If sqrt{y + x} + sqrt{y - x} = c, then frac{d | vedclass

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(A) Given $\sqrt{y + x} + \sqrt{y - x} = c$. Squaring both sides: $(y + x) + (y - x) + 2\sqrt{y^2 - x^2} = c^2 \implies 2y + 2\sqrt{y^2 - x^2} = c^2 \

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$\frac{y}{x}$

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(A) Given $\sqrt{y + x} + \sqrt{y - x} = c$. Squaring both sides: $(y + x) + (y - x) + 2\sqrt{y^2 - x^2} = c^2 \implies 2y + 2\sqrt{y^2 - x^2} = c^2 \implies \sqrt{y^2 - x^2} = \frac{c^2}{2} - y$.Squaring again: $y^2 - x^2 = \frac{c^4}{4} - c^2y + y^2 \implies -x^2 = \frac{c^4}{4} - c^2y \implies c^2y = x^2 + \frac{c^4}{4}$.Differentiating with respect to $x$: $c^2 \frac{dy}{dx} = 2x \implies \frac{dy}{dx} = \frac{2x}{c^2}$.From $\sqrt{y^2 - x^2} = \frac{c^2}{2} - y$, we have $\frac{c^2}{2} = y + \sqrt{y^2 - x^2}$.Thus, $\frac{dy}{dx} = \frac{2x}{2(y + \sqrt{y^2 - x^2})} = \frac{x}{y + \sqrt{y^2 - x^2}} = \frac{x}{y + y\sqrt{1 - (x/y)^2}} = \frac{x/y}{1 + \sqrt{1 - (x/y)^2}}$.Rationalizing the denominator: $\frac{dy}{dx} = \frac{x/y (1 - \sqrt{1 - (x/y)^2})}{1 - (1 - (x/y)^2)} = \frac{x/y (1 - \sqrt{1 - (x/y)^2})}{(x/y)^2} = \frac{1 - \sqrt{1 - (x/y)^2}}{x/y} = \frac{y}{x} - \sqrt{(y/x)^2 - 1}$.Comparing with $\frac{dy}{dx} = f(x) - \sqrt{[f(x)]^2 - 1}$, we get $f(x) = \frac{y}{x}$.

If $\sqrt{y + x} + \sqrt{y - x} = c$, then $\frac{dy}{dx} = f(x) - \sqrt{[f(x)]^2 - 1}$. Find $f(x)$.

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