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: 6


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

Work Step by Step

$(a)$. Use implicit differentiation to find $\frac{dy}{dx}$ for $x=e^y$ Taking the derivative implicitly we get: $$1=e^y\frac{dy}{dx}$$ Solve for $\frac{dy}{dx}$ $$\frac{dy}{dx}=\frac{1}{e^y}=e^{-y}$$ $(b)$. Finding slope of tangent line at the point $(2,\ln2)$ for the above function. We do this by plugging in the point $(2,\ln2)$ into the derivative from part $(a)$. Hence: $$\frac{dy}{dx}=e^{-(\ln2)}=e^{(\ln2^{-1})}=2^{-1}=\frac{1}{2}$$
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