For the curve sqrt{x} + sqrt{y} = 1,the value | vedclass

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(C) Given the curve equation: $\sqrt{x} + \sqrt{y} = 1$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(\sqrt{x}) + \frac{d}{dx}(\sqrt{

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$-1$

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(C) Given the curve equation: $\sqrt{x} + \sqrt{y} = 1$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(\sqrt{x}) + \frac{d}{dx}(\sqrt{y}) = \frac{d}{dx}(1)$ $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0$ Rearranging the terms to solve for $\frac{dy}{dx}$: $\frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = -\frac{1}{2\sqrt{x}}$ $\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$ Now,substitute the point $\left( \frac{1}{4}, \frac{1}{4} \right)$ into the derivative: $\left( \frac{dy}{dx} \right)_{\left( \frac{1}{4}, \frac{1}{4} \right)} = -\frac{\sqrt{1/4}}{\sqrt{1/4}} = -\frac{1/2}{1/2} = -1$.

For the curve $\sqrt{x} + \sqrt{y} = 1$,the value of $\frac{dy}{dx}$ at the point $\left( \frac{1}{4}, \frac{1}{4} \right)$ is:

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