( x \right)}^n}} \right] = ng{\left( x \right)^{n - 1}}\frac{d}{{dx}}\left[ {g'\left( x \right)} \right]{\text{ then}} \cr & \frac{{dy}}{{dx}} = 4{x^2}\left( 5 \right){\left( {3x - 2} \right)^4}\frac{d}{{dx}}\left[ {3x - 2} \right] + {\left( {3x - 2} \right)^5}\frac{d}{{dx}}\left[ {4{x^2}} \right] \cr & {\text{find derivatives}} \cr & \frac{{dy}}{{dx}} = 4{x^2}\left( 5 \right){\left( {3x - 2} \right)^4}\left( 3 \right) + {\left( {3x - 2} \right)^5}\left( {8x} \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = 60{x^2}{\left( {3x - 2} \right)^4} + 8x{\left( {3x - 2} \right)^5} \cr & {\text{factoring}} \cr & \frac{{dy}}{{dx}} = 4x{\left( {3x - 2} \right)^4}\left[ {15x + 2\left( {3x - 2} \right)} \right] \cr & \frac{{dy}}{{dx}} = 4x{\left( {3x - 2} \right)^4}\left( {15x + 6x - 4} \right) \cr & \frac{{dy}}{{dx}} = 4x{\left( {3x - 2} \right)^4}\left( {21x - 4} \right) \cr} $$
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