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: 5


$(a)$. $$\frac{dy}{dx}=\frac{-x^3}{y^3}$$ $(b)$. Slope of tangent line is $m=1$

Work Step by Step

$(a).$ Use implicit differentiation to find $\frac{dy}{dx}$ for $x^4+y^4=2$ Taking the derivative implicitly for the above function we get: $$4x^3+4y^3\frac{dy}{dx}=0$$ Solve for $\frac{dy}{dx}$ $$\frac{dy}{dx}=\frac{-x^3}{y^3}$$ $(b).$ Find the slope of the curve tangent to the point $(1,-1)$ for the above function The slope of our tangent line is determined by plugging in $(1,-1)$ into the derivative from part $(a)$. Hence: $$\frac{dy}{dx}=\frac{-(1)^3}{(-1)^3}=\frac{-1}{-1}=1$$
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