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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

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