## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 3: Derivatives - Practice Exercises - Page 178: 48

#### Answer

$\frac{dy}{dx}=(\frac{1-x}{1+x})^{1/2}\times(\frac{1}{2y(1-x)^2})$

#### Work Step by Step

Rewrite the equation:$y^2=(\frac{1+x}{1-x})^{1/2}$ Take the derivative of the equation on each side separately. Apply chain rule when differentiating the "y" variables since we are differentiating with respect to x: $2y\frac{dy}{dx}=\frac{1}{2}(\frac{1+x}{1-x})^{-1/2}\times(\frac{(1-x)(1)-(1+x)(-1)}{(1-x)^2})$ $2y\frac{dy}{dx}=\frac{1}{2}(\frac{1+x}{1-x})^{-1/2}\times(\frac{1-x+1+x}{(1-x)^2})$ $2y\frac{dy}{dx}=\frac{1}{2}(\frac{1+x}{1-x})^{-1/2}\times(\frac{2}{(1-x)^2})$ $2y\frac{dy}{dx}=(\frac{1-x}{1+x})^{1/2}\times(\frac{1}{(1-x)^2})$ Move all terms with dy/dx to one side of the equation, and isolate dy/dx: $\frac{dy}{dx}=(\frac{1-x}{1+x})^{1/2}\times(\frac{1}{2y(1-x)^2})$

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