Calculus with Applications (10th Edition)

$$2y^{3}(x-3)+x\sqrt {y}=3; \quad\quad x=3$$ The equation of the tangent line at the given point is: $$y =-2x+7$$
$$2y^{3}(x-3)+x\sqrt {y}=3; \quad\quad x=3$$ find the$y$-value : \begin{aligned} 2y^{3}((3)-3)+(3)\sqrt {y}&=3\\ 2y^{3}(0)+(3)\sqrt {y}&=3\\ (3)\sqrt {y}&=3\\ \sqrt {y}&=1\\ \Rightarrow\quad\quad\quad\quad\quad\\ y &=1.\\ \end{aligned} So, the given point is $(3,1)$ . Now, we can calculate $\frac{dy}{dx}$ by implicit differentiation, use the product rule and the chain rule as follows: \begin{aligned} 2 y^{3}(1)+6 y^{2}(x-3) \frac{d y}{d x} & +x\left(\frac{1}{2}\right) y^{-1 / 2} \frac{d y}{d x}+\sqrt{y} &=0 \\ 6 y^{2}(x-3) \frac{d y}{d x}+\frac{x}{2 \sqrt{y}} \frac{d y}{d x} &=-2 y^{3}-\sqrt{y} \\ \left(6 y^{2}(x-3)+\frac{x}{2 \sqrt{y}}\right) \frac{d y}{d x} &=-2 y^{3}-\sqrt{y} \\ \frac{d y}{d x} &=\frac{-2 y^{3}-\sqrt{y}}{6 y^{2}(x-3)+\frac{x}{2 \sqrt{y}}} \\ \frac{d y}{d x} &=\frac{-4 y^{7 / 2}-2 y}{12 y^{5 / 2}(x-3)+x} \end{aligned} To find the slope of the tangent line at the point $(3,1)$, let $x=3$ and $y=1.$ The slope is \begin{aligned} \frac{d y}{d x} &=\frac{-4 y^{7 / 2}-2 y}{12 y^{5 / 2}(x-3)+x} \\ &=\frac{-4 (1)^{7 / 2}-2 (1)}{12 (1)^{5 / 2}((3)-3)+(3)}\\ &=\frac{-6}{3}\\ &=-2. \end{aligned} The equation of the tangent line is then found by using the point-slope form of the equation of a line. \begin{aligned} (y-y_{1}) &=m(x-x_{1})\\ (y-(1)) &=(-2)(x-(3))\\ y-1 &=-2x+6\\ y &=-2x+7\\ \end{aligned} Therefore, the equation of the tangent line at the given point is: $$y =-2x+7$$