Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.8 Implicit Differentiation - 3.8 Exercises - Page 200: 10


$(a)$ $$\frac{dy}{dx}=\frac{2x^{-1/2}}{-y^{-1/2}}$$ $(b)$ The slope of the tangent line to the curve is $m=-1$

Work Step by Step

$(a)$ Use implicit differentiation to determine $\frac{dy}{dx}$ for $\sqrt{x}-2\sqrt{y}=0$ Taking the derivative implicitly we get: $$\frac{1}{2}x^{-1/2}-y^{-1/2}\frac{dy}{dx}=0$$ solve for $\frac{dy}{dx}$ $$\frac{dy}{dx}=\frac{2x^{-1/2}}{-y^{-1/2}}$$ $(b)$ Find the slope of the tangent line to the curve at $(4,1)$ We plug in the point $(4,1)$ into the derivative from part $(a)$. Hence: $$\frac{dy}{dx}=\frac{2(4)^{-1/2}}{-(1)^{-1/2}}=\frac{2(1/2)}{-1}=\frac{1}{-1}=-1$$
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