Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 34
Determine $y''$ of $\sqrt{x} + \sqrt{y} = 1$ by using implicit differentiation. Solving for 1st Derivative $ \begin{equation} \begin{aligned} \frac{d}{dx} (\sqrt{x}) + \frac{d}{dx} (\sqrt{y}) =& \frac{d}{dx} (1) \\ \\ \frac{d}{dx} (x)^{\frac{1}{2}} + \frac{d}{dx} (y)^{\frac{1}{2}} =& \frac{d}{dx} (1) \\ \\ \frac{1}{2} (x)^{\frac{-1}{2}} + \frac{1}{2} (y)^{\frac{-1}{2}} \frac{d}{dx} =& 0 \\ \\ \frac{1}{2 (y)^{\frac{1}{2}}} \frac{dy}{dx} =& \frac{-1}{2(x)^{\frac{1}{2}}} \\ \\ \frac{dy}{dx} =& - \frac{\cancel{2} (y)^{\frac{1}{2}}}{\cancel{2}(x)^{\frac{1}{2}}} \\ \\ \frac{dy}{dx} =& - \frac{(y)^{\frac{1}{2}}}{(x)^{\frac{1}{2}}} \end{aligned} \end{equation} $ Solving for 2nd Derivative $ \begin{equation} \begin{aligned} \frac{d^2 y}{dx^2} =& - \frac{\displaystyle (x)^{\frac{1}{2}} \frac{d}{dx}(y)^{\frac{1}{2}} - (y)^{\frac{1}{2}} \frac{d}{dx} (x)^{\frac{1}{2}} }{[(x)^{\frac{1}{2}}]^2} \\ \\ \\ \\ \frac{d^2 y}{dx^2} =& - \frac{\displaystyle (x)^{\frac{...
Comments
Post a Comment