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: 12


$(a)$ $$\frac{dy}{dx}=\frac{1-y\sec^2(xy)}{x\sec^2(xy)-1}$$ $(b)$ 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 $\tan xy=x+y$ Taking the derivative implicitly we get (using chain rule for $\tan xy$): $$\sec^2(xy)\left((1)y+x\frac{dy}{dx}\right)=1+\frac{dy}{dx}$$ solve for $\frac{dy}{dx}$ by getting both $\frac{dy}{dx}$ on the same side $$\frac{dy}{dx}=\frac{1-y\sec^2(xy)}{x\sec^2(xy)-1}$$ $(b)$ Find the slope of the tangent line to the curve at $(0,0)$ We plug in the point $(0,0)$ into the derivative from part $(a)$. Hence: $$\frac{dy}{dx}=\frac{1-(0)\sec^2((0)(0))}{(0)\sec^2((0)(0))-1}=\frac{1}{-1}=-1$$
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