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(a) Find $ y' $ by implicit differentiation.(b) Solve the equation explicitly for y and differentiate to get $ y' $ in terms of $ x. $(c) Check that your solutions to part (a) and (b) are consistent by substituting the expression for $ y $ into your solution for part (a).

$ \sqrt{x} + \sqrt{y} = 1 $

a. $y^{\prime}=-\frac{\sqrt{y}}{\sqrt{x}}$b, $y^{\prime}=-\frac{1}{\sqrt{x}}+1$c. From part $(\mathrm{a}), y^{\prime}=-\frac{\sqrt{y}}{\sqrt{x}}=-\frac{1-\sqrt{x}}{\sqrt{x}} \quad[\text { from part }(\mathrm{b})]=-\frac{1}{\sqrt{x}}+1,$ which agrees with part

01:38

Frank L.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 5

Implicit Differentiation

Derivatives

Differentiation

Oregon State University

Baylor University

Boston College

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