Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.5 Derivatives of Trigonometric Functions - Exercises Set 2.5: 13

Answer

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

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 $a=\cot x$ $a'=-\csc^2 x$ $b= 1+\csc x$ $b'=-\csc x.\cot x$ Then, substitute in the formula: $f'(x)= \dfrac{ (-\csc^2 x) (1+\csc x) - (\cot x) (-\csc x.\cot x) }{(1+\csc x)^2}$ Simplify and get the answer: $f'(x) = \dfrac{-\csc^2 x - \csc^3 x + \csc x.\cot ^2 x}{(1+\csc x)^2}$
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