Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises - Page 192: 76

Answer

\[\frac{dy}{dx}=\frac{f'\left( x \right)g\left( x \right)+f\left( x \right)g'\left( x \right)}{2\sqrt{f\left( x \right)g\left( x \right)}}\]

Work Step by Step

\[\begin{align} & y=\sqrt{f\left( x \right)g\left( x \right)} \\ & \text{Rewrite} \\ & y={{\left[ f\left( x \right)g\left( x \right) \right]}^{1/2}} \\ & \text{Differentiate both sides with respect to }x \\ & \frac{dy}{dx}=\frac{d}{dx}\left[ {{\left( f\left( x \right)g\left( x \right) \right)}^{1/2}} \right] \\ & \frac{dy}{dx}=\frac{1}{2}{{\left[ f\left( x \right)g\left( x \right) \right]}^{-1/2}}\frac{d}{dx}\left[ f\left( x \right)g\left( x \right) \right] \\ & \text{Use the product rule} \\ & \frac{dy}{dx}=\frac{1}{2}{{\left[ f\left( x \right)g\left( x \right) \right]}^{-1/2}}\left( f'\left( x \right)g\left( x \right)+f\left( x \right)g'\left( x \right) \right) \\ & \text{Simplifying} \\ & \frac{dy}{dx}=\frac{f'\left( x \right)g\left( x \right)+f\left( x \right)g'\left( x \right)}{2\sqrt{f\left( x \right)g\left( x \right)}} \\ \end{align}\]
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