## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 3 - Differentiation - Chapter Review Exercises - Page 163: 41

#### Answer

$$y' =\frac{1+\frac{1+\frac{1}{2\sqrt{x}}}{2\sqrt{x+\sqrt{x}}}}{2\sqrt{x+\sqrt{x+\sqrt{x}}}} .$$

#### Work Step by Step

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

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