Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.4 Exercises: 64

Answer

$y'=\dfrac{-\pi(\cos{(\tan{\pi x})})(\sec^2{\pi x})(\sin{\sqrt{\sin{(\tan{\pi x})}}})}{2\sqrt{\sin{(\tan{\pi x})}}}$

Work Step by Step

Using the Chain Rule: $u=\sqrt{\sin{(\tan{\pi x})}}=\sqrt{v}$ $v=\sin{(\tan{\pi x})}=\sin{z}$ $z=\tan{\pi x}$; $\dfrac{dz}{dx}=\pi\sec^2{\pi x}$ $\dfrac{dv}{dx}=\dfrac{dv}{dz}\times \dfrac{dz}{dx}=(\cos{(\tan{\pi x})})(\pi\sec^2{\pi x})$ $\dfrac{du}{dx}=\dfrac{du}{dv}\times\dfrac{dv}{dx}$ $=(\dfrac{1}{2\sqrt{\sin{(\tan{\pi x})}}})((\cos{(\tan{\pi x})})(\pi\sec^2{\pi x}))$ $=\dfrac{(\cos{(\tan{\pi x})})(\pi\sec^2{\pi x})}{2\sqrt{\sin{(\tan{\pi x})}}}$ $\dfrac{dy}{du}=-\sin{u}$ $\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}$ $=(\dfrac{(\cos{(\tan{\pi x})})(\pi\sec^2{\pi x})}{2\sqrt{\sin{(\tan{\pi x})}}})(-\sin{\sqrt{\sin{(\tan{\pi x})}}})$ $=\dfrac{-\pi(\cos{(\tan{\pi x})})(\sec^2{\pi x})(\sin{\sqrt{\sin{(\tan{\pi x})}}})}{2\sqrt{\sin{(\tan{\pi x})}}}$
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