Explanation
We have,
$$\sqrt{x}+\sqrt{y}=1$$
On differentiating this with respect to x and we get,
$$ \dfrac{1}{2\sqrt{x}}+\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}=0 $$
$$ \Rightarrow \dfrac{dy}{dx}=-\dfrac{\sqrt{x}}{\sqrt{y}} $$
At the point $$\left( \dfrac{1}{4},\dfrac{1}{4} \right)$$
So,
$$ \dfrac{dy}{dx}=-\dfrac{\sqrt{\dfrac{1}{4}}}{\sqrt{\dfrac{1}{4}}} $$
$$ \dfrac{dy}{dx}=-1 $$
Hence, this is the answer.
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