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: 54

Answer

\[\frac{dy}{dx}=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}\left( 1+\frac{1}{2\sqrt{x+\sqrt{x}}}\left( 1+\frac{1}{2\sqrt{x}} \right) \right)\]

Work Step by Step

\[\begin{align} & y=\sqrt{x+\sqrt{x+\sqrt{x}}} \\ & \text{Rewrite the function} \\ & y={{\left( x+{{\left( x+{{x}^{1/2}} \right)}^{1/2}} \right)}^{1/2}} \\ & \frac{dy}{dx}=\frac{d}{dx}\left[ {{\left( x+{{\left( x+{{x}^{1/2}} \right)}^{1/2}} \right)}^{1/2}} \right] \\ & \text{Differentiate by using the chain rule, then} \\ & \frac{dy}{dx}=\frac{1}{2}{{\left( x+{{\left( x+{{x}^{1/2}} \right)}^{1/2}} \right)}^{1/2-1}}\frac{d}{dx}\left[ x+{{\left( x+{{x}^{1/2}} \right)}^{1/2}} \right] \\ & \frac{dy}{dx}=\frac{1}{2}{{\left( x+{{\left( x+{{x}^{1/2}} \right)}^{1/2}} \right)}^{-1/2}}\left( 1+\frac{1}{2}{{\left( x+{{x}^{1/2}} \right)}^{-1/2}}\frac{d}{dx}\left[ x+{{x}^{1/2}} \right] \right) \\ & \text{Compute the derivative} \\ & \frac{dy}{dx}=\frac{1}{2}{{\left( x+{{\left( x+{{x}^{1/2}} \right)}^{1/2}} \right)}^{-1/2}}\left( 1+\frac{1}{2}{{\left( x+{{x}^{1/2}} \right)}^{-1/2}}\left( 1+\frac{1}{2\sqrt{x}} \right) \right) \\ & \text{Simplifying} \\ & \frac{dy}{dx}=\frac{1}{2{{\left( x+{{\left( x+{{x}^{1/2}} \right)}^{1/2}} \right)}^{1/2}}}\left( 1+\frac{1}{2{{\left( x+{{x}^{1/2}} \right)}^{1/2}}}\left( 1+\frac{1}{2\sqrt{x}} \right) \right) \\ & \text{Write the fractional exponents as radicals} \\ & \frac{dy}{dx}=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}\left( 1+\frac{1}{2\sqrt{x+\sqrt{x}}}\left( 1+\frac{1}{2\sqrt{x}} \right) \right) \\ \end{align}\]
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