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)
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\frac{d}{dx} (x)^{\frac{1}{2}} + \frac{d}{dx} (y)^{\frac{1}{2}} =& \frac{d}{dx} (1)
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\frac{1}{2} (x)^{\frac{-1}{2}} + \frac{1}{2} (y)^{\frac{-1}{2}} \frac{d}{dx} =& 0
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\frac{1}{2 (y)^{\frac{1}{2}}} \frac{dy}{dx} =& \frac{-1}{2(x)^{\frac{1}{2}}}
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\frac{dy}{dx} =& - \frac{\cancel{2} (y)^{\frac{1}{2}}}{\cancel{2}(x)^{\frac{1}{2}}}
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\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}
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\frac{d^2 y}{dx^2} =& - \frac{\displaystyle (x)^{\frac{1}{2}} \left( \frac{1}{2} \right) (y)^{\frac{-1}{2}} \frac{dy}{dx} - (y)^{ \frac{1}{2}} \left( \frac{1}{2} \right) (x)^{\frac{-1}{2}} }{x}
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\frac{d^2 y}{dx^2} =& - \frac{\displaystyle \frac{(x)^{\frac{1}{2}}}{2 (y)^{\frac{1}{2}}} \frac{dy}{dx} - \frac{(y)^{\frac{1}{2}}}{2(x)^{\frac{1}{2}}} }{x}
\qquad \qquad \text{We know that $\large \frac{dy}{dx} = - \frac{(y)^{\frac{1}{2}}}{(x)^{\frac{1}{2}}}$}
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\frac{d^2 y}{dx^2} =& - \frac{\displaystyle \left[ \frac{\cancel{(x)^{\frac{1}{2}}}}{2 \cancel{(y)^{\frac{1}{2}}}} \right] \left[ - \frac{\cancel{(y)^{\frac{1}{2}}}}{\cancel{(x)^{\frac{1}{2}}}} \right] - \frac{(y)^{\frac{1}{2}}}{2(x)^{\frac{1}{2}}} }{x}
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\frac{d^2 y}{dx^2} =& - \frac{\displaystyle \left( \frac{-1}{2} \right) - \frac{(y)^{\frac{1}{2}}}{2 (x)^{\frac{1}{2}}}}{x}
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\frac{d^2 y}{dx^2} =& \frac{\displaystyle \frac{1}{2} - \frac{(y)^{\frac{1}{2}}}{2(x)^{\frac{1}{2}}}}{x}
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\frac{d^2 y}{dx^2} =& \frac{\displaystyle \frac{1}{2} + \frac{(y)^{\frac{1}{2}}}{2 (x)^{\frac{1}{2}}}}{x}
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\frac{d^2 y}{dx^2} =& \frac{\displaystyle \frac{(x)^{\frac{1}{2}} + (y)^{\frac{1}{2}}}{2(x)^{\frac{1}{2}}}}{x}
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\frac{d^2 y}{dx^2} =& \frac{(x)^{\frac{1}{2}} + (y)^{\frac{1}{2}} }{2x (x)^{\frac{1}{2}}}
\qquad \qquad \text{We know that $\sqrt{x} + \sqrt{y} = 1$ or $(x)^{\frac{1}{2}} + (y)^{\frac{1}{2}} = 1$}
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\frac{d^2 y}{dx^2} =& \frac{1}{2x \sqrt{x}} \text{ or } y'' = \frac{1}{2x \sqrt{x}}


\end{aligned}
\end{equation}
$