Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 21


$$ f'(x ) =\frac{-2\cos x}{(\sin x -1)^2} .$$

Work Step by Step

Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(\sin x)'=\cos x$. Recall that $(\cos x)'=-\sin x$. Since $ f(x )= \frac{\sin x +1 }{\sin x -1}$, then using the quotient rule, the derivative is given by $$ f'(x )=\frac{(\sin x -1)(\sin x +1)'- (\sin x +1)(\sin x -1)'}{(\sin x -1)^2}\\ =\frac{(\sin x -1)(\cos x)- (\sin x +1)(\cos x)}{(\sin x -1)^2}\\ =\frac{\sin x\cos x -\cos x- \sin x\cos x -\cos x}{(\sin x -1)^2} \\ =\frac{-2\cos x}{(\sin x -1)^2} .$$
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