## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 3: Derivatives - Section 3.5 - Derivatives of Trigonometric Functions - Exercises 3.5 - Page 141: 12

#### Answer

The Derivative is: $y'=-\frac{1}{(1+\sin x)}$

#### Work Step by Step

$y=\frac{\cos x}{1+\sin x}$ Applying Derivative Rules: $y'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g^2(x)}$ $y'=\frac{\frac{d}{dx}(\cos x)\cdot (1+\sin x) - (\cos x)\cdot \frac{d}{dx}(1+\sin x)}{(1+\sin x)^2}$ $y'=\frac{(-\sin x)(1+\sin x)-(\cos x)(\cos x)}{(1+\sin x)^2}$ $y'=\frac{-\sin x-\sin^2 x -\cos^2 x}{(1+\sin x)^2}$ $y'=\frac{(-1)(\sin x+\sin^2 x+\cos^2 x)}{(1+\sin x)^2}$ Applying trigonometric identities $\sin^2 x+\cos^2 x=1$ $y'=\frac{-(1 +\sin x)}{(1+\sin x)^2}$ $y'=-\frac{1}{(1+\sin x)}$

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