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Show y has fixed sign if x^(1/2) + y^(1/2) = 5

by RoRi

Consider the equation

    \[ x^{\frac{1}{2}} + y^{\frac{1}{2}} = 5 \qquad 0 < x < 5. \]

Assuming y' exists, show that y' has a fixed sign without solving for y.

We differentiate with respect to x, keeping in mind that y is a function of x so we must use the chain rule to differentiate y^{\frac{1}{2}}.

    \begin{align*} x^{\frac{1}{2}} + y^{\frac{1}{2}} = 5 && \implies && \frac{1}{2}x^{-\frac{1}{2}} + \frac{1}{2} \left( y^{-\frac{1}{2}} \right)y' = 0 \\ && \implies && \frac{1}{2 \sqrt{x}} + \frac{y'}{2 \sqrt{y}} = 0. \end{align*}

Since \sqrt{x} > 0, we have \frac{1}{2 \sqrt{x}} > 0 implies \frac{y'}{2 \sqrt{y}} < 0. Hence, y' < 0.

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