## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 2 - The Derivative - 2.5 Derivatives of Trigonometric Functions - Exercises Set 2.5: 14

#### Answer

$f'(x) = \dfrac{\sec x. \tan x + \sec x .\tan^2 x - \sec^3 x}{(1+\tan x)^2}$

#### Work Step by Step

In order to derivate this function, you have to apply the quotient rule $\dfrac{d}{dx}(\dfrac{a}{b})=\dfrac{a'b-ab'}{b^2}$ So, let's identify a and b and derivate them $a=\sec x$ $a'=\sec x. \tan x$ $b=1+\tan x$ $b'=\sec^2 x$ Then, substitute in the formula: $f'(x) = \dfrac{(\sec x. \tan x)(1+\tan x)-(\sec x)(\sec^2 x)}{(1+\tan x)^2}$ Simplify and get the answer: $f'(x) = \dfrac{\sec x. \tan x + \sec x .\tan^2 x - \sec^3 x}{(1+\tan x)^2}$

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