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Implicit Differentiation

If √ x + √ ...

Question

If $\sqrt{x} + \sqrt{y} = \sqrt{a},$ prove that $\frac{d y}{d x} = - \sqrt{\frac{y}{x}}$

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Solution

$\begin{matrix} \sqrt{x} + \sqrt{y} = \sqrt{a} D i f f e r e n t i a t e w i t h r e s p e c t t o x \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = 0 \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = - \frac{1}{2 \sqrt{x}} \frac{d y}{d x} = - \sqrt{\frac{y}{x}} \end{matrix}$

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\sqrt{x} + \sqrt{y} = \sqrt{a}

\frac{d y}{d x} = - \sqrt{\frac{y}{x}}

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