## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 68

#### Answer

$$y =-\frac{3}{2}(4+x)^{-5/2}\sec(1+(4+x)^{-3/2})\tan(1+(4+x)^{-3/2}).$$

#### Work Step by Step

Since $y=\sec(1+(4+x)^{-3/2})$, by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$ and recalling that $(\sec x)'=\sec x\tan x$, the derivative $y'$ is given by $$y'=\sec(1+(4+x)^{-3/2})\tan(1+(4+x)^{-3/2})(1+(4+x)^{-3/2})' \\=\sec(1+(4+x)^{-3/2})\tan(1+(4+x)^{-3/2})(-\frac{3}{2}(4+x)^{-5/2})\\ =-\frac{3}{2}(4+x)^{-5/2}\sec(1+(4+x)^{-3/2})\tan(1+(4+x)^{-3/2}).$$

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