Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 152: 29

$$y'= \frac{1}{4} $$

Given $$(x+2)^{2}-6(2 y+3)^{2}=3,\ \ \ ( 1,-1) $$ Differentiate with respect to $x$ \begin{align*} 2(x+2)-12(2 y+3)(2y')&=0\\ y'&= \frac{2(x+2)}{24(2 y+3)}\\ &= \frac{x+2}{12(2 y+3)} \end{align*} Then \begin{aligned} \left.y^{\prime}\right|_{(1,-1)} &=\frac{1+2}{12(-2 +3)} \\ &= \frac{1}{4} \end{aligned}