## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 3: Derivatives - Practice Exercises - Page 178: 53

#### Answer

a. $\frac{-2xy^3-2x^4}{y^5}$. b.$\frac{-2xy^2-1}{x^4y^3}$.

#### Work Step by Step

a. Given $x^3+y^3=1$, we have $3x^2dx+3y^2dy=0$ which gives $\frac{dy}{dx}=-\frac{x^2}{y^2}=-x^2y^{-2}$. Differentiating again, we have $\frac{d^2y}{dx^2}=-(2xy^{-2}-2x^2y^{-3}\frac{dy}{dx})=-2xy^{-2}+2x^2y^{-3}(-x^2y^{-2})=-2xy^{-2}-2x^4y^{-5})=\frac{-2xy^3-2x^4}{y^5}$. b. Given $y^2=1-\frac{2}{x}=1-2x^{-1}$, we have $2ydy=2x^{-2}dx$ which gives $\frac{dy}{dx}=x^{-2}y^{-1}$ Differentiating again, we have $\frac{d^2y}{dx^2}=-2x^{-3}y^{-1}-x^{-2}y^{-2}\frac{dy}{dx}=-2x^{-3}y^{-1}-x^{-2}y^{-2}(x^{-2}y^{-1})=-2x^{-3}y^{-1}-x^{-4}y^{-3}=\frac{-2xy^2-1}{x^4y^3}$.

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