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प्रश्न
If \[\sqrt{y + x} + \sqrt{y - x} = c, \text {show that } \frac{dy}{dx} = \frac{y}{x} - \sqrt{\frac{y^2}{x^2} - 1}\] ?
उत्तर
\[\text{ Here,} \] \[ \sqrt{y + x} + \sqrt{y - x} = c\]
Differentiating with respect to x,
\[\Rightarrow \frac{d}{dx}\left( \sqrt{y + x} \right) + \frac{d}{dx}\sqrt{y - x} = \frac{d}{dx}\left( c \right)\]
\[ \Rightarrow \frac{1}{2\sqrt{y + x}}\frac{d}{dx}\left( y + x \right) + \frac{1}{2\sqrt{y - x}}\frac{d}{dx}\left( y - x \right) = 0 \]
\[ \Rightarrow \frac{1}{2\sqrt{y + x}}\left( \frac{dy}{dx} + 1 \right) + \frac{1}{2\sqrt{y - x}}\left( \frac{dy}{dx} - 1 \right) = 0\]
\[ \Rightarrow \frac{dy}{dx}\left( \frac{1}{2\sqrt{y + x}} \right) + \frac{dy}{dx}\left( \frac{1}{2\sqrt{y - x}} \right) = \frac{1}{2\sqrt{y - x}} - \frac{1}{2\sqrt{y + x}}\]
\[ \Rightarrow \frac{dy}{dx} \times \frac{1}{2}\left[ \frac{1}{\sqrt{y + x}} + \frac{1}{\sqrt{y - x}} \right] = \frac{1}{2}\left[ \frac{\sqrt{y + x} - \sqrt{y - x}}{\sqrt{y - x}\sqrt{y + x}} \right]\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{\sqrt{y - x} + \sqrt{y + x}}{\sqrt{y + x}\sqrt{y - x}} \right] = \left[ \frac{\sqrt{y + x} - \sqrt{y - x}}{\sqrt{y - x}\sqrt{y + x}} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sqrt{y + x} - \sqrt{y - x}}{\sqrt{y + x} + \sqrt{y - x}} \times \frac{\sqrt{y + x} - \sqrt{y - x}}{\sqrt{y + x} - \sqrt{y - x}} \left[ \text{ rationalizing the denominator } \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( y + x \right) + \left( y - x \right) - 2\sqrt{y + x}\sqrt{y - x}}{y + x - y + x}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2y - 2\sqrt{y^2 - x^2}}{2x}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2y}{2x} - \frac{2\sqrt{y^2 - x^2}}{2x}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y}{x} - \sqrt{\frac{y^2 - x^2}{x^2}}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y}{x} - \sqrt{\frac{y^2}{x^2} - 1}\]
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