## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 3 - Differentiation - 3.5 Higher Derivatives - Exercises - Page 135: 25

#### Answer

$$h''(1)= \frac{1}{8 }$$

#### Work Step by Step

Given $$h(w)=\frac{1}{\sqrt{w}+1}$$ Since \begin{align*} h'(w)&= \frac{d}{dw}\left(\left(\sqrt{w}+1\right)^{-1}\right)\\ &= \frac{-1}{2\sqrt{w}\left(\sqrt{w}+1\right)^2}\\ h''(w)&= \frac{3w+4\sqrt{w}+1}{4w\sqrt{w}\left(\sqrt{w}+1\right)^4} \end{align*} Then $$h''(1)= \frac{1}{8 }$$

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