Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 70

Answer

$$ y'= \frac{ 1}{4 \sqrt{1+x}\sqrt{\sqrt{1+x}+1}}.$$

Work Step by Step

Since $ y=\sqrt{\sqrt{1+x}+1}$, then by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$, the derivative $ y'$ is given by $$ y'=\frac{(\sqrt{1+x}+1)'}{2\sqrt{\sqrt{1+x}+1}} \\=\frac{\frac{1}{2\sqrt{1+x}}}{2\sqrt{\sqrt{1+x}+1}}=\frac{ 1}{4 \sqrt{1+x}\sqrt{\sqrt{1+x}+1}}.$$
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