# Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 154: 57

a) $$\frac{-1}{3}$$ b) $$\frac{10}{27}$$

#### Work Step by Step

We are given $$x y^{2}+y-2=0$$ a) We implicitly derive both sides: \begin{align*} x \cdot 2 y y^{\prime}+y^{2} \cdot 1+y^{\prime}&=0\\ y'&=-\frac{y^{2}}{2 x y+1}\\ y'\bigg|_{(1,1)}&=\frac{-1}{3} \end{align*} b) Now we take the second derivative: \begin{align*} y''&= -\frac{(2 x y+1)\left(2 y y^{\prime}\right)-y^{2}\left(2 x y^{\prime}+2 y\right)}{(2 x y+1)^{2}}\\ y''\bigg|_{(1,1)}&=-\frac{(3)\left(-\frac{2}{3}\right)-(1)\left(-\frac{2}{3}+2\right)}{3^{2}}\\ &=-\frac{-6+2-6}{27}\\ &=\frac{10}{27} \end{align*}

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