Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 42

Answer

$y'=\dfrac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}(1+\dfrac{1}{2\sqrt{x+\sqrt{x}}})(1+\dfrac{1}{2\sqrt{x}})$

Work Step by Step

$y=\sqrt{x+\sqrt{x+\sqrt{x}}}$ Let's rewrite the function like this: $y=[x+(x+x^{1/2})^{1/2}]^{1/2}$ Differentiate using the chain rule: $y'=\dfrac{1}{2}[x+(x+x^{1/2})^{1/2}]^{-1/2}[x+(x+x^{1/2})^{1/2}]'=...$ $...=\dfrac{1}{2}[x+(x+x^{1/2})^{1/2}]^{-1/2}[1+[(x+x^{1/2})^{1/2}]']=...$ Apply the chain rule one more time to find $[(x+x^{1/2})^{1/2}]'$: $...=\dfrac{1}{2}[x+(x+x^{1/2})^{1/2}]^{-1/2}[1+[\dfrac{1}{2}(x+x^{1/2})^{-1/2}(x+x^{1/2})']]=...$ $...=\dfrac{1}{2}[x+(x+x^{1/2})^{1/2}]^{-1/2}[1+[\dfrac{1}{2}(x+x^{1/2})^{-1/2}(1+\dfrac{1}{2}x^{-1/2})]]$ Use algebra to simplify: $...=\dfrac{1}{2[x+(x+x^{1/2})^{1/2}]^{1/2}}[1+[\dfrac{1}{2(x+x^{1/2})^{1/2}}][1+\dfrac{1}{2x^{1/2}}]]=$ $...=\dfrac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}(1+\dfrac{1}{2\sqrt{x+\sqrt{x}}})(1+\dfrac{1}{2\sqrt{x}})$
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