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Q)

For the curve $\sqrt x + \sqrt y=1,\large \frac{dy}{dx}$ at $\bigg(\large\frac{1}{4},\frac{1}{4}\bigg)$ is ____________.

$\begin{array}{1 1}(a)\;1\\(b)\;0\\(c)\;2\\(d)\;3\end{array}$

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A)
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  • Chain Rule :If $z=f(y)$ and $y=g(x)$,then $\large\frac{dz}{dx}=\frac{dz}{dy}-\frac{dy}{dx}$
$\sqrt x+\sqrt y=1$
Differentiating on both sides w.r.t $x$ we get,
$\large\frac{1}{2\sqrt x}+\frac{1}{2\sqrt y}\frac{dy}{dx}=$$0$
$\Rightarrow \large\frac{dy}{dx}=-\large\frac{\sqrt y}{\sqrt x}$
$\large\frac{dy}{dx}$ at $\big(\large\frac{1}{4},\frac{1}{4}\big)$ is written as
$\large\frac{dy}{dx}=-\frac{\sqrt{\large\frac{1}{4}}}{\sqrt{\large\frac{1}{4}}}$
$\qquad=-1$
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